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### Course: Precalculus>Unit 3

Lesson 8: Multiplying and 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 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: $\left(1+i\sqrt{3}{\right)}^{3}$‍

Take the number $z=1+i\sqrt{3}$, which has absolute value $\sqrt{{1}^{2}+\left(\sqrt{3}{\right)}^{2}}=2$, 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 $\left(1+i\sqrt{3}{\right)}^{3}=-8$.
We can also see this using algebra as follows:

### Example 2: $\left(1+i{\right)}^{8}$‍

Next, suppose we multiply everything on the plane by $\left(1+i\right)$ 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 $\left(\sqrt{2}{\right)}^{8}={2}^{4}=16$.
Since the angle of $\left(1+i\right)$ 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 $\left(1+i{\right)}^{8}=16$.
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 $\mathrm{cos}\left({72}^{\circ }\right)+i\mathrm{sin}\left({72}^{\circ }\right)$, 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 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 $\sqrt[6]{\phantom{A}27}=\sqrt{3}$, 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
$\begin{array}{rl}\sqrt{3}\left(\mathrm{cos}\left({30}^{\circ }\right)+i\mathrm{sin}\left({30}^{\circ }\right)\right)& =\sqrt{3}\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\\ & =\frac{3}{2}+i\frac{\sqrt{3}}{2}\end{array}$
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 $w$:
$w=r\left(\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)\right)$
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
$z=\sqrt[n]{\phantom{A}r}\cdot \left(\mathrm{cos}\left(\frac{\theta }{n}\right)+i\mathrm{sin}\left(\frac{\theta }{n}\right)\right)$
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
$z=\sqrt[n]{\phantom{A}r}\cdot \left(\mathrm{cos}\left(\frac{\theta +2k\pi }{n}\right)+i\mathrm{sin}\left(\frac{\theta +2k\pi }{n}\right)\right)$
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 $\frac{\theta +2n\pi }{n}=\frac{\theta }{n}+2\pi$ 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 $n-1$.

## Want to join the conversation?

• In example 4, z^6 = -27 , it says that it is suppose to rotate 180 degrees because of the negative sign? Can someone explain this further?
• Yes! Because -27 is rotated by 180. All negative numbers are. If you are at 1 to get to -1, you need to rotate 180 degrees, Same for 27.
• In the powers of complex numbers question section their is a question asking to : Find the solution of the equation whose argument is strictly between 180 and 270
1) z^5 = -243i

Now in this math slide it says if their is a negative in the equation then the argument will be 180 + k(360) ... however in the question when I ask for a hint it says the argument is 270 + k(360) . Can some explain how they got 270?
• So we have z^5 = -243.
This is the same as z^5 = 243 (0 - 1*i).
Let x be the argument.
When is cos(x) = 0 and sin(x) = -1?
(With help from the unit circle) this is when x = 270 + k * 360 degrees.
Therefore, the argument is 270 + k * 360 degrees and we can write:
z^5 = -243i = 243 (cos(270 + k * 360) + i * sin(270 + k * 360))
and continue to solve the problem.
• In the example 2, can those points make a spiral?
Can you describe Fibocnacci with complex powers?
https://en.wikipedia.org/wiki/Golden_spiral
https://en.wikipedia.org/wiki/Logarithmic_spiral

So what I got from it is that in the Example 2, you could make a curve that connects the dots and that curve would be called a logarithmic spiral.

You can also construct a "Golden" logarithmic spiral using the Golden ratio as your "growth constant" (which is just the ratio between the angle and the magnitude in the example).

However, "Golden" spiral is only an approximation of a Fibonacci spiral (which is an illustration of Fibonacci series). This is because the ratio of adjacent terms in the Fibonacci series is NOT equal to the Golden Ratio, instead it approaches it as your terms approach infinity.
• I'm having trouble understanding a specific piece of solving for the powers of complex numbers when it asks to solve for the argument in a specific range.

Example: Find the solution of the following equation whose argument is strictly between 225​∘ and 315∘. Round your answer to the nearest thousandth.
z^7=128i

Under the hints, it says the argument is 90 + (360)K, K being any integer. How do I determine what to add to (360)K? Any further explanation is greatly appreciated!
• It sounds like you're asking how they got the 90 in that problem. If that's not your question, please let me know.

The 90 is the angle (in degrees) of 128𝑖, because 128𝑖 is directly above the origin on the complex plane, or 90° clockwise from the positive real axis.

If the equation to solve were 𝓏⁷ = -128, you would use the angle of ‒128, which is 180°. If the equation were 𝓏⁷ = 128 ‒ 128𝑖, you would use its angle, 315°.
• I don't believe that this was covered in the videos or articles, but if you are confused (as I was) about why the angles can just be added when exponentiating (cos(x) + isin(x)), try using FOIL on it by squaring it.

Using double-angle identities yields the result: (cos(2x) + isin(2x)).
• i need an explanation to solve this: z^10 = i

i understand every step, but i don't understand, solving for theta :(
ex:
10 ⋅ θ = 90 + k ⋅ 360​
θ = 9 + k ⋅ 36
Remember that θ is strictly between 120, degree and 180, degree.
so 120 - 9 = 111 and 180 - 9 = 171. This multiply is 144 so θ = 153.

someone can help me understand these steps?
why it is 144??
​​
thanks
Byeee :)
• To say that θ is between 120 and 180 degrees, is the same as saying:

120 < θ < 180, but you know that θ = 9 + k ⋅ 36, so we could say:

120 < 9 + k ⋅ 36 < 180, we can solve this as we usually solve equations (to find k):

120 - 9 < k ⋅ 36 < 180 - 9
111 < k ⋅ 36 < 171
111 / 36 < k < 171 / 36
3.083 ... < k < 4.75, since k is an integer we must have k = 4

Then we have: θ = 9 + k ⋅ 36 = 9 + 4 ⋅ 36 = 9 + 144 = 153.

Hope it helps! :)
• Can someone explain why if we are cubing Cos (60) we get 180 degrees? In the problem the radius of 2 is cubed (2^3). If you have cos (60)^3, you get 1800 degrees?
• in this exercise the hint explains how to find the angle, but i don't understand why it subtract 90/7 from 225. I am lost.
here is the exercise. and a piece of the Hint that confuses me.

Find the solution of the following equation whose argument is strictly between 225 and 315.
z^7 = 128i

HINT
Remember that θ is between 225 and 315 degree.
Therefore, we need to find the multiple of 360/7 that is strictly within the range of
225-90/7 = 1485/7 and 315-90/7=2115/7.
This multiple is simply 1800/7, so θ = 270

why 225-90/7 ? and how do you know that the multiple is 1800/7 ?

• I don't see this exercise on this page. Where does it come from?
• Is it just me or is that the Fibonacci spiral for (1+i)^8?
• That is something that is known as the golden spiral, which is (to quote Wikipedia) "a logarithmic spiral whose growth ratio is the golden ratio", and the golden ratio does in fact have a relationship to the fibonacci sequence.
• In example 2, it says:

16(cos(360)+i sin(360))
=16

Even though cos(360) = 1,
and i*sin(360) = 1i

Therefore shouldn't 16(cos(360)+i sin(360))
=16(1+1i)
=16+16i?

Sorry if I'm asking a dumb question :)