According to the Fundamental Theorem of Algebra, every nonzero, single variable polynomial of degree n with real coefficients must have exactly n roots, counting multiplicity. However, in some cases, it is not possible to describe all of these roots using real numbers. For example, the polynomial $x^{2 }\+ 1 \=0$ must have two roots, since it has real coefficients and degree 2, but there are no real numbers which satisfy this equation. So, how do we solve for these nonreal roots? We use complex numbers!

Define $\mathrm{I}$ to be $\sqrt{-1}$. Within the real number system, we cannot take the square root of a negative number, so I must not be a real number and is therefore known as the imaginary unit. Using the set of all numbers of the form $a\+b \mathrm{I}$ , called the complex numbers, we can obtain the two roots of $x^2\+1\=0$.  (What are they?)

The set of all real numbers, =, is a subset of the set of all complex numbers, C, because every real number has the form $z \= a \+ 0 \mathrm{I}$. A number of the form $\mathrm{z} \= 0\+b\mathit{ }\mathrm{I}$ is called purely imaginary because it has no real part.

Converting between Cartesian and Polar Coordinates

When points in the complex plane are described using polar coordinates: $z \= \left(r\, \mathrm{\θ}\right)$, r is called the modulus or magnitude of z and $\mathrm{\θ}$ is the argument or phase of z.   We define the modulus as $r \= \left| z \right| \= \sqrt{z \mathrm{z\&conjugate0;} }$, where $\mathrm{z\&conjugate0;} \= a - b \mathrm{I}$ is the complex conjugate of z. So, $r \= \sqrt{\left(a \+ b \mathrm{I}\right)\left(a - b\mathrm{I}\right)} \= \sqrt{a^2 - a b \mathrm{I} \+ a b \mathrm{I} - b^2 {\mathrm{I}}^2} \= \sqrt{a^2 - b^2\left(-1\right)} \= \sqrt{a^2 \+ b^2}$. The argument, $\mathrm{\θ}$, measures (in radians) the angle between the positive Re(z) axis and the line segment connecting the point z to the origin. To switch back to a regular Cartesian point from polar form, we can use Euler's formula: $z\mathbf{}\= r {\ⅇ}^{\mathrm{I}\mathbf{}\mathrm{\θ}} \= r \mathrm{cis}\left(\mathrm{\theta }\right)\= r \left(\cos \left(\mathrm{\θ}\right) \+ \mathrm{I} \sin \left(\mathrm{\θ}\right)\right)$.