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### Course: Precalculus>Unit 3

Lesson 8: Multiplying and dividing complex numbers in polar form

# Complex number polar form review

Review the polar form of complex numbers, and use it to multiply, divide, and find powers of complex numbers.

## What is polar form?

$r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)$
The polar form of complex numbers emphasizes their graphical attributes: $\text{absolute value}$ (the distance of the number from the origin in the complex plane) and $\text{angle}$ (the angle that the number forms with the positive Real axis). These are also called $\text{modulus}$ and $\text{argument}$.
Note that if we expand the parentheses in the polar representation, we get the number's rectangular form:
Want to learn more about complex number polar form? Check out this video.
Want to learn more about the different forms of complex numbers? Check out this article.
Want to learn more about converting between rectangular and polar forms? Check out this article.

## Practice set 1: Multiplying and dividing in polar form

Polar form is really useful for multiplying and dividing complex numbers:
$\begin{array}{rl}{z}_{1}& ={r}_{1}\left(\mathrm{cos}{\theta }_{1}+i\mathrm{sin}{\theta }_{1}\right)\\ {z}_{2}& ={r}_{2}\left(\mathrm{cos}{\theta }_{2}+i\mathrm{sin}{\theta }_{2}\right)\\ & ⇓\\ {z}_{1}{z}_{2}& ={r}_{1}{r}_{2}\left[\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}+{\theta }_{2}\right)\right]\\ \\ \frac{{z}_{1}}{{z}_{2}}& =\frac{{r}_{1}}{{r}_{2}}\left[\mathrm{cos}\left({\theta }_{1}-{\theta }_{2}\right)+i\mathrm{sin}\left({\theta }_{1}-{\theta }_{2}\right)\right]\end{array}$
Want to learn more about multiplication and division in polar form? Check out this video.
Problem 1.1
${w}_{1}=5\left[\mathrm{cos}\left({15}^{\circ }\right)+i\mathrm{sin}\left({15}^{\circ }\right)\right]$
${w}_{2}=3\left[\mathrm{cos}\left({45}^{\circ }\right)+i\mathrm{sin}\left({45}^{\circ }\right)\right]$
${w}_{1}\cdot {w}_{2}=$

Your answer should be in polar form. The angle should be given in degrees.

Want to try more problems like this? Check out this exercise.

## Practice set 2: Powers of complex numbers in polar form

$\begin{array}{rl}{z}_{1}& ={r}_{1}\left(\mathrm{cos}{\theta }_{1}+i\mathrm{sin}{\theta }_{1}\right)\\ & ⇓\\ \left({z}_{1}{\right)}^{n}& =\left({r}_{1}{\right)}^{n}\left[\mathrm{cos}\left(n\cdot {\theta }_{1}\right)+i\mathrm{sin}\left(n\cdot {\theta }_{1}\right)\right]\end{array}$

### Example 1

Let's evaluate $\left(1+\sqrt{3}i{\right)}^{6}$. First, we convert to polar form:
$\left(1+\sqrt{3}i\right)=2\left(\mathrm{cos}{60}^{\circ }+i\mathrm{sin}{60}^{\circ }\right)$
Now we use the rule from above:
$\begin{array}{rl}& \phantom{=}\left[2\left(\mathrm{cos}{60}^{\circ }+i\mathrm{sin}{60}^{\circ }\right){\right]}^{6}\\ \\ & =\left(2{\right)}^{6}\left[\mathrm{cos}\left(6\cdot {60}^{\circ }\right)+i\mathrm{sin}\left(6\cdot {60}^{\circ }\right)\right]\\ \\ & =64\left(\mathrm{cos}{360}^{\circ }+i\mathrm{sin}{360}^{\circ }\right)\\ \\ & =64\left(1+i\cdot 0\right)\\ \\ & =64\end{array}$

### Example 2

Let's find the solutions to the equation ${z}^{3}=27$. First, we define $r$ and $\theta$ to be the absolute value and angle of $z$. So ${z}^{3}$ is ${r}^{3}\left[\mathrm{cos}\left(3\cdot \theta \right)+i\mathrm{sin}\left(3\cdot \theta \right)\right]$.
The number $27$ can be written as $27\left[\mathrm{cos}\left(k\cdot {360}^{\circ }\right)+i\mathrm{sin}\left(k\cdot {360}^{\circ }\right)\right]$.
We obtain two equations from the main equation ${z}^{3}=27$:
${r}^{3}=27$
$3\cdot \theta =k\cdot {360}^{\circ }$
The solution of the first equation is $r=3$. The solution of the second equation is $\theta =k\cdot {120}^{\circ }$, which has three distinct solutions: ${0}^{\circ }$, ${120}^{\circ }$, and ${240}^{\circ }$. These correspond to the following three solutions:
$\begin{array}{rl}{z}_{1}& =3\\ \\ {z}_{2}& =-\frac{3}{2}+\frac{3\sqrt{3}}{2}i\\ \\ {z}_{3}& =-\frac{3}{2}-\frac{3\sqrt{3}}{2}i\end{array}$
Problem 2.1
$\left(\sqrt{2}+\sqrt{2}i{\right)}^{6}=$

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• Hello, so I have a question related to this topic. It is on the subject "Powers of complex numbers" The problems that are like this: "(√​2+√​2i)^6" I understand. No problem. The problems like this: "Find the solution of the following equation whose argument is strictly between 225 degrees and 315 degrees: z^7=128i" I do not get.

I look at the hints, and this is the part that has me confused: The number
128i has a modulus of 128. (The argument of 128i can be 90 degrees plus any multiple of 360 degrees.)

The part of the hint in the parentheses has me in a bind. How do I figure out the degree, or degree range, in this problem?

Any help would be greatly appreciated.

Thank you.
(23 votes)
• When it comes to finding the modulus of a number with just 'i' like 128i it comes down to logic and a good understanding of what a modulus means and what an argument means. Modulus is just a fancy way of saying the distance from the origin, so if you were to plot 128i as just a simple dot, it would be 128 units away from the origin. Therefore the modulus is 128. Now for the argument, thats also another fancy way for saying the angle from the x-axis to your point. Well isnt 128i on the positive Im axis? So from the x-axis its just simply 90 degrees away to the positive Im axis. I hope this helped, because I was really confused like you with this.
(36 votes)
• I am beyond confused with the following (seen in the quiz):
z^6 = i

Under hints you say:
"Remember that theta is between 90 and 180 degrees. Therefore, we need to find the multiple of 60 degrees that is strictly within the range of 90 - 15 = 75 and 180 - 15 = 165. This multiple is simply 120, so theta = 135."

1st Question: Why do we need to find the multiple of 60 degrees that is within the range? Why can't it be a different number?
2nd Q: Why are we subtracting 15 degrees from the range at all? Where is 15 degrees coming from?
3rd Q: Given the solution from 2nd Question, why is the multiple "simply" 120? I have no idea where this number came from. Ain't nothing "simple" about it.
4th Q (and most confusing): How in the world do you get that theta equals 165 from all of this?

Please help!
(17 votes)
• 1. Why do we need to find the multiple of 60 degrees that is within the range? Why can't it be a different number?

When you simplify z^6=i to polar form, you get z^6= cos(90)+isin(90). Simplify that using the properties we learned, and you get
z=cos((90+360k)/6)+isin((90+360k)/6)

z=cos(15+60k)+isin(15+60k)

We were given the constraint that theta has to be greater than 90 and less than 180. With the information we have now, that can be written as 90<15+60k<180. We are now thinking about what multiples of 60 will help us satisfy that inequality.

2. Why are we subtracting 15 degrees from the range at all? Where is the 15 degrees coming from?

To answer this question, we go back to the inequality 90<15+60k<180. To simplify this inequality, you subtract 15 from 90, 15+60k and 180. This leaves you with 75<60k<165. The 15 degrees was subtracted to simplify the inequality further.

3. Given the solution from 2nd question, why is the multiple "simply" 120? I have no idea where this number came from.

This question also requires us to go back to the inequality 90<15+60k<180. The inequality is asking you, "What multiple of 60 satisfies this inequality?". If you solve the inequality, you will find that the k=2 which means that 60*2, (120), is the only multiple of 60 that satisfies the inequality.

4. How in the world do you get theta equals 165 from all of this?

I am going to assume that the 165 is a typo for 135. In the previous answer, we found that k=2. Plug that into 15+60k and you get 135. This means that theta equals 135.
(4 votes)
• It seems that I need to enter the degree symbol to earn a green check mark. I am afraid that I don't know how to get that symbol entered. I do not think that the symbol is part of the shortcuts available and all keyboard commands that I have tried have not worked. (Mac/Chrome). Thanks for your thoughts.

Follow up - thank you everyone for your thoughts. It looks like I was using brackets [ ] and that may have been what caused my problem.
(7 votes)
• I've never, ever entered a degree symbol for any type of problem. I wouldn't know how.

When they say the answer must be given in degrees, they mean "as opposed to radians".

I DO need to write the multiplicand information inside parentheses - it doesn't know what to do with brackets. But something like this: 4(cos(20)+isin(20)) has always been enough.
(4 votes)
• So I am a bit confused with a couple of problems on the quiz.

Problem -Find the solution of the following equation whose argument is strictly between 180 and 270, degree.
Round your answer to the nearest thousandth.
z^5=-243iz

In the solution it states that the number -243i, has a modulus of 243. The argument of -243i can be 270 degree plus any multiple of 360.

In a similar problem it stated that -512i can be thought of as 180 degrees + k * 360.

My question is how does one know what to add to the degrees( 180 or 270)? I understand that a negative value can be thought of as that value plus 180, but why suddenly the switch to 270?
(8 votes)
• "Find the solution of the following equation whose argument is strictly between 270(degree) and 360.

'z^3=-512' (z power 3)

Some of the explanation:

"[...]z^3 =−512 (z power 3)
The number -512 has a modulus of 512. The argument of −512 can be 180(deg) plus any multiple of 360(deg), so we can write it as 180(deg)+k*360(deg)[...]"

I wonder how to know to add 180 instead of 0 (zero)? Where is the 180(deg)?

Like example 2 (practice set 2), why you don't add any degree in "z**3 = 27"?

Thanks
(3 votes)
• We add 180 when it's a negative real number (you go 180 deg from the positive real axis to get to a point on the negative real axis).

Zero is for when it's a positive real number, since the positive real numbers are already on the positive real axis and thus make a zero deg "angle" with it.

In the same way, you go 90 deg from the + real axis to get to the + imaginary axis. So all positive pure imaginary numbers (like i, 3i, 64i) have a 90 deg angle.

Finally, you go 270 deg from the + real axis to get to the - imaginary axis. So all negative pure imaginary numbers (like -i, -3i, -64i) have a 270 deg angle.
(11 votes)
• Hello, at the moment I have a problem with the practice: Powers of complex numbers. The equation is:
z^7=128i
After about 30min, I had not been able to solve it and looked at the hints. Until 4/6, I had been able to understand, but the step 4/6 is just completely mind-boggling.
So:
z^7=128i
modulus=2
Solving for theta:
z^7=7r(cos(7theta)+isin(7theta)=128(cos(90+360k)+isin(90+360k)).
Now here's the confusing part for me, they then go on to solve for theta through a strange procedure that I cannot really understand why.
225*7<7theta<315*7=90+360k
225<theta<315=90/7+360k/7
225-90/7<theta-90/7<315-90/7=360k/7
Then it just goes to say, we need to find a multiple of 360/7 and that multiple is 1800/7 so theta=270.
It doesn't explain how it got there, the meaning of the procedures, how it works, just subtracted 90/7 both sides and got the answer 270. I understand the equalities, I just really don't understand the point of that procedure and how it works.
Sorry if this was a bit long, I hope it wasn't a disruption
(7 votes)
• How do you express a complex number in trig form when only given a negative or positive integer, such as, -2?
(4 votes)
• If your number is real, then your angle is either 0 (for a positive number) or π (for a negative number). The magnitude is just the absolute value of the number.

So -2 in polar form is 2(cos(π)+isin(π))
(6 votes)
• I have a question regarding complex number polar form. I have a problem already in polar form that stats off like find the four fourth roots of the complex number 16(cos(240°)+isin(240°)) I understand that there are supposed to be four parts of the problem or four answers I just don't understand how to solve it.
(4 votes)
• The questions that are on Khan give you a domain to solve the four roots, for example:
"Find the answer that lies in between 90 and 180 degrees"
or something like that.

If you want to find all four roots, you ignore the constraint, that is, ignore the "90 and 180 degrees" part. Instead, you want to find answers for all possible k values.

Consider a problem where you eventually get something like: 15 + 90k. Then you will set k equal to 1, 2, 3, and 4 to find the four roots.
(4 votes)
• In example 2 above, (z^3=27) I understand everything except " The solution of the second equation is theta = k*360 degrees has three distinct solutions: 0 degree, 120 degree and 240 degrees. I don't understand how those three possibilities are determined.
Any help would be greatly appreciated.
Thanks in advance.
(4 votes)
• When we cube a complex number, we triple its angle. 27 is real, so its angle is 0, or 360º, or 720º, or 1080º, etc.

So the solutions to this equation will have angles that are 1/3 of these numbers, which are 0, 120º, 240º, 360º, etc. 0 and 360º are coterminal, so this list has begun to generate repeat solutions, so the only genuinely different angles are 0, 120º, and 240º.
(3 votes)
• I have a little question. Is modulus of 75 just 75 then?
(2 votes)
• Yes. The modulus of any positive real number is just the number itself.
(5 votes)
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