Complex Number Roots

Question

When I am solving to find the root of a complex number what exactly am I finding? Does it relate somehow to the complex plane? What would be it's geometrical representation if it has one?

Answer 1

Background

Multiplication of complex numbers can be thought of as scaling and rotating. Any (nonzero) complex number has a unique representation in polar form, i.e. of the form $re^{i\theta}$ with $r>0$ a positive real number and $0 \le \theta < 2\pi$.

If $z=re^{i\theta}$ and $w=se^{i\phi}$ then $zw = rse^{i(\theta+\phi)}$. The effect on $z$ of multiplying by $w$ is to scale it by $s$ and rotate it by $\phi$ about the origin. So for example when we go from $w = se^{i\phi}$ to $w^2=s^2e^{2i\phi}$, we square the magnitude and rotate by $\phi$ about the origin.

Answer

Suppose $z=re^{i\theta}$ and we're looking for $w=se^{i\phi}$ such that $w^2=z$. This will be a square root of $z$. The only possibility for $s$ is $\sqrt{r}$, since this is the only positive real number that squares to give $r$. Now you'd think that the only possibility for $\phi$ is $\frac{\theta}{2}$, but this isn't so. Because $e^{2i\pi}=1$, we also have $$e^{2i(\phi+\pi)} = e^{2i\phi}e^{2i\pi}=e^{2i\phi}$$ so both $\frac{\theta}{2}$ and $\frac{\theta}{2}+\pi$ work. So the two square roots of $z=re^{i\theta}$ are $$w_0 = \sqrt{r}e^{i\frac{\theta}{2}} \qquad \text{and} \qquad w_1 = \sqrt{r}e^{i(\frac{\theta}{2}+\pi)}$$

More generally, the $n^{\text{th}}$ roots of $z=re^{i\theta}$ are $w_0, \dots, w_{n-1}$, where $$w_k = \sqrt[n]{r}e^{i(\frac{\theta}{n} + \frac{2k\pi}{n})}$$ Why? Because $(\sqrt[n]{r})^n=r$ and so the magnitudes work out correctly, and for each $k$ we have $$n\left(\frac{\theta}{n} + \frac{2k\pi}{n}\right) = \theta + k \cdot 2\pi$$ so that $$\left(e^{i\left(\frac{\theta}{n} + \frac{2k\pi}{n}\right)}\right)^n = e^{i\theta}(e^{2i\pi})^n = e^{i\theta}$$

Sorry that the notation is so ugly.

Illustration

Take $-1$ as an example. Written in polar form, this is just $1 \cdot e^{i\pi}$. The two roots are therefore $$1 \cdot e^{i\frac{\pi}{2}} \qquad \text{and} \qquad 1 \cdot e^{i\frac{3\pi}{2}}$$ which turn out to be $i$ and $-i$. Now $i^2$ is what you get when you scale $i$ by $|i|=1$ -- which does nothing -- and rotate it by $\arg i = \frac{\pi}{2}$, which clearly gives you $-1$. Likewise, $(-i)^2$ is what you get when you scale $-i$ by $|i|=1$ -- which does nothing -- and rotate it by $\arg(-i) = \frac{3\pi}{2}$ about the origin, which again clearly gives you $-1$.

Answer 2

Let's suppose you have a complex number $z$. A square root of $z$ is a complex number $w$ such that $w^2 = z$, obviously. Note that I wrote "a square root", not "the square root". So, now let's think about what squaring does to a complex number. This is easier if you think of the complex number in "polar" form. Say $w = re^{i\theta}$, where r is the modulus of $w$ and $\theta$ is its argument. Then we know that $w^2 = r^2e^{2i\theta}$. In other words, squaring a complex number squares its modulus and doubles its argument. Or looking at it the other way around, you can get a square root of a complex number by taking the square root of its modulus and halving its argument.

For a fabulously well-written geometric treatment of complex numbers, I highly recommend this book by Tristan Needham.