# Visualizing complex number powers

Learn how powers of complex numbers behave when you look at their graphical effect on the complex plane.

## Connection between $i^2 = -1$ and where $i$ lives

We began our study of complex numbers by inventing a number $i$ that satisfies $i^2 = -1$, and later visualized it by placing it outside the number line, one unit

*above*$0$. With the visualizations offered in the last article, we can now see why that point in space is such a natural home for a number whose square is $-1$.You see, multiplication by $i$ gives a $90^\circ$ rotation about the origin:

You can think about this either because $i$ has absolute value $1$ and angle $90^\circ$, or because this rotation is the only way to move the grid around (fixing $0$) which places $1$ on the spot where $i$ started off.

So what happens if we multiply everything in the plane by $i$ twice?

It is the same as a $180^\circ$ rotation about the origin, which is multiplication by $-1$. This of course makes sense, because multiplying by $i$ twice is the same as multiplying by $i^2$, which should be $-1$.

It is interesting to think about how if we had tried to place $i$ somewhere else while still maintaining its characteristic quality that $i^2 = -1$, we could not have had such a clean visualization for complex multiplication.

## Powers of complex numbers

Let's play around some more with repeatedly multiplying by some complex number.

### Example 1: $(1 + i\sqrt{3})^3$

Take the number $z = 1 + i\sqrt{3}$, which has absolute value $\sqrt{1^2 + (\sqrt{3})^2} = 2$, and angle $60^\circ$. What happens if we multiply everything on the plane by $z$ three times in a row?

Everything is stretched by a factor of $2$ three times, and so is ultimately stretched by a factor of $2^3 = 8$. Likewise everything is rotated by $60^\circ$ three times in a row, so is ultimately rotated by $180^\circ$. Hence, at the end it's the same as multiplying by $-8$, so $(1 + i\sqrt{3})^3 = -8$.

We can also see this using algebra as follows:

### Example 2: $(1 + i)^8$

Next, suppose we multiply everything on the plane by $(1 + i)$ eight successive times:

Since the magnitude of $1 + i$ is

$|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$,

everything is stretched by a factor of $\sqrt{2}$ eight times, and hence is ultimately stretched by a factor of $(\sqrt{2})^8 = 2^4 = 16$.

Since the angle of $(1 + i)$ is $45^\circ$, everything is ultimately rotated by $8 \cdot 45^\circ = 360^\circ$, so in total it's as if we didn't rotate at all. Therefore $(1 + i)^8 = 16$.

Alternatively, the way to see this with algebra is

### Example 3: $z^5 = 1$

Now let's start asking the reverse question: Is there a number $z$ such that after multiplying everything in the plane by $z$ five successive times, things are back to where they started? In other words, can we solve the equation $z^5 = 1$? One simple answer is $z = 1$, but let's see if we can find any others.

First off, the magnitude of such a number would have to be $1$, since if it were more than $1$, the plane would keep stretching, and if it were less than $1$, it would keep shrinking. Rotation is a different animal, though, since you can get back to where you started after repeating certain rotations. In particular, if you rotate $\dfrac{1}{5}$ of the way around, like this

then doing this $5$ successive times will bring you back to where you started.

The number which rotates the plane in this way is $\cos(72^\circ) + i\sin(72^\circ)$, since $\dfrac{360^\circ}{5} = 72^\circ$.

There are also other solutions, such as rotating $\dfrac{2}{5}$ of the way around:

or $\dfrac{1}{5}$ of the way around the other way:

In fact, beautifully, the numbers which solve the equation form a perfect pentagon on the unit circle:

### Example 4: $z^6 = -27$

Looking at the equation $z^6 = -27$, it is asking us to find a complex number $z$ such that multiplying by this number $6$ successive times will

**stretch by a factor of $27$**, and**rotate by $180^\circ$**, since the negative indicates $180^\circ$ rotation.Something which will stretch by a factor of $27$ after $6$ applications must have magnitude $\sqrt[6]{27} = \sqrt{3}$, and one way to rotate which gives $180^\circ$ after $6$ applications is to rotate by $\dfrac{180^\circ}{6} = 30^\circ$. Therefore one number that solves this equation $z^6 = -27$ is

However, there are also other answers! In fact, those answers form a perfect hexagon on the circle with radius $\sqrt{3}$:

Can you see why?

## Solving $z^n= w$ in general

Let's generalize the last two examples. If you are given values $w$ and $n$, and asked to solve for $z$, as in the last example where $n=6$ and $w = -27$, you first find the polar representation of $w$:

This means the angle of $z$ must be $\dfrac{\theta}{n}$, and its magnitude must be $\sqrt[n]{r}$, since this way multiplying by $z$ a total of $n$ successive times will in effect rotate by $\theta$ and scale by $r$, just as $w$ does, so

To find the other solutions, we keep in mind that the angle $\theta$ could have been thought of as $\theta + 2\pi$, or $\theta + 4\pi$, or $\theta + 2k\pi$ for any integer $k$, since these are all really the same angle. The reason this matters is because it can affect the value of $\dfrac{\theta}{n}$ if we replace $\theta$ with $\theta + 2\pi k$ before dividing. Hence all the answers will be of the form

for some integer value of $k$. These values will be different as $k$ ranges from $0$ to $n-1$, but once $k=n$ we can note that the angle $\dfrac{\theta + 2n\pi}{n} = \dfrac{\theta}{n} + 2\pi$ is really the same as $\dfrac{\theta}{n}$, since they differ by one full rotation. Therefore one sees all the answers just by considering values of $k$ ranging from $0$ to $n-1$.