So, we can do basic algebra with complex numbers, take powers of them, and apply trig and exponential functions to them. There isn’t much left that we might want to do, but taking powers of them is very important and also pretty easy. Let’s say we wanted to calculate the solution to the equation: $z^2=-1$

Well, we know what $z$ is for this, because that’s what got us into this mess in the first place! We know that there are two solutions and they are $\pm i$. That is to say that if you multiply either of these numbers by themselves, you get -1, by definition. How about $z^2=i$. The first thing to do whenever you have to take the root of a complex number, or a real number (here the second root of $i$) is to convert that number into modulus argument form – it will be infinitely easier like that. We know that $i=re^{i\theta}$ where $r=1$ and $\theta=\frac{\pi}{2}+2n\pi$. So we can write: $z^2=e^{\frac{i\pi}{2}+2n\pi i}$

or just: $z=e^{\frac{i\pi}{4}+n\pi i}$ $n$ again goes over the integers, but we have to ask whether it’s going to make a difference to what $z$ is. When $n=0$ this number is $e^{\frac{i\pi}{4}}$ and you can work out that this is $\frac{1+i}{\sqrt{2}}$, when $n=1$ we have $\frac{-1-i}{\sqrt{2}}$ which is different, but when $n=2$ we get back to the same answer as when $n=0$, and for $n=3$ we get the answers for when $n=1$. We can see very quickly that there are actually only two unique solutions to this equation which are $\pm\frac{1+i}{\sqrt{2}}$. Plot these in the Argand plane and you will see how they relate to $i$. It is pretty clear that when you multiply these numbers by themselves that they will give you $i$ (remember, multiplying two complex numbers together is done by adding the arguments and multiplying the magnitudes. In this case the magnitudes are 1 and the phases add up to give $\frac{\pi}{4}$ (modulo $2\pi$ in both cases)).

How about a slightly more complicated question. How about: $z^5=(1+i)$

Again, we start by converting from cartesian to mod/arg form which gives: $z^5=\sqrt{2}e^{\frac{i\pi}{4}+2n\pi i}$

so: $z=2^{\frac{1}{10}}e^{\frac{i\pi}{20}+\frac{2n}{5}\pi i}$

You can see quickly that when $n=0,1,2,3$ and $4$ you will get unique results, but $n=5$ will give the same as $n=0$, and $n=6$ will give the same as $n=1$, etc. We can plot these in the Argand plane and see what they look like: The five fifth roots of $1+i$. The point $1+i$ is shown in blue, the five roots are shown in black. You can see that when you multiply each one of them by itself five times you will get $1+i$ – this is the definition of the fifth root.

It’s always nice to have a general procedure when we’re trying to solve a set of problems, and in this case we can find a very simple general procedure. You should have noticed one thing by now, which is that the $n^{th}$ root of a number (be it real or imaginary) will have $n$ solutions. That might seem strange that there are 7 solutions to the equation $z^7=1$ but check it – it is true, and this is because there are seven angles, that when you add them to themselves 7 times, give you integer multiples of $2\pi$ – think about it…

So, what if we want to solve: $z^n=q$

We start by writing $q$ in mod/arg form, calling its magnitude $\rho$ and its angle $\phi$: $z^n=\rho e^{i\phi+2k\pi i}$

now: $z=\rho^{\frac{1}{n}} e^{\frac{i\phi}{n}+2\frac{k}{n}\pi i}$

where now $k=0,1,...(n-1)$. so again we have $n$ solutions for an $n^{th}$ root.

We should note something rather important here. When we write $\rho^{\frac{1}{n}}$ this really is the $n^{th}$ root of a real number, whereas when we write $z^n=\rho$, the two are not inverse of each other. You can see this by the fact that you write $x^2=1$ which has two solutions but writing $x=\sqrt{1}$ is just one of them, the negative solution is encoded in the angular part of the solution. ie. in this case you would write $x=\sqrt{1}e^{in\pi}$ where $n=0, 1$ which is just the statement that $\pm 1$ are the two solutions to the equation $x^2=1$.

 How clear is this post?