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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?

start color #e07d10, r, end color #e07d10, left parenthesis, cosine, start color #aa87ff, theta, end color #aa87ff, plus, i, sine, start color #aa87ff, theta, end color #aa87ff, right parenthesis
The polar form of complex numbers emphasizes their graphical attributes: start color #e07d10, start text, a, b, s, o, l, u, t, e, space, v, a, l, u, e, end text, end color #e07d10 (the distance of the number from the origin in the complex plane) and start color #aa87ff, start text, a, n, g, l, e, end text, end color #aa87ff (the angle that the number forms with the positive Real axis). These are also called start color #e07d10, start text, m, o, d, u, l, u, s, end text, end color #e07d10 and start color #aa87ff, start text, a, r, g, u, m, e, n, t, end text, end color #aa87ff.
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:
z1=r1(cosθ1+isinθ1)z2=r2(cosθ2+isinθ2)z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]z1z2=r1r2[cos(θ1θ2)+isin(θ1θ2)]\begin{aligned} z_1&=\goldD{r_1}(\cos\purpleC{\theta_1}+i\sin\purpleC{\theta_1}) \\ z_2&=\goldD{r_2}(\cos\purpleC{\theta_2}+i\sin\purpleC{\theta_2}) \\ &\Downarrow \\ z_1z_2&=\goldD{r_1r_2}[\cos(\purpleC{\theta_1+\theta_2})+i\sin(\purpleC{\theta_1+\theta_2})] \\\\ \dfrac{z_1}{z_2}&=\dfrac{\goldD{r_1}}{\goldD{r_2}}[\cos(\purpleC{\theta_1-\theta_2})+i\sin(\purpleC{\theta_1-\theta_2})] \end{aligned}
Want to learn more about multiplication and division in polar form? Check out this video.
Problem 1.1
w, start subscript, 1, end subscript, equals, 5, open bracket, cosine, left parenthesis, 15, degrees, right parenthesis, plus, i, sine, left parenthesis, 15, degrees, right parenthesis, close bracket
w, start subscript, 2, end subscript, equals, 3, open bracket, cosine, left parenthesis, 45, degrees, right parenthesis, plus, i, sine, left parenthesis, 45, degrees, right parenthesis, close bracket
w, start subscript, 1, end subscript, dot, w, start subscript, 2, end subscript, equals

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

z1=r1(cosθ1+isinθ1)(z1)n=(r1)n[cos(nθ1)+isin(nθ1)]\begin{aligned} z_1&=\goldD{r_1}(\cos\purpleC{\theta_1}+i\sin\purpleC{\theta_1}) \\ &\Downarrow \\ (z_1)^n&=(\goldD{r_1})^n[\cos(n\cdot\purpleC{\theta_1})+i\sin(n\cdot\purpleC{\theta_1})] \end{aligned}

Example 1

Let's evaluate left parenthesis, 1, plus, square root of, 3, end square root, i, right parenthesis, start superscript, 6, end superscript. First, we convert to polar form:
left parenthesis, 1, plus, square root of, 3, end square root, i, right parenthesis, equals, start color #e07d10, 2, end color #e07d10, left parenthesis, cosine, start color #aa87ff, 60, degrees, end color #aa87ff, plus, i, sine, start color #aa87ff, 60, degrees, end color #aa87ff, right parenthesis
Now we use the rule from above:
=[2(cos60+isin60)]6=(2)6[cos(660)+isin(660)]=64(cos360+isin360)=64(1+i0)=64\begin{aligned} &\phantom{=}[\goldD{2}(\cos\purpleC{60^\circ}+i\sin\purpleC{60^\circ})]^6 \\\\ &=(\goldD 2)^6[\cos(6\cdot\purpleC{60^\circ})+i\sin(6\cdot\purpleC{60^\circ})] \\\\ &=64(\cos360^\circ+i\sin360^\circ) \\\\ &=64(1+i\cdot 0) \\\\ &=64 \end{aligned}

Example 2

Let's find the solutions to the equation z, cubed, equals, 27. First, we define r and theta to be the absolute value and angle of z. So z, start superscript, start color #11accd, 3, end color #11accd, end superscript is r, start superscript, start color #11accd, 3, end color #11accd, end superscript, open bracket, cosine, left parenthesis, start color #11accd, 3, end color #11accd, dot, theta, right parenthesis, plus, i, sine, left parenthesis, start color #11accd, 3, end color #11accd, dot, theta, right parenthesis, close bracket.
The number 27 can be written as 27, open bracket, cosine, left parenthesis, k, dot, 360, degrees, right parenthesis, plus, i, sine, left parenthesis, k, dot, 360, degrees, right parenthesis, close bracket.
We obtain two equations from the main equation z, cubed, equals, 27:
r, start superscript, start color #11accd, 3, end color #11accd, end superscript, equals, 27
start color #11accd, 3, end color #11accd, dot, theta, equals, k, dot, 360, degrees
The solution of the first equation is r, equals, 3. The solution of the second equation is theta, equals, k, dot, 120, degrees, which has three distinct solutions: 0, degrees, 120, degrees, and 240, degrees. These correspond to the following three solutions:
z1=3z2=32+332iz3=32332i\begin{aligned} z_1&=3 \\\\ z_2&=-\dfrac{3}{2}+\dfrac{3\sqrt 3}{2}i \\\\ z_3&=-\dfrac{3}{2}-\dfrac{3\sqrt 3}{2}i \end{aligned}
Problem 2.1
left parenthesis, square root of, 2, end square root, plus, square root of, 2, end square root, i, right parenthesis, start superscript, 6, end superscript, equals

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

Want to join the conversation?

  • blobby green style avatar for user Sean McGinnis
    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)
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    • male robot hal style avatar for user John Kameas
      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.
      (34 votes)
  • piceratops ultimate style avatar for user jcvack
    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)
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    • piceratops ultimate style avatar for user ANB
      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.
      (3 votes)
  • spunky sam blue style avatar for user chris.meono
    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)
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    • starky ultimate style avatar for user Jesse
      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)
  • blobby green style avatar for user lvltcode
    "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)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Aria
      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)
  • hopper jumping style avatar for user Brandon
    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?
    (6 votes)
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  • blobby green style avatar for user Jacqueline Cyr
    How do you express a complex number in trig form when only given a negative or positive integer, such as, -2?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • boggle blue style avatar for user MostDangerousMagpie
    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
    (5 votes)
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  • mr pants pink style avatar for user Emma
    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.
    (3 votes)
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    • duskpin sapling style avatar for user elainechen311
      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)
  • blobby green style avatar for user alemberikman
    Can anyone please explain how to get the equation for the argument? For example for r^2 = 27, the argument equation was 90 + 270k. Where did the 90 come from and where did the 270 come from?
    Or for z^2 = -81i, the argument's equation was 90 + 180k. For z^6 = i, the argument equation was 90 + 360k.
    I have no idea where these numbers are coming from. Can someone please explain?
    (3 votes)
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  • leaf blue style avatar for user Charles
    How does [cos(x)+i*sin(x)]^6 = [cos(6*x)+i*sin(6*x)]?
    (2 votes)
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