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Visualizing complex number powers

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

Connection between i2=1 and where i lives

We began our study of complex numbers by inventing a number i that satisfies i2=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 rotation about the origin:
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You can think about this either because i has absolute value 1 and angle 90, 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?
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It is the same as a 180 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 i2, 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 i2=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

Take the number z=1+i3, which has absolute value 12+(3)2=2, and angle 60. What happens if we multiply everything on the plane by z three times in a row?
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Everything is stretched by a factor of 2 three times, and so is ultimately stretched by a factor of 23=8. Likewise everything is rotated by 60 three times in a row, so is ultimately rotated by 180. Hence, at the end it's the same as multiplying by 8, so (1+i3)3=8.
We can also see this using algebra as follows:
=(2(cos(60)+isin(60)))3=23(cos(60+60+60)+isin(60+60+60) =8(cos(180)+isin(180))=8

Example 2: (1+i)8

Next, suppose we multiply everything on the plane by (1+i) eight successive times:
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Since the magnitude of 1+i is
|1+i|=12+12=2,
everything is stretched by a factor of 2 eight times, and hence is ultimately stretched by a factor of (2)8=24=16.
Since the angle of (1+i) is 45, everything is ultimately rotated by 845=360, 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
=(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

Example 3: z5=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 z5=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 15 of the way around, like this
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then doing this 5 successive times will bring you back to where you started.
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The number which rotates the plane in this way is cos(72)+isin(72), since 3605=72.
There are also other solutions, such as rotating 25 of the way around:
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or 15 of the way around the other way:
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In fact, beautifully, the numbers which solve the equation form a perfect pentagon on the unit circle:

Example 4: z6=27

Looking at the equation z6=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, since the negative indicates 180 rotation.
Something which will stretch by a factor of 27 after 6 applications must have magnitude A276=3, and one way to rotate which gives 180 after 6 applications is to rotate by 1806=30. Therefore one number that solves this equation z6=27 is
3(cos(30)+isin(30))=3(32+i12)=32+i32
However, there are also other answers! In fact, those answers form a perfect hexagon on the circle with radius 3:
Can you see why?

Solving zn=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:
w=r(cos(θ)+isin(θ))
This means the angle of z must be θn, and its magnitude must be Arn, since this way multiplying by z a total of n successive times will in effect rotate by θ and scale by r, just as w does, so
z=Arn(cos(θn)+isin(θn))
To find the other solutions, we keep in mind that the angle θ could have been thought of as θ+2π, or θ+4π, or θ+2kπ for any integer k, since these are all really the same angle. The reason this matters is because it can affect the value of θn if we replace θ with θ+2πk before dividing. Hence all the answers will be of the form
z=Arn(cos(θ+2kπn)+isin(θ+2kπn))
for some integer value of k. These values will be different as k ranges from 0 to n1, but once k=n we can note that the angle θ+2nπn=θn+2π is really the same as θn, since they differ by one full rotation. Therefore one sees all the answers just by considering values of k ranging from 0 to n1.