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### Course: Algebra (all content) > Unit 16

Lesson 10: Multiplying & dividing complex numbers in polar form# 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 $0$ . 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 $-1$ .

*above*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

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 $\frac{1}{5}$ 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 $\mathrm{cos}({72}^{\circ})+i\mathrm{sin}({72}^{\circ})$ , since $\frac{{360}^{\circ}}{5}}={72}^{\circ$ .

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

or $\frac{1}{5}$ 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 ${180}^{\circ}$ rotation.

**stretch by a factor of**$27$ , and**rotate by**${180}^{\circ}$ , since the negative indicatesSomething which will stretch by a factor of $27$ after $6$ applications must have magnitude $\sqrt[6]{\phantom{A}27}=\sqrt{3}$ , and one way to rotate which gives ${180}^{\circ}$ after $6$ applications is to rotate by $\frac{{180}^{\circ}}{6}}={30}^{\circ$ . 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 $\frac{\theta}{n}$ , and its magnitude must be $\sqrt[n]{\phantom{A}r}$ , 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 $\frac{\theta}{n}$ 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 $\frac{\theta +2n\pi}{n}}={\displaystyle \frac{\theta}{n}}+2\pi $ is really the same as $\frac{\theta}{n}$ , since they differ by one full rotation. Therefore one sees all the answers just by considering values of $k$ ranging from $0$ to $n-1$ .

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