Chapter 8 - Maple Help

All Products Maple MapleSim

Home : Support : Online Help : Education : Study Guides : Precalculus : Chapter 8

Precalculus Study Guide

Chapter 8: Roots and Rational Powers

Introduction

The algebraic notation $x^{n}$ , where the integer $n$ is a power, means that $n$ copies of the letter $x$ are multiplied together.

Thus, $x^{2} = x \cdot x$ is called $x$ -squared, while $x^{3} = x \cdot x \cdot x$ is called $x$ -cubed.

The inverse of the $n$ th-power function is the $n$ th-root function, written $x^{1 / n}$ .

Thus, $x^{1 / 2}$ = $\sqrt{x}$ is called the square-root function, while $x^{1 / 3}$ is called the cube-root function.

Unfortunately, when $n$ is an even integer, the power function $x^{n}$ is not one-to-one. It is many-one, since, for example, both $x = - 2$ and $x = 2$ square to 4. Hence, the inverse function is not well defined, and the convention is that $\sqrt{x}$ means the positive square root.

Even if $x$ is real, $x^{1 / n}$ admits $n$ values, some of which can be complex numbers. If $n = 3$ , there is one real cube root, but two complex cube roots. Unfortunately, the principal cube root, the analog of defining $\sqrt{x}$ to be the positive root, is a complex number when $x$ is negative.

Hence, even for integer $n$ , computing $n$ th-roots of real numbers $x$ requires the care devoted to it in this chapter.

In addition, it turns out that computing $x^{n / m}$ , where both $n$ and $m$ are integers, requires additional care. Interpreting the $m$ th-root as the principal root means that the root must be computed before the power. Computing the power first, then taking the root does not yield the same value as would be obtained by use of DeMoivre's laws, the rules that in effect, provide a definition of $x^{n / m}$ .

Because of the complications introduced by complex numbers, this chapter provides an introduction to the arithmetic of complex numbers before it faces the issue of roots and powers of both real and complex quantities.

Chapter Glossary

The following terms in Chapter 8 are linked to the Maple Math Dictionary.

absolute value

Argand diagram

argument

binomial

Cartesian plane

complex conjugate

complex number

complex plane

cube

cube root

DeMoivre's laws

even

exponent

exponentiate

floating-point

hypotenuse

imaginary

imaginary axis

imaginary part

integer

length

magnitude

many-one

modulus

negative

one-to-one

open interval

positive

power

Pythagoras' theorem

quotient

rational number

real axis

real number

real part

root

square root

symmetry

Typical Problems

In Problems 8.1 - 8.8, implement the indicated calculations for the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$ .

8.1. Obtain the sum $z + w$ .

8.2. Obtain the difference $z - w$ .

8.3. Obtain the product $z w$ .

8.4. Obtain the quotient $\frac{z}{w}$ .

8.5. Write $z$ and $w$ in polar form.

8.6. Use DeMoivre's law to compute $z^{5}$ and $w^{7}$ .

8.7. Compute $\sqrt{z}$ and $\sqrt{w}$ .

8.8. Compute $z^{1 / 3}$ and $w^{1 / 3}$ .

8.9. If $a$ is a rational number (the ratio of two integers), exponentiation is defined by

$z^{a} = {|z|}^{a} cis (a theta)$ = ${|z|}^{a} (\cos (a theta) + i \sin (a theta))$

where $theta$ is the argument of $z$ .

For example, if $z = - 2 + 3 i$ and $a = \frac{2}{3}$ , compute $z^{a}$ by this definition; then compute ${(z^{2})}^{1 / 3}$ and ${(z^{1 / 3})}^{2}$ , where the principal cube root is taken in each case, and determine which of these two agrees with the definition of $z^{a}$ . (Both of these particular calculations can be implemented in the Study Guide's Complex Arithmetic Tutor, as demonstrated below in the solution to Problem 8.9.)

8.10. Repeat Problem 9 for $z = - 2 - 3 i$ and $a = \frac{3}{2}$ . Thus, compute $z^{a}$ by the definition, then compute ${(z^{3})}^{1 / 2}$ and ${(z^{1 / 2})}^{3}$ , where the principal square root is taken in each case. Determine which of these two agrees with the definition of $z^{a}$ .

8.11. Graph the following functions.

(a) $f (x) = {(x^{2} + 1)}^{2 / 3}$

(b) $g (x) = {(x^{2} - 1)}^{3 / 2}$

In particular, what is the value of $g (0)$ ?

8.12. Which of the following two calculations is correct?

(a) $- 1 = {(- 1)}^{2 / 2}$ = ${({(- 1)}^{1 / 2})}^{2} = i^{2}$ = $- 1$

(b) $- 1 = {(- 1)}^{2 / 2}$ = ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1

Maple Initializations

Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.

In Maple, the default symbol for $\sqrt{- 1}$ , the imaginary unit, is the letter I. In keeping with the typical textbook, our initialization sets the imaginary unit to the letter "i" throughout this chapter.

Our initialization also defines the function $cis (θ) = \cos (θ) = i \sin (θ)$ .

Solutions

Problem 8.1

8.1 - Mathematical Solution

The sum of the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$ is obtained in the tableau at the right.

The real part of the sum is the sum of the real parts of $z$ and $w$ .

The imaginary part of the sum is the sum of the imaginary parts of $z$ and $w$ .

$2 + 3 i$

$5 - 2 i$

___________

$7 + i$

8.1 - Maplet Solution

The sum of the complex numbers

$z = 2 + 3 i$

and

$w = 5 - 2 i$

can be found with the Study Guide's Complex Arithmetic Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.1.1.

The complex numbers $z$ and $w$ are entered by separately giving their real and imaginary parts.

(The imaginary part of $a + b i$ is the real number $b$ .)

Figure 8.1.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

To launch the Study Guide's Complex Arithmetic Tutor, click the link: Complex Arithmetic Tutor

8.1 - Interactive Solution

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

•	Type $z + w$

•	Context Panel: Evaluate and Display Inline

8.1 - Programmatic Solution

Enter the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$

>	$z ≔ 2 + 3 i &semi; w ≔ 5 - 2 i$

Obtain the sum $z + w$

$z + w$

Complex numbers of the form $z = a + b i$ , where $a$ and $b$ are themselves real numbers, are said to be in rectangular form.

The real part of $z$ is the real number $a$ , while the imaginary part of $z$ is the real number $b$ .

The real part of the sum of two complex numbers in rectangular form is obtained by adding the real parts; the imaginary part, by adding the imaginary parts.

Thus, the sum of $z_{1} = a + b i$ and $z_{2} = c + d i$ is $z_{1} + z_{2} = (a + c) + (b + d) i$ , obtained in Table 8.1.1. The sum of $z = 2 + 3 i$ and $w = 5 - 2 i$ is given in Table 8.1.2.

$a + b i$

$c + d i$

________________

$(a + c) + (b + d) i$

$2 + 3 i$

$5 - 2 i$

___________

$7 + i$

Table 8.1.1 Sum of $z_{1}$ and $z_{2}$

Table 8.1.2 Sum of $z$ and $w$

Problem 8.2

8.2 - Mathematical Solution

The difference of the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$ is obtained in the tableau at the right.

The real part of the difference is the difference of the real parts of $z$ and $w$ .

The imaginary part of the difference is the difference of the imaginary parts of $z$ and $w$ .

$2 + 3 i$

$5 - 2 i$

___________

$- 3 + 5 i$

8.2 - Maplet Solution

The difference of the complex numbers

$z = 2 + 3 i$

and

$w = 5 - 2 i$

can be found with the Study Guide's Complex Arithmetic Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.2.1.

The complex numbers $z$ and $w$ are entered by separately giving their real and imaginary parts.

(The imaginary part of $a + b i$ is the real number $b$ .)

Figure 8.2.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

To launch the Study Guide's Complex Arithmetic Tutor, click the link: Complex Arithmetic Tutor

8.2 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

•	Type $z - w$

•	Context Panel: Evaluate and Display Inline

8.2 - Programmatic Solution

Enter the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$

>	$z ≔ 2 + 3 i &semi; w ≔ 5 - 2 i$

Obtain the difference $z - w$

$z - w$

Complex numbers of the form $z = a + b i$ , where $a$ and $b$ are themselves real numbers, are said to be in rectangular form.

The real part of $z$ is the real number $a$ , while the imaginary part of $z$ is the real number $b$ .

The real part of the difference of two complex numbers in rectangular form is obtained by subtracting the real parts; the imaginary part, by subtracting the imaginary parts.

Thus, the difference of $z_{1} = a + b i$ and $z_{2} = c + d i$ is $z_{1} - z_{2} = (a - c) + (b - d) i$ , obtained in Table 8.2.1. The difference of $z = 2 + 3 i$ and $w = 5 - 2 i$ is given in Table 8.2.2.

$a + b i$

$c + d i$

________________

$(a - c) + (b - d) i$

$2 + 3 i$

$5 - 2 i$

___________

$- 3 + 5 i$

Table 8.2.1 Difference of $z_{1}$ and $z_{2}$

Table 8.2.2 Difference of $z$ and $w$

Problem 8.3

8.3 - Mathematical Solution

The product of $z_{1} = a + b i$ and $z_{2} = c + d i$ is $z_{1} z_{2} = (a - c) + (b - d) i$ , obtained in the following tableau where the numbers are multiplied as if they were two binomials.

$a + b i$

$c + d i$

______________________

$(ac - bd) + (bc + ad) i$

The product of the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$ is obtained in the following tableau.

$2 + 3 i$

$5 - 2 i$

______________________________________________

$[(2) \cdot (5) + (3) \cdot (- 2) i^{2}] + i [(2) \cdot (- 2) + (5) \cdot (3)]$

Since $i^{2} = - 1$ by definition, this product reduces to $[10 + 6] + i [15 - 4]$ = $16 - 11 i$ .

8.3 - Maplet Solution

The product of the complex numbers

$z = 2 + 3 i$

and

$w = 5 - 2 i$

can be found with the Study Guide's Complex Arithmetic Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.3.1.

The complex numbers $z$ and $w$ are entered by separately giving their real and imaginary parts.

(The imaginary part of $a + b i$ is the real number $b$ .)

Figure 8.3.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

To launch the Study Guide's Complex Arithmetic Tutor, click the link: Complex Arithmetic Tutor

8.3 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

•	Using a space for multiplication, type $z w$

•	Context Panel: Evaluate and Display Inline

8.3 - Programmatic Solution

Enter the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$

>	$z ≔ 2 + 3 i &semi; w ≔ 5 - 2 i$

Using a space for multiplication, obtain the product $z w$

$z w$

Complex numbers of the form $z = a + b i$ , where $a$ and $b$ are themselves real numbers, are said to be in rectangular form.

The real part of $z$ is the real number $a$ , while the imaginary part of $z$ is the real number $b$ .

The product of two complex numbers in rectangular form is obtained by multiplying the complex numbers as two binomials.

Thus, the product of $z_{1} = a + b i$ and $z_{2} = c + d i$ is $z_{1} z_{2} = (a - c) + (b - d) i$ , obtained in Table 8.3.1. The product of $z = 2 + 3 i$ and $w = 5 - 2 i$ is given in Table 8.3.2.

$a + b i$

$c + d i$

______________________

$(ac - bd) + (bc + ad) i$

$2 + 3 i$

$5 - 2 i$

______________________________________________

$[(2) \cdot (5) + (3) \cdot (- 2) i^{2}] + i [(2) \cdot (- 2) + (5) \cdot (3)]$

Table 8.3.1 Product of $z_{1}$ and $z_{2}$

Table 8.3.2 Product of $z$ and $w$

Since $i^{2} = - 1$ by definition, the product $z w$ reduces to $[10 + 6] + i [15 - 4]$ = $16 - 11 i$ .

Problem 8.4

8.4 - Mathematical Solution

The quotient of two complex numbers in rectangular form is obtained by multiplying the numerator and denominator of the quotient by the complex conjugate of the denominator.

Thus, the quotient of $z_{1} = a + b i$ and $z_{2} = c + d i$ is

$\frac{z_{1}}{z_{2}} = \frac{a + b i}{c + d i}$ $\frac{c - d i}{c - d i} = \frac{(a + b i) (c - d i)}{c^{2} + d^{2}}$

The quotient of $z$ and $w$ is therefore

$\frac{2 + 3 i}{5 - 2 i} = \frac{2 + 3 i}{5 - 2 i}$ $\frac{5 + 2 i}{5 + 2 i} = \frac{4 + 19 i}{29}$ = $\frac{4}{29} + \frac{19}{29}$ $i$

The product of the complex number $z_{2} = c + d i$ by its complex conjugate ${\overset{&conjugate0;}{z}}_{2} = c - d i$ is given by the tableau

$c + d i$

$c - d i$

________________

$c^{2} - d^{2} i^{2} + i (c d - c d)$

Since $i^{2} = - 1$ by definition, this product reduces to $c^{2} + d^{2}$ .

8.4 - Maplet Solution

The quotient of the complex numbers

$z = 2 + 3 i$

and

$w = 5 - 2 i$

can be found with the Study Guide's Complex Arithmetic Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.4.1.

The complex numbers $z$ and $w$ are entered by separately giving their real and imaginary parts.

(The imaginary part of $a + b i$ is the real number $b$ .)

Figure 8.4.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

To launch the Study Guide's Complex Arithmetic Tutor, click the link: Complex Arithmetic Tutor

8.4 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

•	Using the slash (/) for division, type $\frac{z}{w}$

•	Context Panel: Evaluate and Display Inline

8.4 - Programmatic Solution

Enter the complex numbers $z = 2 + 3 i$ and $w = 5 - 2 i$

>	$z ≔ 2 + 3 i &semi; w ≔ 5 - 2 i$

Obtain the quotient $\frac{z}{w}$

>	$\frac{z}{w}$

Complex numbers of the form $z = a + b i$ , where $a$ and $b$ are themselves real numbers, are said to be in rectangular form.

The real part of $z$ is the real number $a$ , while the imaginary part of $z$ is the real number $b$ .

Thus, the quotient of $z_{1} = a + b i$ and $z_{2} = c + d i$ is

$\frac{z_{1}}{z_{2}} = \frac{a + b i}{c + d i}$ $\frac{c - d i}{c - d i} = \frac{(a + b i) (c - d i)}{c^{2} + d^{2}}$

The quotient of $z$ and $w$ is therefore

$\frac{2 + 3 i}{5 - 2 i} = \frac{2 + 3 i}{5 - 2 i}$ $\frac{5 + 2 i}{5 + 2 i} = \frac{4 + 19 i}{29}$ = $\frac{4}{29} + \frac{19}{29}$ $i$

The product of the complex number $z_{2} = c + d i$ by its complex conjugate ${\overset{&conjugate0;}{z}}_{2} = c - d i$ is given by the tableau

$c + d i$

$c - d i$

________________

$c^{2} - d^{2} i^{2} + i (c d - c d)$

Since $i^{2} = - 1$ by definition, this product reduces to $c^{2} + d^{2}$ .

Problem 8.5

8.5 - Mathematical Solution

The polar form of the complex number $z = 2 + 3 i$ is determined by the magnitude

$ρ = \sqrt{2^{2} + 3^{2}}$ = $\sqrt{13}$

and the argument

$θ = \arctan (\frac{3}{2})$

The geometry of this calculation is best seen from Figure 8.5.1, the Argand diagram for $z$ , a plot of $z$ as a point in the complex plane.

The horizontal axis is called the real axis, and the vertical, the imaginary axis. The complex number $z = 2 + 3 i$ is plotted as the point $(2, 3)$ in the real Cartesian plane, but the interpretation is as a point in the complex plane.

Figure 8.5.1 Argand diagram for $z = 2 + 3 i$

The assignment of the number $\sqrt{a^{2} + b^{2}} = \sqrt{2^{2} + 3^{2}}$ = $\sqrt{13}$ as the length or magnitude of the complex number $z$ follows from the theorem of Pythagoras - it's just the length of the hypotenuse of the right triangle formed by the real and imaginary parts of the complex number. In Figure 8.5.1, the real part corresponds to the red segment, the imaginary part, to the green, and the hypotenuse, to the black.

The polar form of the complex number $w = 5 - 2 i$ is determined by the magnitude

$ρ = \sqrt{5^{2} + {(- 2)}^{2}}$ = $\sqrt{29}$

and the argument

$θ = - \arctan (\frac{2}{5})$

A typical range for the angle $θ$ , measured from the positive real axis, is $- π < θ \leq π$ .

The Argand diagram for $w = 5 - 2 i$ is given by Figure 8.5.2.

Figure 8.5.2 Argand diagram for $w = 5 - 2 i$

The rectangular form of the complex number whose polar form is given by $ρ$ and $θ$ is

$ρ (\cos (θ) + i \sin (θ))$

Some texts abbreviate this as

$ρ cis (θ)$

where the letters "cis" stand for "cosine $i$ sine."

The "cis" function is not native to Maple. However, it is easy to define, and the Initialize button in the Initialization section implements the definition $cis ≔ x \to \cos (x) + i \sin (x)$

8.5 - Maplet Solution

The polar form of the complex number

$z = 2 + 3 i$

can be found with the Study Guide's Complex Arithmetic Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.5.3.

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

The button labeled Polar Form gives $|z| = ρ$ = $\sqrt{13}$ , with the argument being $θ = \arctan (\frac{3}{2})$ .

Figure 8.5.3 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

The polar form of the complex number

$w = 5 - 2 i$

can also be found with the Study Guide's Complex Arithmetic Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.5.4. However, it has to be entered as if its name were $z$ .

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

The button labeled Polar Form gives $ρ = \sqrt{29}$ , with the argument being $θ = - \arctan (\frac{2}{5})$ .

Figure 8.5.4 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

8.5 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

Enter the Data

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

Polar Form of $z$

•	Type $z$ and press the Enter key.

•	Context Panel: Modulus (or Absolute Value)

•	Type $z$ and press the Enter key.

•	Context Panel: Argument

Polar Form of $w$

•	Type $w$ and press the Enter key.

•	Context Panel: Modulus (or Absolute Value)

•	Type $w$ and press the Enter key.

•	Context Panel: Argument

Convert Polar Form of $z$ to Rectangular Form

•	Using $ρ$ and $θ$ computed for $z$ , enter $ρ cis (θ)$

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify

Convert Polar Form of $w$ to Rectangular Form

•	Using $ρ$ and $θ$ computed for $w$ , enter $ρ cis (θ)$

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify

8.5 - Programmatic Solution

The polar form of the complex number $z = a + b i$ consists of the two numbers, $ρ$ and $θ$ , where $ρ$ is called the magnitude (or modulus, or length, or absolute value) of $z$ , and $θ$ is called the argument of $z$ . The argument of $z$ is the angle from the positive real axis to the radial line from the origin to the complex number $z$ . It lies in the range $(- π, π]$ .

Maple can convert the rectangular form of $z$ to the polar form. Maple's representation of the polar form of the complex number

>	$z ≔ 2 + 3 i$

>	$convert (z, polar)$

whereas for the complex number

>	$w ≔ 5 - 2 i$

the polar form is

>	$convert (w, polar)$

In each case, the first number is the magnitude of the complex number, a number Maple can separately calculate as

$abs (z)$

for $z$ , and

$abs (w)$

for $w$ .

The second number in the polar representation is the argument, which is an angle that can be separately calculated in Maple by

>	$argument (z)$

for $z$ , and

>	$argument (w)$

for $w$ .

The Argand diagram in Figure 8.5.1 illustrates the geometry of the magnitude and argument of $z = 2 + 3 i$ , whereas the Argand diagram in Figure 8.5.2 illustrates the geometry of the magnitude and argument of $w = 5 - 2 i$ .

The rectangular form of the complex number whose polar form is given by $ρ$ and $θ$ is

$ρ (\cos (θ) + i \sin (θ))$

Some texts abbreviate this as

$ρ cis (θ)$

where the letters "cis" stand for "cosine $i$ sine." Thus, the rectangular form of $z = 2 + 3 i$ is recovered from $ρ = \sqrt{13}$ and $θ = \arctan (3 / 2)$ by executing

>	$simplify (\sqrt{13} cis (\arctan (3 / 2)))$

>	$simplify (\sqrt{13} (\cos (\arctan (3 / 2)) + i \sin (\arctan (3 / 2))))$

Similarly, the rectangular form of $w = 5 - 2 i$ can be recovered from $ρ = \sqrt{29}$ and $θ = - \arctan (\frac{2}{5})$ by executing

>	$simplify (\sqrt{29} cis (- \arctan (2 / 5)))$

>	$simplify (\sqrt{29} (\cos (- \arctan (2 / 5)) + i \sin (- \arctan (2 / 5))))$

Problem 8.6

8.6 - Mathematical Solution

DeMoivre's law (also spelled de Moivre, Demoivre, etc.) shows how to compute integer powers of $z$ in terms of its polar representation. Suppose $z$ has its polar representation given by $ρ$ and $θ$ . Then $z^{n}$ , where $n$ is an integer (positive or negative), is given by

$z^{n} = ρ^{n} (\cos (n θ) + i \sin (n θ)) = ρ^{n} cis (n θ)$

The polar form of the complex number $z = 2 + 3 i$ consists of $ρ = \sqrt{13}$ and $θ = \arctan (\frac{3}{2})$ . To raise $z$ to the power 5, compute

$z^{5} = {(\sqrt{13})}^{5} (\cos (5 \arctan (\frac{3}{2})) + i \sin (5 \arctan (\frac{3}{2})))$ = $122 - 597 i$

The simplification to final form can be done numerically with a computing device such as a calculator. Alternatively, the trig formulas

$\cos (5 x) = 16 {\cos (x)}^{5} - 20 {\cos (x)}^{3} + 5 \cos (x)$

and

$\sin (5 x) = 16 {\sin (x)}^{5} - 20 {\sin (x)}^{3} + 5 \sin (x)$

could be used in conjunction with the identities

$\cos (\arctan (\frac{3}{2})) = \frac{2}{\sqrt{13}}$

and

$\sin (\arctan (\frac{3}{2})) = \frac{3}{\sqrt{13}}$

to reduce $z^{5}$ to the exact value given above.

The polar form of the complex number $w = 5 - 2 i$ consists of $ρ = \sqrt{29}$ and $θ = - \arctan (\frac{2}{5})$ . To raise $w$ to the power 7, compute

$w^{7} = {(\sqrt{29})}^{7} (\cos (- 7 \arctan (\frac{2}{5})) + i \sin (- 7 \arctan (\frac{2}{5})))$ = $- 116615 - 60422 i$

The simplification to final form can be done numerically with a computing device such as a calculator. Alternatively, the trig formulas

$\cos (7 x) = 64 {\cos (x)}^{7} - 112 {\cos (x)}^{5} + 56 {\cos (x)}^{3} - 7 \cos (x)$

and

$\sin (7 x) = - 64 {\sin (x)}^{7} + 112 {\sin (x)}^{5} - 56 {\sin (x)}^{3} + 7 \sin (x)$

could be used in conjunction with the identities

$\cos (\arctan (\frac{2}{5})) = \frac{5}{\sqrt{29}}$

and

$\sin (\arctan (\frac{2}{5})) = \frac{2}{\sqrt{29}}$

to reduce $w^{7}$ to the exact value given above.

8.6 - Maplet Solution

The Complex Arithmetic Tutor implements DeMoivre's law for the computation of $z^{5}$ for the complex number

$z = 2 + 3 i$

Clicking the above maplet link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.6.1.

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

In the section Exponentiation of complex numbers, enter $n = 5, m = 1$ and click the button labeled Float opposite the symbol $Z^{n / m}$ . The floating-point form of the solution

${(2 + 3 i)}^{5} = 122 - 597 i$

Figure 8.6.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

is returned.

The Complex Arithmetic Tutor implements DeMoivre's law for the computation of $w^{7}$ for the complex number

$w = 5 - 2 i$

However, it has to be entered as if its name were $z$ .

Clicking the above maplet link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.6.2.

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

In the section Exponentiation of complex numbers, enter $n = 7, m = 1$ and click the button labeled Float opposite the symbol $Z^{n / m}$ .

Figure 8.6.2 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

The floating-point form of the solution, to four significant figures, is

${(5 - 2 i)}^{7} = - (116615 + 60422 i) ≐ - 0.1166 \times 10^{6} - 0.6042 \times 10^{5} i$

To launch the Study guide's Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

8.6 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

$z^{5}$

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $z$ and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Az$

•	Type ${\| z \|}^{5} cis (5 Az)$ and press the Enter key.

•	Context Panel: Simplify≻Normalize Expanded

•	Type $z^{5}$

•	Context Panel: Evaluate and Display Inline

$w^{7}$

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

•	Type $z$ and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Aw$

•	Type ${\|w\|}^{7} cis (7 Aw)$ and press the Enter key.

•	Context Panel: Simplify≻Normalize Expanded

•	Type $w^{7}$

•	Context Panel: Evaluate and Display Inline

8.6 - Programmatic Solution

$z^{n} = ρ^{n} (\cos (n θ) + i \sin (n θ)) = ρ^{n} cis (n θ)$

$z^{5}$

Enter $z = 2 + 3 i$

>	$z ≔ 2 + 3 i$

Obtain polar form of $z$

>	$convert (z, polar)$

Obtain $z^{5}$ by applying DeMoivre's law.

>	$qz ≔ sqrt {(13)}^{5} cis (5 \arctan (3 / 2))$

Expand $ρ^{5} cis (5 θ)$

>	$expand (qz)$

Compare to $z^{5}$ computed directly in Maple.

$z^{5}$

$w^{7}$

Enter $w = 5 - 2 i$

>	$w ≔ 5 - 2 i$

Obtain polar form of $w$

>	$convert (w, polar)$

Obtain $w^{7}$ by applying DeMoivre's law.

>	$qw ≔ sqrt {(29)}^{7} cis (7 (- \arctan (2 / 5)))$

Expand $ρ^{7} cis (7 θ)$

>	$expand (qw)$

Compare to $w^{7}$ computed directly in Maple.

$w^{7}$

Problem 8.7

8.7 - Mathematical Solution

It can be shown that a consequence of DeMoivre's law for integer powers is the related law for fractional powers

$z^{1 / m} = ρ^{1 / m} (\cos (\frac{θ}{m} + \frac{2 k π}{m}) + i \sin (\frac{θ}{m} + \frac{2 k π}{m})), k = 0, 1, .. &period;, m - 1$

For convenience, we will also refer to this law as DeMoivre's law for fractional powers.

The complex number $z = 2 + 3 i$ has polar form determined by $ρ = \sqrt{13}$ and $θ = \arctan (\frac{3}{2})$ . There are two values for $z^{1 / 2}$ . Given by DeMoivre's law for fractional powers, they are

$z_{1}$	$= {(\sqrt{13})}^{1 / 2} (\cos (\frac{\arctan (\frac{3}{2})}{2}) + i \sin (\frac{\arctan (\frac{3}{2})}{2}))$
	$= \frac{1}{2}$ $(\sqrt{4 + 2 \sqrt{13}} + i \sqrt{2 \sqrt{13} - 4})$
	$= 1.732050808 + 1.732050808 i$

and

$z_{2}$	$= {(\sqrt{13})}^{1 / 2} (\cos (\frac{\arctan (\frac{3}{2})}{2} + π) + i \sin (\frac{\arctan (\frac{3}{2})}{2} + π))$
	$= - \frac{1}{2}$ $(\sqrt{4 + 2 \sqrt{13}} + i \sqrt{2 \sqrt{13} - 4})$
	$= - 1.732050808 - 1.732050808 i$

The real part of $z_{1}$ is obtained by using the identity

$\cos (\arctan (\frac{3}{2})) = \frac{2}{\sqrt{13}}$

and the half-angle formula

$\cos (\frac{x}{2}) = \sqrt{\frac{1 + \cos (x)}{2}}$

to yield

$\sqrt{\sqrt{13}} \sqrt{\frac{1 + \frac{2}{\sqrt{13}}}{2}}$ = $\sqrt{\frac{\sqrt{13} + 2}{2}} = \sqrt{\frac{2 \sqrt{13} + 4}{4}}$ = $\frac{1}{2}$ $\sqrt{4 + 2 \sqrt{13}}$

Similarly, the imaginary part of $z_{1}$ is obtained by using the half-angle formula

$\sin (\frac{x}{2}) = \sqrt{\frac{1 - \cos (x)}{2}}$

to yield

$\sqrt{\sqrt{13}} \sqrt{\frac{1 - \frac{2}{\sqrt{13}}}{2}}$ = $\sqrt{\frac{\sqrt{13} - 2}{2}} = \sqrt{\frac{2 \sqrt{13} - 4}{4}}$ = $\frac{1}{2}$ $\sqrt{2 \sqrt{13} - 4}$

The floating-point value for $z_{1}$ is obtained with a computing device such as a calculator. The calculations for $z_{2}$ proceed along the same lines, except that the trig identities

$\cos (x + π) = \cos (x) \cos (π) - \sin (x) \sin (π)$ = $- \cos (x)$

and

$\sin (x + π) = \sin (x) \cos (π) + \cos (x) \sin (π)$ = $- \sin (x)$

must be used first.

The complex number $w = 5 - 2 i$ has polar form determined by $ρ = \sqrt{29}$ and $θ = - \arctan (\frac{2}{5})$ . There are two values for $w^{1 / 2}$ . Given by DeMoivre's law for fractional powers, they are

$w_{1}$	$= {(\sqrt{29})}^{1 / 2} (\cos (- \frac{\arctan (\frac{2}{5})}{2}) + i \sin (- \frac{\arctan (\frac{2}{5})}{2}))$
	$= \frac{1}{2}$ $(\sqrt{2 \sqrt{29} + 10} - i \sqrt{2 \sqrt{29} - 10})$
	$= 2.27872385417085 - 0.438842116902252 i$

and

$w_{2}$	$= {(\sqrt{29})}^{1 / 2} (\cos (- \frac{\arctan (\frac{2}{5})}{2} + π) + i \sin (- \frac{\arctan (\frac{2}{5})}{2} + π))$
	$= - \frac{1}{2}$ $(\sqrt{2 \sqrt{29} + 10} - i \sqrt{2 \sqrt{29} - 10})$
	$= - 2.27872385417085 + 0.438842116902252 i$

The calculations for $w_{1}$ and $w_{2}$ proceed along the same lines as those for $z_{1}$ and $z_{2}$ , except that they begin with use of the trig identities

$\cos (- x) = \cos (x)$

and

$\sin (- x) = - \sin (x)$

8.7 - Maplet Solution

The Study Guide's Complex Arithmetic Tutor can be used for the computation of $\sqrt{z}$ for the complex number $z = 2 + 3 i$ .

Clicking the above maplet link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.7.1.

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

In the section Exponentiation of complex numbers, enter $n = 1, m = 2$ and click the button labeled Float opposite the symbol $Z^{n / m}$ . The floating-point form of the solution

$\sqrt{2 + 3 i} = 1.674 + 0.8960 i$

is returned.

Figure 8.7.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

Alternatively, setting $k = 2$ in the section Roots of complex numbers, and clicking the button labeled Float to the right of the button labeled kth principal root yields the same result.

The Study Guide's Complex Arithmetic Tutor can be used for the computation of $\sqrt{w}$ for the complex number $w = 5 - 2 i$ .

Clicking the above maplet link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.7.2.

However, it has to be entered as if its name were $z$ .

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

In the section Exponentiation of complex numbers, enter $n = 1, m = 2$ and click the button labeled Float opposite the symbol $Z^{n / m}$ . The floating-point form of the solution, to four significant figures, is $\sqrt{5 - 2 i} = 2.279 - 0.4388 i$ .

Figure 8.7.2 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

Alternatively, setting $k = 2$ in the section Roots of complex numbers, and clicking the button labeled Float to the right of the button labeled kth principal root yields the same result.

To launch the Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

8.7 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

$\sqrt{z}$

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $z$ and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Az$

•	Type $\sqrt{\|z\|} cis (Az / 2)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type $\sqrt{\|z\|} cis (Az / 2 + π)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type $\sqrt{z}$ and press the Enter key.

•	Context Panel: Approximate≻10

$\sqrt{w}$

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

•	Type and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Aw$

•	Type $\sqrt{\|w\|} cis (Aw / 2)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type $\sqrt{\|w\|} cis (Aw / 2 + π)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type $\sqrt{w}$ and press the Enter key.

•	Context Panel: Approximate≻10

8.7 - Programmatic Solution

It can be shown that a consequence of DeMoivre's law for integer powers is the related law for fractional powers

$z^{1 / m} = ρ^{1 / m} (\cos (\frac{θ}{m} + \frac{2 k π}{m}) + i \sin (\frac{θ}{m} + \frac{2 k π}{m})), k = 0, 1, .. &period;, m - 1$

For convenience, we will also refer to this law as DeMoivre's law for fractional powers.

Maple gives the principal square root of

>	$z ≔ 2 + 3 i$

>	$zp ≔ evalc (sqrt (z))$

whose floating-point form is

>	$evalf (zp)$

To apply DeMoivre's law for fractional powers, we first single out the magnitude and argument of $z$ with

>	$rz ≔ abs (z) &semi; tz ≔ argument (z)$

Then, the two square roots of $z$ are

>	$z1 ≔ sqrt (rz) cis (tz / 2) &semi; z2 ≔ sqrt (rz) cis (tz / 2 + Pi)$

Careful inspection shows that these two roots simply differ by a factor of $- 1$ , as the following conversion to floating-point form reveals.

>	$evalf (z1) &semi; evalf (z2)$

Maple struggles to convert the trigonometric form of $z_{1}$ to the radical form of $z_{p}$ , the principal square root. We can settle for the equivalence of the floating-point forms, or we can impose the simplifications

>	$z1a ≔ evalc (simplify (radnormal (convert (z1, \exp))))$

Maple cannot easily be coerced into factoring out 13 from each of the larger radicals. Were it able to do that, the equivalence with $z_{p}$ would be obvious. Instead, we must use

>	$simplify (z1a - zp)$

Maple gives the principal square root of

>	$w ≔ 5 - 2 i$

>	$wp ≔ evalc (sqrt (w))$

whose floating-point form is

>	$evalf (wp)$

To apply DeMoivre's law for fractional powers, we first single out the magnitude and argument of $z$ with

>	$rw ≔ abs (w) &semi; tw ≔ argument (w)$

Then, the two square roots of $z$ are

>	$w1 ≔ sqrt (rw) cis (tw / 2) &semi; w2 ≔ sqrt (rw) cis (tw / 2 + Pi)$

Careful inspection shows that these two roots simply differ by a factor of $- 1$ , as the following conversion to floating-point form reveals.

>	$evalf (w1) &semi; evalf (w2)$

Maple struggles to convert the trigonometric form of $w_{1}$ to the radical form of $w_{p}$ , the principal square root. We can settle for the equivalence of the floating-point forms, or we can impose the simplifications

>	$w1a ≔ evalc (simplify (radnormal (convert (w1, \exp))))$

Maple cannot easily be coerced into factoring out 29 from each of the larger radicals. Were it able to do that, the equivalence with $w_{p}$ would be obvious. Instead, we must use

>	$simplify (w1a - wp)$

to complete the demonstration of exact equivalence.

Problem 8.8

8.8 - Mathematical Solution

It can be shown that a consequence of DeMoivre's law for integer powers is the related law for fractional powers

$z^{1 / m} = ρ^{1 / m} (\cos (\frac{θ}{m} + \frac{2 k π}{m}) + i \sin (\frac{θ}{m} + \frac{2 k π}{m})), k = 0, 1, .. &period;, m - 1$

For convenience, we will also refer to this law as DeMoivre's law for fractional powers.

The complex number $z = 2 + 3 i$ has three cube roots given by DeMoivre's law for fractional powers with $m = 3$ . Since the polar form of $z$ is determined by $ρ = \sqrt{13}$ and $θ = \arctan (\frac{3}{2})$ , these three cube roots are

$z_{1}$	$= λ (\cos (\frac{\arctan (\frac{3}{2})}{3}) + i \sin (\frac{\arctan (\frac{3}{2})}{3}))$	= $1.451856618 + 0.4934035341 i$
$z_{2}$	$= λ (\cos (\frac{\arctan (\frac{3}{2}) + 2 π}{3}) + i \sin (\frac{\arctan (\frac{3}{2}) + 2 π}{3}))$	$= - 1.153228304 + 1.010642947 i$
$z_{3}$	$= λ (\cos (\frac{\arctan (\frac{3}{2}) + 4 π}{3}) + i \sin (\frac{\arctan (\frac{3}{2}) + 4 π}{3}))$	$= - 0.2986283143 - 1.504046481 i$

where $λ = {(\sqrt{13})}^{1 / 3} = 13^{1 / 6}$ .

The complex number $w = 5 - 2 i$ has three cube roots given by DeMoivre's law for fractional powers with $m = 3$ . Since the polar form of $z$ is determined by $ρ = \sqrt{29}$ and $θ = - \arctan (\frac{2}{5})$ , these three cube roots are

$w_{1}$	$= λ (\cos (- \frac{\arctan (\frac{2}{5})}{3}) + i \sin (- \frac{\arctan (\frac{2}{5})}{3}))$	$= 1.738722582 - 0.2217219179 i$
$w_{2}$	$= λ (\cos (\frac{- \arctan (\frac{2}{5}) + 2 π}{3}) + i \sin (\frac{- \arctan (\frac{2}{5}) + 2 π}{3}))$	$= - 0.677344478 + 1.61663889 i$
$w_{3}$	$= λ (\cos (\frac{- \arctan (\frac{2}{5}) + 4 π}{3}) + i \sin (\frac{- \arctan (\frac{2}{5}) + 4 π}{3}))$	$= - 1.061378105 - 1.394916967 i$

The principal cube roots are $z_{1}$ and $w_{1}$ , respectively. The cube roots are not expressed in terms of radicals because the trig formulas for $\cos (\frac{x}{3})$ and $\sin (\frac{x}{3})$ are exceedingly complex. (They involve cube roots of expressions containing square roots.) Use of a computing device such as a calculator is the more expedient course of action.

8.8 - Maplet Solution

The Study Guide's Complex Arithmetic Tutor can be used for the computation of $z^{1 / 3}$ for the complex number

$z = 2 + 3 i$

Clicking the above maplet link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.8.1.

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

In the section Exponentiation of complex numbers, enter $n = 1, m = 3$ and click the button labeled Float opposite the symbol $Z^{n / m}$ . The floating-point form of the solution

${(2 + 3 i)}^{1 / 3} = 1.452 + 0.4933 i$

Figure 8.8.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

is returned. Alternatively, setting $k = 3$ in the section Roots of complex numbers, and clicking the button labeled Float to the right of the button labeled kth principal root yields the same result.

The Study Guide's Complex Arithmetic Tutor can be used for the computation of $w^{1 / 3}$ for the complex number

$w = 5 - 2 i$

Clicking the above maplet link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.8.2.

However, it has to be entered as if its name were $z$ .

Enter the complex number by separately giving its real and imaginary part. (The imaginary part of $a + b i$ is the real number $b$ .)

Figure 8.8.2 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

In the section Exponentiation of complex numbers, enter $n = 1, m = 3$ and click the button labeled Float opposite the symbol $Z^{n / m}$ . The floating-point form of the solution, to four significant figures, is

${(5 - 2 i)}^{1 / 3} = 1.739 - 0.2217 i$

Alternatively, setting $k = 3$ in the section Roots of complex numbers, and clicking the button labeled Float to the right of the button labeled kth principal root yields the same result.

To launch the Study Guide's Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

8.8 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

$z^{1 / 3}$

•	Type $z = 2 + 3 i$

•	Context Panel: Assign Name

•	Type $z$ and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Az$

•	Type ${\|z\|}^{1 / 3} cis (Az / 3)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type ${\|z\|}^{1 / 3} cis (Az / 3 + 2 π / 3)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type ${\|z\|}^{1 / 3} cis (Az / 3 + 4 π / 3)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type $z^{1 / 3}$ and press the Enter key.

•	Context Panel: Approximate≻10

$w^{1 / 3}$

•	Type $w = 5 - 2 i$

•	Context Panel: Assign Name

•	Type and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Aw$

•	Type ${\|w\|}^{1 / 3} cis (Aw / 3)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type ${\|w\|}^{1 / 3} cis (Aw / 3 + 2 π / 3)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type ${\|w\|}^{1 / 3} cis (Aw / 3 + 4 π / 3)$ and press the Enter key.

•	Context Panel: Approximate≻10

•	Type $w^{1 / 3}$ and press the Enter key.

•	Context Panel: Approximate≻10

8.8 - Programmatic Solution

It can be shown that a consequence of DeMoivre's law for integer powers is the related law for fractional powers

$z^{1 / m} = ρ^{1 / m} (\cos (\frac{θ}{m} + \frac{2 k π}{m}) + i \sin (\frac{θ}{m} + \frac{2 k π}{m})), k = 0, 1, .. &period;, m - 1$

For convenience, we will also refer to this law as DeMoivre's law for fractional powers.

Maple gives the principal cube root of

>	$z ≔ 2 + 3 i$

>	$zp ≔ evalc (z^{1 / 3})$

whose floating-point form is

>	$evalf (zp)$

To apply DeMoivre's law for fractional powers, we first single out the magnitude and argument of $z$ with

>	$rz ≔ abs (z) &semi; tz ≔ argument (z)$

Then, the three cube roots of $z$ are

>	$z1 ≔ {rz}^{1 / 3} cis (tz / 3) &semi; z2 ≔ {rz}^{1 / 3} cis (tz / 2 + 2 π / 3) &semi; z3 ≔ {rz}^{1 / 3} cis (tz / 3 + 4 π / 3)$

The floating-point form of these numbers, namely,

>	$evalf (z1) &semi; evalf (z2) &semi; evalf (z3)$

shows that the first is the principle cube root.

Maple gives the principal cube root of

>	$w ≔ 5 - 2 i$

>	$wp ≔ evalc (w^{1 / 3})$

whose floating-point form is

>	$evalf (wp)$

To apply DeMoivre's law for fractional powers, we first single out the magnitude and argument of $z$ with

>	$rw ≔ abs (w) &semi; tw ≔ argument (w)$

Then, the three cube roots of $z$ are

>	$w1 ≔ {rw}^{1 / 3} cis (tw / 3) &semi; w2 ≔ {rw}^{1 / 3} cis (tw / 3 + 2 π / 3) &semi; w3 ≔ {rw}^{1 / 3} cis (tw / 3 + 4 π / 3)$

The floating-point form of these numbers, namely,

>	$evalf (w1) &semi; evalf (w2) &semi; evalf (w3)$

shows the first is the principal cube root.

Problem 8.9

8.9 - Mathematical Solution

If $a = \frac{2}{3}$ and $z = - 2 + 3 i$ , with polar form determined by $ρ = \sqrt{13}$ and $θ = π - \arctan (\frac{3}{2})$ , then

$z^{a} = {|z|}^{a} cis (a θ)$ = ${|z|}^{a} (\cos (a θ) + i \sin (a θ))$

evaluates to

$λ (\cos (\frac{2}{3} (π - \arctan (\frac{3}{2}))) + i \sin (\frac{2}{3} (π - \arctan (\frac{3}{2})))) = 0.3085363542 + 2.331004104 i$

where $λ = {(\sqrt{13})}^{2 / 3} = 13^{1 / 3}$ .

Since $z^{2} = - 5 - 12 i$ , with polar form determined by $ρ = 13$ and $θ = \arctan (\frac{12}{5}) - π$ , we then have

${(z^{2})}^{1 / 3}$	$= 13^{1 / 3} (\cos (\frac{\arctan (\frac{12}{5}) - Pi}{3}) + i \sin (\frac{\arctan (\frac{12}{5}) - Pi}{3}))$
	$=$ $1.864440593 - 1.432702373 i$

On the other hand, $z^{1 / 3}$ has polar form determined by $ρ = {(\sqrt{13})}^{1 / 3}$ = $13^{1 / 6}$ and $θ = \frac{1}{3} (π - \arctan (\frac{3}{2}))$ , so that

${(z^{1 / 3})}^{2}$	$= {(13^{1 / 6})}^{2}$ $(\cos (\frac{2}{3} (π - \arctan (\frac{3}{2}))) + i \sin (\frac{2}{3} (π - \arctan (\frac{3}{2}))))$
	$= 0.3085363556 + 2.331004104 i$

Calculating the root before the power gives the same number as the principal value of $z^{a}$ , where $a$ is a rational number.

8.9 - Maplet Solution

For $z = - 2 + 3 i$ , the three complex numbers

$z^{2 / 3} = 0.3084 + 2.331 i$

${(z^{2})}^{1 / 3} = 1.864 - 1.432 i$

${(z^{1 / 3})}^{2} = 0.3084 + 2.331 i$

are computed with the Study Guide's Complex Arithmetic Tutor .

Clicking this maplet link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.9.1.

Figure 8.9.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

The experimental evidence shows that the correct way to raise a complex number to a rational exponent is to compute the root before the power.

To launch the Study Guide's Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

8.9 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

•	Type $z = - 2 + 3 i$

•	Context Panel: Assign Name

Maple's value for $z^{2 / 3}$

•	Type $z^{2 / 3}$ and press the Enter key.

•	Context Panel: Conversions≻ $a + b I$

•	Context Panel: Approximate≻10

By definition, compute $z^{2 / 3} = {|z|}^{2 / 3} cis (\frac{2}{3} Arg (z))$

•	Type $z$ and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Az$

•	Type ${\|z\|}^{2 / 3} cis (\frac{2}{3} Az)$ and press the Enter key.

•	Context Panel: Approximate≻10

Compute ${(z^{2})}^{1 / 3}$

•	Type $z^{2}$ and press the Enter key.

•	Context Panel: Assign to a Name≻ $Z2$

•	Type ${Z2}^{1 / 3}$ and press the Enter key.

•	Context Panel: Approximate≻10

Compute ${(z^{1 / 3})}^{2}$

•	Type $z^{1 / 3}$ and press the Enter key.

•	Context Panel: Conversions≻ $a + b I$

•	Context Panel: Assign to a Name≻ $Z3$

•	Type ${Z3}^{2}$ and press the Enter key.

•	Context Panel: Conversions≻ $a + b I$

•	Context Panel: Approximate≻10

8.9 - Programmatic Solution

Let $z$ be the complex number

>	$z ≔ - 2 + 3 i$

whose polar form is

>	$convert (z, polar)$

Thus, the magnitude and argument are, respectively,

>	$rz ≔ abs (z) &semi; tz ≔ argument (z)$

Using the definition $z^{a} = {|z|}^{a} cis (a θ)$ , the value of $z^{2 / 3}$ is

>	$za ≔ simplify ({rz}^{2 / 3} (cis (2 tz / 3))) &colon; za = evalf (za)$

which happens to agree with what Maple computes for this number, as we see from

>	$zaM ≔ simplify (evalc (z^{2 / 3})) &colon; zaM = evalf (zaM)$

Now, compute ${(z^{2})}^{1 / 3}$ as follows.

>	$N ≔ abs {(z^{2})}^{1 / 3} cis (argument (z^{2}) / 3) &colon; N = evalf (N)$

Surprisingly, squaring first and then taking the cube root does not give the result that the definition gives.

Alternatively, compute ${(z^{1 / 3})}^{2}$ as follows.

>	$M ≔ {(abs {(z)}^{1 / 3} cis (argument (z) / 3))}^{2} &colon; simplify (M) = evalf (M)$

Taking the principal cube root first, then squaring, gives the same result as the definition.

Problem 8.10

8.10 - Mathematical Solution

If $a = \frac{3}{2}$ and $z = - 2 - 3 i$ , with polar form determined by $ρ = \sqrt{13}$ and $θ = \arctan (\frac{3}{2}) - π$ , then

$z^{a} = {|z|}^{a} cis (a θ)$ = ${|z|}^{a} (\cos (a θ) + i \sin (a θ))$

evaluates to

$λ (\cos (\frac{3}{2} (\arctan (\frac{3}{2}) - π)) + i \sin (\frac{3}{2} (\arctan (\frac{3}{2}) - π))) = - 6.814402636 + 0.6603660268 i$

where $λ = {(\sqrt{13})}^{3 / 2}$ .

Since $z^{3} = 46 - 9 i$ , with polar form determined by $ρ = 13 \sqrt{13}$ and $θ = - \arctan (\frac{9}{46})$ , we then have

${(z^{3})}^{1 / 2}$	$= {(13 \sqrt{13})}^{1 / 2} (\cos (- \frac{1}{2} \arctan (\frac{9}{46})) + i \sin (- \frac{1}{2} \arctan (\frac{9}{46})))$
	$=$ $6.814402636 - 0.6603660276 i$

On the other hand, $z^{1 / 2}$ has polar form determined by $ρ = {(\sqrt{13})}^{1 / 2}$ = $13^{1 / 4}$ and $θ = - \arctan (\frac{\sqrt{2 \sqrt{13} + 4}}{\sqrt{2 \sqrt{13} - 4}})$ = $- \arctan (\frac{2 + \sqrt{13}}{3})$ , so that

${(z^{1 / 2})}^{3}$	$= {((13^{1 / 4}))}^{3}$ $(\cos (- 3 \arctan (\frac{2 + \sqrt{13}}{3})) + i \sin (- 3 \arctan (\frac{2 + \sqrt{13}}{3})))$
	$= - 6.814402636 + 0.6603660286 i$

Calculating the root before the power gives the same number as the principal value of $z^{a}$ , where $a$ is a rational number.

8.10 - Maplet Solution

For $z = - 2 - 3 i$ , the three complex numbers

$z^{3 / 2} = - 6.814 + 0.6604 i$

${(z^{3})}^{1 / 2} = 6.814 - 0.6604 i$

${(z^{1 / 2})}^{3} = - 6.814 + 0.6604 i$

are computed with the Study Guide's Complex Arithmetic Tutor .

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.10.1.

Figure 8.10.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

The experimental evidence shows that the correct way to raise a complex number to a rational exponent is to compute the root before the power.

To launch the Study Guide's Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

8.10 - Interactive Solution

Re-initialize Maple by clicking the button to the right.

•	Type $z = - 2 - 3 i$

•	Context Panel: Assign Name

Maple's value for $z^{3 / 2}$

•	Type $z^{3 / 2}$ and press the Enter key.

•	Context Panel: Conversions≻ $a + b I$

•	Context Panel: Approximate≻10

By definition, compute $z^{3 / 2} = {|z|}^{3 / 2} cis (\frac{3}{2} Arg (z))$

•	Type $z$ and press the Enter key.

•	Context Panel: Argument

•	Context Panel: Assign to a Name≻ $Az$

•	Type ${\|z\|}^{3 / 2} cis (\frac{3}{2} Az)$ and press the Enter key.

•	Context Panel: Approximate≻10

Compute ${(z^{3})}^{1 / 2}$

•	Type $z^{3}$ and press the Enter key.

•	Context Panel: Assign to a Name≻ $Z3$

•	Type ${Z3}^{1 / 2}$ and press the Enter key.

•	Context Panel: Approximate≻10

Compute ${(z^{1 / 2})}^{3}$

•	Type $z^{1 / 2}$ and press the Enter key.

•	Context Panel: Conversions≻ $a + b I$

•	Context Panel: Assign to a Name≻ $Z2$

•	Type ${Z2}^{3}$ and press the Enter key.

•	Context Panel: Conversions≻ $a + b I$

•	Context Panel: Approximate≻10

8.10 - Programmatic Solution

Let $z$ be the complex number

>	$z ≔ - 2 - 3 i$

whose polar form is

>	$convert (z, polar)$

Thus, the magnitude and argument are, respectively,

>	$rz ≔ abs (z) &semi; tz ≔ argument (z)$

Using the definition $z^{a} = {|z|}^{a} cis (a θ)$ , the value of $z^{3 / 2}$ is

>	$za ≔ simplify ({rz}^{3 / 2} (cis (3 tz / 2))) &colon; za = evalf (za)$

which happens to agree with what Maple computes for this number, as we see from

>	$zaM ≔ simplify (evalc (z^{3 / 2})) &colon; zaM = evalf (zaM)$

Unfortunately, Maple uses different algorithms for these two computations, and it is difficult to get Maple to convert the trig form of the number to radicals. We can settle for the equivalence of the floating-point forms, or we can impose the following simplifications.

>	$za_exp ≔ radnormal (evalc (simplify (convert (za, \exp))))$

Maple does not readily factor 13 from each of the larger radicals. Were it to do so, the exact equivalence would be obvious. Instead, we show this equivalence via

>	$simplify (za_exp - zaM)$

Now, compute ${(z^{3})}^{1 / 2}$ as follows.

>	$N ≔ abs {(z^{3})}^{1 / 2} cis (argument (z^{3}) / 2) &colon; N = evalf (N)$

Surprisingly, squaring first and then taking the cube root does not give the result that the definition gives.

Alternatively, compute ${(z^{1 / 2})}^{3}$ as follows.

>	$M ≔ {(abs {(z)}^{1 / 2} cis (argument (z) / 2))}^{3} &colon; simplify (M) = evalf (M)$

Taking the principal cube root first, then squaring, gives the same result as the definition.

Problem 8.11

8.11 - Mathematical Solution

8.11 (a) - Mathematical Solution

Without using a graphing device of some sort, the graph of $f (x) = {(x^{2} + 1)}^{2 / 3}$ would have to be obtained painstakingly with a table of values such as the one in Table 8.11.1. Then, using a sheet of graph paper and the even symmetry exhibited by this function, a graph such as the one in Figure 8.11.1 could be drawn.

$[\begin{array}{c} x & f (x) \\ 0. & 1. \\ 0.2 & 1.026491978 \\ 0.4 & 1.104007421 \\ 0.6 & 1.227512544 \\ 0.8 & 1.390686478 \\ 1.0 & 1.587401052 \\ 1.2 & 1.812424333 \end{array}]$	$[\begin{array}{c} x & f (x) \\ 1.4 & 2.061552856 \\ 1.6 & 2.331488792 \\ 1.8 & 2.619653821 \\ 2.0 & 2.924017738 \\ 2.2 & 3.242962286 \\ 2.4 & 3.575179054 \\ 2.6 & 3.919594571 \end{array}]$	$[\begin{array}{c} x & f (x) \\ 2.8 & 4.275315572 \\ 3.0 & 4.641588834 \\ 3.2 & 5.017771427 \\ 3.4 & 5.403308444 \\ 3.6 & 5.797716104 \\ 3.8 & 6.200568800 \\ 4.0 & 6.611489018 \end{array}]$
Table 8.11.1 Values of $f (x) = {(x^{2} + 1)}^{2 / 3}$

Figure 8.11.1 Graph of $f (x) = {(x^{2} + 1)}^{2 / 3}$

8.11 (b) - Mathematical Solution

Without using a graphing device of some sort, the graph of $g (x) = {(x^{2} - 1)}^{3 / 2}$ would have to be obtained painstakingly with a table of values such as the one in Table 8.11.2. Note that for $|x^{2}| < 1$ , $g (x)$ is complex-valued. In particular, the value of $g (x)$ at $x = 0$ is $g (0) = - i$ . This is obtained if the square root is taken before the power is applied. The square root of $- 1$ is $i$ , and $i^{3} = - i$ .

Using a sheet of graph paper and the even symmetry exhibited by this function, a graph such as the one in Figure 8.11.2 could then be drawn.

$[\begin{array}{c} x & g (x) \\ 1.0 & 0. \\ 1.2 & 0.2918629816 \\ 1.4 & 0.9406040612 \\ 1.6 & 1.948439377 \\ 1.8 & 3.352525020 \\ 2.0 & 5.196152427 \end{array}]$

Table 8.11.2 Values of $g (x) = {(x^{2} - 1)}^{3 / 2}$

Figure 8.11.2 Graph of $g (x) = {(x^{2} - 1)}^{3 / 2}$

8.11 - Maplet Solution

8.11 (a) - Maplet Solution

The function $f (x) = {(x^{2} + 1)}^{2 / 3}$ can be graphed in Maple with a variety of "point-and-click" approaches. One way is to execute a secondary click on

${(x^{2} + 1)}^{2 / 3}$

In the Context Panel that pops up, select Plots, and 2-D Plot. Maple will automatically generate a graph similar to the one in Figure 8.11.3.

Alternatively, from the Context Panel, select Plots and Plot Builder. In the Plot Builder pane that opens, select "2-D plot" and interactively apply appropriate options.

The Plot Builder can also be accessed from the Tools/Assistants menu at the top of the worksheet interface.

In the window that opens, click Add, and type in the expression to be graphed. Then, click Accept and OK. The result is the same as described when the Plot Builder is launched from the Context Panel.

Figure 8.11.3 Graph of $f (x) = {(x^{2} + 1)}^{2 / 3}$

8.11 (b) - Maplet Solution

A graph of $g (x) = {(x^{2} - 1)}^{3 / 2}$ , similar to the one shown in Figure 8.11.4, can be obtained by the techniques described in Part (a). In particular, execute a secondary click on

${(x^{2} - 1)}^{3 / 2}$

In the Context Panel that pops up, select Plots, and 2-D Plot. Maple will automatically generate a graph similar to the one in Figure 8.11.4.

Alternatively, from the Context Panel, select Plots and Plot Builder. In the Plot Builder pane that opens, select "2-D plot" and interactively apply appropriate options.

The Plot Builder can also be accessed from the Tools/Assistants menu at the top of the worksheet interface.

In the window that opens, click Add, and type in the expression to be graphed. Then, click Accept and OK. The result is the same as described when the Plot Builder is launched from the Context Panel.

Figure 8.11.4 Graph of $g (x) = {(x^{2} - 1)}^{3 / 2}$

To obtain the value of $g (0) = {(- 1)}^{3 / 2}$ , the Study Guide's Complex Arithmetic Tutor can be used.

Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.11.5.

To launch the Study Guide's Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

Figure 8.11.5 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

8.11 - Interactive Solution

8.11 (a) - Interactive Solution

•	Type ${(x^{2} + 1)}^{2 / 3}$

•	Context Panel: Plots≻Plot Builder Select "2-D plot" Change the range to $- 4 \leq x \leq 4$ Change the view to $0 \leq y \leq 7$

8.11 (b) - Interactive Solution

Re-initialize Maple by clicking the button to the right.

•	Type $g (x) = {(x^{2} - 1)}^{3 / 2}$

•	Context Panel: Assign Function

•	Type $g (x)$ and press the Enter key.

•	Context Panel: Plots≻Plot Builder Select "2-D plot" Change the range to $- 2 \leq x \leq 2$ Change the view to $0 \leq y \leq 5$

•	Type $g (0)$

•	Context Panel: Evaluate and Display Inline

8.11 - Programmatic Solution

8.11 (a) - Programmatic Solution

The graph of the function

>	$f ≔ {(x^{2} + 1)}^{2 / 3}$

appears in Figure 8.11.6.

>	$plot (f, x = - 3 .. 3, y = 0 .. 5)$

Figure 8.11.6 Graph of $f (x) = {(x^{2} + 1)}^{2 / 3}$

Because the expression $x^{2} + 1$ is always positive, computation of the cube root presents no particular problems. Maple computes the real cube root, squares it, and draws the graph.

8.11 (b) - Programmatic Solution

The graph of the function

>	$g ≔ {(x^{2} - 1)}^{3 / 2}$

appears in Figure 8.11.7.

>	$plot (g, x = - 3 .. 3)$

Figure 8.11.6 Graph of $g (x) = {(x^{2} - 1)}^{3 / 2}$

For $x$ in the open interval $(- 1, 1)$ , the expression $x^{2} - 1$ is negative. The cube of this negative number is negative, and its square root is then imaginary. Hence, the graph has a gap between $x = - 1$ and $x = 1$ .

In particular, the function value $g (0)$ is the imaginary number $- i$ , as evidenced by

>	$eval (g, x = 0)$

For positive numbers, computing a rational power such as $3 / 2$ poses no particular problem. However, for a negative number, there appears to be two different ways to proceed. The root can be computed first, then the power, or the power can be computed first, and then the root taken. Each of these calculations leads to different results, only one of which is correct.

When $x = 0$ , the function $g (x)$ requires the calculation of the complex number ${(- 1)}^{3 / 2}$ .

If the root is taken first, the computation amounts to $i^{3}$ which is $- i$ .

If the power is taken first, the computation requires $- 1$ to be cubed first, which yields $- 1$ . Then, taking the square root, we get $\sqrt{- 1} = i$ .

The first calculation, the one whose result is consistent with the Maple outcome, is the correct one. The student who remembers the adage

Roots before Powers

will always get the correct answer in these calculations. Moreover, the student who reasons that one must be firmly rooted before one can exert power, will more easily remember the maxim that "roots must be taken before powers."

Problem 8.12

8.12 - Mathematical Solution

	8.12 (a) - Mathematical Solution
	The calculation $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(−1)}^{1 / 2})}^{2} = i^{2}$ = $- 1$ is correct because the root was taken before the power was applied.

	8.12 (b) - Mathematical Solution
	The calculation $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1 is not correct because the power was applied before the root was taken. Thus, the statement is best written as $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ $\neq$ ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1

8.12 - Maplet Solution

8.12 (a) - Maplet Solution

The Study Guide's Complex Arithmetic Tutor can be used to verify that the calculation

${(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(- 1)}^{1 / 2})}^{2} = i^{2}$ = $- 1$

is correct because the root was taken before the power was applied.

Clicking the above link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.12.1.

In the section Exponentiation of complex numbers, the result opposite the symbol $Z^{n / m}$ is obtained by the definition given in Problem 8.9. This agrees with the result opposite the symbol ${(Z^{1 / m})}^{n}$ , again consistent with the rule "roots before powers."

Figure 8.12.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

To launch the Study Guide's Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

8.12 (b) - Maplet Solution

The Study Guide's Complex Arithmetic Tutor can be used to verify that the calculation

$- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1

is not correct because the power was applied before the root was taken. Thus, the statement is best written as

$- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ $\neq$ ${({(−1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1

Clicking the above link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 8.12.2.

Figure 8.12.1 Thumbnail image of the Study Guide's Complex Arithmetic Tutor

In the section Exponentiation of complex numbers, the result opposite the symbol $Z^{n / m}$ is obtained by the definition given in Problem 9. This does not agrees with the result opposite the symbol ${(Z^{n})}^{1 / m}$ , again demonstrating that violating the rule "roots before powers" leads to incorrect results.

To launch the Study Guide's Complex Arithmetic Tutor, click the following link: Complex Arithmetic Tutor

8.12 - Interactive Solution

	8.12 (a) - Interactive Solution
	The calculation $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(- 1)}^{1 / 2})}^{2} = i^{2}$ = $- 1$ is correct because the root was taken before the power was applied.

	8.12 (b) - Interactive Solution
	The calculation $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1 is not correct because the power was applied before the root was taken. Thus, the statement is best written as $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ $\neq$ ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1

8.12 - Programmatic Solution

	8.12 (a) - Programmatic Solution
	The calculation $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(- 1)}^{1 / 2})}^{2} = i^{2}$ = $- 1$ is correct because the root was taken before the power was applied.

	8.12 (b) - Programmatic Solution
	The calculation $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ = ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1 is not correct because the power was applied before the root was taken. Thus, the statement is best written as $- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}}$ $\neq$ ${({(- 1)}^{2})}^{1 / 2} = \sqrt{1}$ = 1

Exercises - Chapter 8

For the two complex numbers $z$ and $w$ given in each of Exercises 8.1 - 8.4, obtain

(a) $z + w$

(b) $z - w$

(d) $\frac{z}{w}$

8.1. $z = 7 - 5 i, w = - 3 - 9 i$

8.2. $z = 4 + i, w = - 6 - 3 i$

8.3. $z = 8 + 7 i, w = - 7 - 8 i$

8.4. $z = - 1 + 4 i, w = - 1 - 3 i$

In Exercises 8.5 - 8.8, obtain the polar form (i.e., find $rho = |z|$ and $theta$ , the argument of $z$ ) for the indicated complex number. In each case, show that $rho cis (theta)$ restores the rectangular form of $z$ .

8.5. $z = 1 + i$

8.6. $z = - 1 + i$

8.7. $z = - 1 - i$

8.8. $z = 1 - i$

In Exercises 8.9 - 8.12, use DeMoivre's law for integer powers to obtain $z^{n}$ for the given $z$ and $n$ .

8.9. $z = 3 + 5 i, n = 3$

8.10. $z = - 3 + 5 i, n = - 2$

8.11. $z = - 3 - 5 i, n = 5$

8.12. $z = 3 - 5 i, n = - 4$

In Exercises 8.13 - 8.16, obtain both square roots of the given complex number. In each case, indicate which is the principal square root.

8.13. $z = 1 + i \sqrt{3}$

8.14. $z = 2 - i \sqrt{3}$

8.15. $z = \sqrt{3} - 2 i$

8.16. $z = \sqrt{3} - i$

In Exercises 8.17 - 8.20, obtain all three cube roots of the given complex number. In each case, indicate which is the principal cube root.

8.17. $z = 1 - i \sqrt{3}$

8.18. $z = 2 + i \sqrt{3}$

8.19. $z = \sqrt{3} + 2 i$

8.20. $z = \sqrt{3} + i$

In Exercises 8.21 - 8.22, a function $f (x) = {u (x)}^{m / n}$ and a value for $x$ are given.

(a) Obtain a graph of $f (x)$ .

(b) At the given value of $x$ , compute both ${[{u (x)}^{1 / n}]}^{m}$ and ${[{u (x)}^{m}]}^{1 / n}$ .

Indicate which of these two values is correct.

8.21. $f (x) = {(x^{4} + 1)}^{2 / 3}$ , $x = 0$

8.22. ${(f (x) = x^{2} - 1)}^{3 / 2}, x = 0$

Go to Chapter 1 2 3 4 5 6 7 9 10 11

Go to Contents

Download Help Document

Maple

Maple Add-Ons

MapleSim

MapleSim Add-Ons

Systems Engineering

Consulting Services

Maple T.A. and Möbius

Education

Industries

Automotive and Aerospace

Robotics

Machine Design & Industrial Automation

Other

Application Areas

Product Pricing

Purchasing

Institutional Student Licensing

Maplesoft Elite Maintenance (EMP)

Support

Product Training

Online Product Help

Webinars & Events

Publications

Content Hubs

Examples & Applications

Community

About Maplesoft

Media Center

User Community

Contact

Online Help

All Products Maple MapleSim

Maple

Powerful math software that is easy to use

Maple Add-Ons

MapleSim

Advanced System Level Modeling

MapleSim Add-Ons

Systems Engineering

Consulting Services

Maple T.A. and Möbius

Education

Industries

Automotive and Aerospace

Robotics

Machine Design & Industrial Automation

Other

Application Areas

Product Pricing

Purchasing

Institutional Student Licensing

Maplesoft Elite Maintenance (EMP)

Support

Product Training

Online Product Help

Webinars & Events

Publications

Content Hubs

Examples & Applications

Community

About Maplesoft

Media Center

User Community

Contact

Online Help

All Products Maple MapleSim