Complex Numbers and Polar Coordinates

Forgot to hit “publish” earlier…

So we’ve seen that the unit complex numbers can be written in the form $e^{i\theta}$ where $\theta$ denotes the (signed) angle between the point on the circle and $1+0i$ . We’ve also seen that this view behaves particularly nicely with respect to multiplication: multiplying two unit complex numbers just adds their angles. Today I want to extend this viewpoint to the whole complex plane.

If we start with any nonzero complex number $z=a+bi$ , we can find its absolute value $\lvert z\rvert=\sqrt{a^2+b^2}$ . This is a positive real number which we’ll also call $r$ . We can factor this out of $z$ to find $z=r\left(\frac{a}{r}+\frac{b}{r}i\right)$ . The complex number in parentheses has unit absolute value, and so we can write it as $e^{i\theta}$ for some $\theta$ between $-\pi$ and $\pi$ . Thus we’ve written our complex number in the form

$\displaystyle z=re^{i\theta}$

where the positive real number $r$ is the absolute value of $z$ , and $\theta$ — a real number in the range $\left(-\pi,\pi\right]$ — is the angle $z$ makes with the reference point $1+0i$ . But this is exactly how we define the polar coordinates $(r,\theta)$ back in high school math courses.

Just like we saw for unit complex numbers, this notation is very well behaved with respect to multiplication. Given complex numbers $r_1e^{i\theta_1}$ and $r_2e^{i\theta_2}$ we calculate their product:

$\displaystyle r_1e^{i\theta_1}r_2e^{i\theta_2}=\left(r_1r_2\right)e^{i\left(\theta_1+\theta_2\right)}$

That is, we multiply their lengths (as we already knew) and add their angles, just like before. This viewpoint also makes division simple:

$\displaystyle\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}=\frac{r_1}{r_2}e^{i\left(\theta_1-\theta_2\right)}$

In particular we see that

$\displaystyle\left(re^{i\theta}\right)^{-1}=\frac{1}{r}e^{-i\theta}=\frac{1}{r^2}\overline{re^{i\theta}}$

so multiplicative inverses are given in terms of complex conjugates and magnitudes as we already knew.

Powers (including roots) are also easy, which gives rise to easy ways to remember all those messy double- and triple-angle formulæ from trigonometry:

$\displaystyle\begin{aligned}\cos(2\theta)+i\sin(2\theta)&=e^{i\left(2\theta\right)}\\&=\left(e^{i\theta}\right)^2\\&=\left(\cos(\theta)+i\sin(\theta)\right)^2\\&=\left(\cos(\theta)^2-\sin(\theta)^2\right)+i\left(2\sin(\theta)\cos(\theta)\right)\end{aligned}$

$\displaystyle\begin{aligned}\cos(3\theta)+i\sin(3\theta)&=e^{i\left(3\theta\right)}\\&=\left(e^{i\theta}\right)^3\\&=\left(\cos(\theta)+i\sin(\theta)\right)^3\\&=\left(\cos(\theta)^3-3\cos(\theta)\sin(\theta)^2\right)+i\left(3\cos(\theta)^2\sin(\theta)-\sin(\theta)^3\right)\\&=\left(\cos(\theta)^3-3\cos(\theta)\left(1-\cos(\theta)^2\right)\right)+i\left(3\left(1-\sin(\theta)^2\right)\sin(\theta)-\sin(\theta)^3\right)\\&=\left(4\cos(\theta)^3-3\cos(\theta)\right)+i\left(3\sin(\theta)-4\sin(\theta)^3\right)\end{aligned}$

Other angle addition formulæ should be similarly easy to verify from this point.

In general, since we consider complex numbers multiplicatively so often it will be convenient to have this polar representation of complex numbers at hand. It will also generalize nicely, as we will see.

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May 29, 2009 - Posted by John Armstrong | Fundamentals, Numbers

3 Comments »

[...] are like nonnegative real numbers. And so this “polar decomposition” is like the polar form of a complex number, where we write a complex number as the product of a unit complex number and a nonnegative real [...]

Pingback by Polar Decomposition « The Unapologetic Mathematician | August 19, 2009 | Reply
please send me any subject from complex numbers and numerical numbers

Comment by atef alarjan | September 9, 2009 | Reply
[...] points in the plane . This is all well and good, but we might want to talk about the function in polar coordinates. To this end, we may define and . These are the component functions describing a transformation [...]

Pingback by Cauchy’s Invariant Rule « The Unapologetic Mathematician | October 8, 2009 | Reply

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