Polar Coordinates for Complex Numbers

A point in the plane can be specified in polar coordinates instead of rectangular coordinates. This is often convenient when the point in question represents a complex number.

Consider the complex number $z = a + bi$, corresponding to point $(a,b)$ in the plane. Its polar coordinates are $r$ and $\theta$, where $r$ is the distance from the origin and $\theta$ is the counterclockwise angle from the positive real axis to the point. We have $a = r \cos \theta$ and $b = r \sin \theta$, so

\begin{displaymath}z = a + b i = r (\cos \theta + i \sin \theta) = r {\rm e}^{i \theta}\end{displaymath}

(by our definition of $\exp$ for complex numbers). Thus a complex number can be written as $z = r \exp(i\theta)$ where $r > 0$ and $\theta$ is real (we can, but don't always choose to, take $\theta$ in the interval from $0$ to $2\pi$). Note that $\displaystyle r = \sqrt{a^2 + b^2} = \vert z\vert$.

For example, if we want to represent $z = -\sqrt{3}+i$ in polar coordinates, we first calculate $\displaystyle a^2 + b^2 = (-\sqrt{3})^2 + 1^2 = 4$ so $r = 2$. Then we want $\sin \theta = b/r = 1/2$ and $\cos \theta = a/r = -\sqrt{3}/2$. The angle with this sine and this cosine in the interval $0$ to $2\pi$ is $5 \pi/6$. So the polar representation is $\displaystyle z = 2 \exp(5 \pi i/6)$.

The polar representation is especially useful for multiplication and division of complex numbers. If $z_1 = r_1 \exp(i \theta_1)$ and $z_2 = r_2 \exp(i \theta_2)$ then

\begin{displaymath} \vcenter{\openup.7ex\mathsurround=0pt
\ialign{\strut\hfil...
...= \frac{r_1}{r_2} {\rm e}^{i(\theta_1 - \theta_2)}\cr\crcr}}  \end{displaymath}

The rule can be summed up as: to multiply complex numbers, multiply the $r$'s and add the $\theta$'s. To divide, divide the $r$'s and subtract the $\theta$'s.

For example, let's divide $z = 1 + i$ by $w = -\sqrt{3} + i = 2 \exp(5 \pi i/6)$. We have $1 + i = \sqrt{2} \exp(\pi i/4)$ since $\cos (\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$. So

\begin{displaymath}z/w = \frac{\sqrt{2}}{2} \exp((5 \pi/6 - \pi/4)i) =
\frac{\sqrt{2}}{2} {\rm e}^{7\pi i/12}\end{displaymath}





Robert Israel
2002-02-17