From Example 9.2.8, the polar basis vectors are

${\mathit{e}}_{\mathit{r}}\=\frac{\partial \mathbf{R}}{\partial r}$ = $\left[\begin{array}{c}\cos \(\mathrm{\θ}\)\\ \sin \(\mathrm{\θ}\)\end{array}\right]$ and ${\mathit{e}}_{\mathbf{\θ}}\=\frac{1}{r}\frac{\partial \mathbf{R}}{\partial \mathrm{\θ}}$ = $\frac{1}{r} \left[\begin{array}{c}-r \sin \(\mathrm{\θ}\)\\ r \cos \(\mathrm{\θ}\)\end{array}\right]\=\left[\begin{array}{c}-\sin \(\mathrm{\θ}\)\\ \cos \(\mathrm{\θ}\)\end{array}\right]$ 

where R is the position vector to the point $\left(x\,y\right)\=\left(r \cos \left(\mathrm{\θ}\right)\,r \sin \left(\mathrm{\θ}\right)\right)$.

Consequently,

$\mathbf{F}\=a {\mathit{e}}_{\mathit{r}}\+b {\mathit{e}}_{\mathbf{\θ}}$ = $a \left[\begin{array}{c}\cos \(\mathrm{\θ}\)\\ \sin \(\mathrm{\θ}\)\end{array}\right]\+b \left[\begin{array}{c}-\sin \(\mathrm{\θ}\)\\ \cos \(\mathrm{\θ}\)\end{array}\right]$ = $a \left[\begin{array}{c}x\/r\\ y\/r\end{array}\right]\+b \left[\begin{array}{c}-y\/r\\ x\/r\end{array}\right]$ 

so the Cartesian form of F is $\frac{a x-b y}{\sqrt{x^2\+y^2}} \mathbf{i}\+\frac{a y\+b x}{\sqrt{x^2\+y^2}} \mathbf{j}$.