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

Connection between i2=1i^2 = -1 and where ii lives

We began our study of complex numbers by inventing a number ii that satisfies i2=1i^2 = -1, and later visualized it by placing it outside the number line, one unit above 00. 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 1-1.
You see, multiplication by ii gives a 9090^\circ rotation about the origin:
You can think about this either because ii has absolute value 11 and angle 9090^\circ, or because this rotation is the only way to move the grid around (fixing 00) which places 11 on the spot where ii started off.
So what happens if we multiply everything in the plane by ii twice?
It is the same as a 180180^\circ rotation about the origin, which is multiplication by 1-1. This of course makes sense, because multiplying by ii twice is the same as multiplying by i2i^2, which should be 1-1.
It is interesting to think about how if we had tried to place ii somewhere else while still maintaining its characteristic quality that i2=1i^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+i3)3(1 + i\sqrt{3})^3

Take the number z=1+i3z = 1 + i\sqrt{3}, which has absolute value 12+(3)2=2\sqrt{1^2 + (\sqrt{3})^2} = 2, and angle 6060^\circ. What happens if we multiply everything on the plane by zz three times in a row?
Everything is stretched by a factor of 22 three times, and so is ultimately stretched by a factor of 23=82^3 = 8. Likewise everything is rotated by 6060^\circ three times in a row, so is ultimately rotated by 180180^\circ. Hence, at the end it's the same as multiplying by 8-8, so (1+i3)3=8(1 + i\sqrt{3})^3 = -8.
We can also see this using algebra as follows:

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

Next, suppose we multiply everything on the plane by (1+i)(1 + i) eight successive times:
Since the magnitude of 1+i1 + i is
1+i=12+12=2|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2},
everything is stretched by a factor of 2\sqrt{2} eight times, and hence is ultimately stretched by a factor of (2)8=24=16(\sqrt{2})^8 = 2^4 = 16.
Since the angle of (1+i)(1 + i) is 4545^\circ, everything is ultimately rotated by 845=3608 \cdot 45^\circ = 360^\circ, so in total it's as if we didn't rotate at all. Therefore (1+i)8=16(1 + i)^8 = 16.
Alternatively, the way to see this with algebra is
=(1+i)8=(2(cos(45)+isin(45))8=(2)8(cos(45++458 times)+isin(45++458 times))=16(cos(360)+isin(360))=16\begin{aligned} &\phantom{=}(1 + i)^8 \\\\ &= \left(\sqrt{2}\cdot(\cos(45^\circ) + i \sin(45^\circ) \right)^8 \\ &= (\sqrt{2})^8 \cdot \left( \cos(\underbrace{45^\circ + \cdots + 45^\circ}_{\text{$8$ times}}) + i\sin(\underbrace{45^\circ + \cdots + 45^\circ}_{\text{$8$ times}}) \right) \\\\ &= 16 \left(\cos(360^\circ) + i\sin(360^\circ) \right) \\\\ &= 16 \end{aligned}

Example 3: z5=1z^5 = 1

Now let's start asking the reverse question: Is there a number zz such that after multiplying everything in the plane by zz five successive times, things are back to where they started? In other words, can we solve the equation z5=1z^5 = 1? One simple answer is z=1z = 1, but let's see if we can find any others.
First off, the magnitude of such a number would have to be 11, since if it were more than 11, the plane would keep stretching, and if it were less than 11, 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 15\dfrac{1}{5} of the way around, like this
then doing this 55 successive times will bring you back to where you started.
The number which rotates the plane in this way is cos(72)+isin(72)\cos(72^\circ) + i\sin(72^\circ), since 3605=72\dfrac{360^\circ}{5} = 72^\circ.
There are also other solutions, such as rotating 25\dfrac{2}{5} of the way around:
or 15\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:
Solutions to z5=1z^5 = 1

Example 4: z6=27z^6 = -27

Looking at the equation z6=27z^6 = -27, it is asking us to find a complex number zz such that multiplying by this number 66 successive times will stretch by a factor of 2727, and rotate by 180180^\circ, since the negative indicates 180180^\circ rotation.
Something which will stretch by a factor of 2727 after 66 applications must have magnitude 276=3\sqrt[6]{27} = \sqrt{3}, and one way to rotate which gives 180180^\circ after 66 applications is to rotate by 1806=30\dfrac{180^\circ}{6} = 30^\circ. Therefore one number that solves this equation z6=27z^6 = -27 is
3(cos(30)+isin(30))=3(32+i12)=32+i32\begin{aligned} \sqrt{3}(\cos(30^\circ) + i\sin(30^\circ)) &= \sqrt{3}\left(\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \\ &= \frac{3}{2} + i \frac{\sqrt{3}}{2} \end{aligned}
However, there are also other answers! In fact, those answers form a perfect hexagon on the circle with radius 3\sqrt{3}:
Solutions to z6=27z^6 = -27
Can you see why?

Solving zn=wz^n= w in general

Let's generalize the last two examples. If you are given values ww and nn, and asked to solve for zz, as in the last example where n=6n=6 and w=27w = -27, you first find the polar representation of ww:
w=r(cos(θ)+isin(θ))w = r(\cos(\theta) + i\sin(\theta))
This means the angle of zz must be θn\dfrac{\theta}{n}, and its magnitude must be rn\sqrt[n]{r}, since this way multiplying by zz a total of nn successive times will in effect rotate by θ\theta and scale by rr, just as ww does, so
z=rn(cos(θn)+isin(θn)) z = \sqrt[n]{r} \cdot \left( \cos\left(\dfrac{\theta}{n}\right) + i\sin\left(\dfrac{\theta}{n}\right) \right)
To find the other solutions, we keep in mind that the angle θ\theta could have been thought of as θ+2π\theta + 2\pi, or θ+4π\theta + 4\pi, or θ+2kπ\theta + 2k\pi for any integer kk, since these are all really the same angle. The reason this matters is because it can affect the value of θn\dfrac{\theta}{n} if we replace θ\theta with θ+2πk\theta + 2\pi k before dividing. Hence all the answers will be of the form
z=rn(cos(θ+2kπn)+isin(θ+2kπn)) z = \sqrt[n]{r} \cdot \left( \cos\left(\dfrac{\theta + 2k\pi}{n}\right) + i\sin\left(\dfrac{\theta + 2k\pi}{n}\right) \right)
for some integer value of kk. These values will be different as kk ranges from 00 to n1n-1, but once k=nk=n we can note that the angle θ+2nπn=θn+2π\dfrac{\theta + 2n\pi}{n} = \dfrac{\theta}{n} + 2\pi is really the same as θn\dfrac{\theta}{n}, since they differ by one full rotation. Therefore one sees all the answers just by considering values of kk ranging from 00 to n1n-1.