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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 i, squared, equals, minus, 1 and where i lives

We began our study of complex numbers by inventing a number i that satisfies i, squared, equals, minus, 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 minus, 1.
You see, multiplication by i gives a 90, degrees rotation about the origin:
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You can think about this either because i has absolute value 1 and angle 90, degrees, 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, degrees rotation about the origin, which is multiplication by minus, 1. This of course makes sense, because multiplying by i twice is the same as multiplying by i, squared, which should be minus, 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, squared, equals, minus, 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 parenthesis, 1, plus, i, square root of, 3, end square root, right parenthesis, cubed

Take the number z, equals, 1, plus, i, square root of, 3, end square root, which has absolute value square root of, 1, squared, plus, left parenthesis, square root of, 3, end square root, right parenthesis, squared, end square root, equals, 2, and angle 60, degrees. 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 2, cubed, equals, 8. Likewise everything is rotated by 60, degrees three times in a row, so is ultimately rotated by 180, degrees. Hence, at the end it's the same as multiplying by minus, 8, so left parenthesis, 1, plus, i, square root of, 3, end square root, right parenthesis, cubed, equals, minus, 8.
We can also see this using algebra as follows:

Example 2: left parenthesis, 1, plus, i, right parenthesis, start superscript, 8, end superscript

Next, suppose we multiply everything on the plane by left parenthesis, 1, plus, i, right parenthesis eight successive times:
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Since the magnitude of 1, plus, i is
vertical bar, 1, plus, i, vertical bar, equals, square root of, 1, squared, plus, 1, squared, end square root, equals, square root of, 2, end square root,
everything is stretched by a factor of square root of, 2, end square root eight times, and hence is ultimately stretched by a factor of left parenthesis, square root of, 2, end square root, right parenthesis, start superscript, 8, end superscript, equals, 2, start superscript, 4, end superscript, equals, 16.
Since the angle of left parenthesis, 1, plus, i, right parenthesis is 45, degrees, everything is ultimately rotated by 8, dot, 45, degrees, equals, 360, degrees, so in total it's as if we didn't rotate at all. Therefore left parenthesis, 1, plus, i, right parenthesis, start superscript, 8, end superscript, equals, 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: z, start superscript, 5, end superscript, equals, 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, start superscript, 5, end superscript, equals, 1? One simple answer is z, equals, 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 start fraction, 1, divided by, 5, end fraction 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 cosine, left parenthesis, 72, degrees, right parenthesis, plus, i, sine, left parenthesis, 72, degrees, right parenthesis, since start fraction, 360, degrees, divided by, 5, end fraction, equals, 72, degrees.
There are also other solutions, such as rotating start fraction, 2, divided by, 5, end fraction of the way around:
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or start fraction, 1, divided by, 5, end fraction 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:
Solutions to z, start superscript, 5, end superscript, equals, 1

Example 4: z, start superscript, 6, end superscript, equals, minus, 27

Looking at the equation z, start superscript, 6, end superscript, equals, minus, 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, degrees, since the negative indicates 180, degrees rotation.
Something which will stretch by a factor of 27 after 6 applications must have magnitude root, start index, 6, end index, equals, square root of, 3, end square root, and one way to rotate which gives 180, degrees after 6 applications is to rotate by start fraction, 180, degrees, divided by, 6, end fraction, equals, 30, degrees. Therefore one number that solves this equation z, start superscript, 6, end superscript, equals, minus, 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 square root of, 3, end square root:
Solutions to z, start superscript, 6, end superscript, equals, minus, 27
Can you see why?

Solving z, start superscript, n, end superscript, equals, 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, equals, 6 and w, equals, minus, 27, you first find the polar representation of w:
w, equals, r, left parenthesis, cosine, left parenthesis, theta, right parenthesis, plus, i, sine, left parenthesis, theta, right parenthesis, right parenthesis
This means the angle of z must be start fraction, theta, divided by, n, end fraction, and its magnitude must be root, start index, n, end index, 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, equals, root, start index, n, end index, dot, left parenthesis, cosine, left parenthesis, start fraction, theta, divided by, n, end fraction, right parenthesis, plus, i, sine, left parenthesis, start fraction, theta, divided by, n, end fraction, right parenthesis, right parenthesis
To find the other solutions, we keep in mind that the angle theta could have been thought of as theta, plus, 2, pi, or theta, plus, 4, pi, or theta, plus, 2, k, 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 start fraction, theta, divided by, n, end fraction if we replace theta with theta, plus, 2, pi, k before dividing. Hence all the answers will be of the form
z, equals, root, start index, n, end index, dot, left parenthesis, cosine, left parenthesis, start fraction, theta, plus, 2, k, pi, divided by, n, end fraction, right parenthesis, plus, i, sine, left parenthesis, start fraction, theta, plus, 2, k, pi, divided by, n, end fraction, right parenthesis, right parenthesis
for some integer value of k. These values will be different as k ranges from 0 to n, minus, 1, but once k, equals, n we can note that the angle start fraction, theta, plus, 2, n, pi, divided by, n, end fraction, equals, start fraction, theta, divided by, n, end fraction, plus, 2, pi is really the same as start fraction, theta, divided by, n, end fraction, since they differ by one full rotation. Therefore one sees all the answers just by considering values of k ranging from 0 to n, minus, 1.

Want to join the conversation?

  • blobby green style avatar for user ejcabanban
    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?
    (30 votes)
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    • piceratops ultimate style avatar for user T H
      Euler's identity: -1 = e^(πi)
      Let z = re^(θi), where r is the absolute value of z and θ is the argument of z.
      -z = -1 * z = e^(πi) * re^(θi) = re^((θ+π)i)
      Now the argument is θ+π. We also know that π rad = 180°. So multiplying by -1 is the same thing as rotating 180°.
      (8 votes)
  • blobby green style avatar for user ejcabanban
    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?
    (15 votes)
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    • aqualine ultimate style avatar for user David
      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.
      (29 votes)
  • duskpin tree style avatar for user Radek
    In the example 2, can those points make a spiral?
    Can you describe Fibocnacci with complex powers?
    (13 votes)
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    • leaf green style avatar for user asd asd
      I have literally no knowledge about this, but a Google search shows these 2 pages from wiki:
      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.
      (12 votes)
  • blobby green style avatar for user msamorin
    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!
    (10 votes)
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    • primosaur ultimate style avatar for user Jordan Cooper
      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°.
      (4 votes)
  • piceratops seed style avatar for user Michelangelo
    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 :)
    (3 votes)
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    • hopper jumping style avatar for user Nils Petter
      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! :)
      (14 votes)
  • blobby green style avatar for user R C
    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?
    (5 votes)
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  • blobby green style avatar for user Andrea Menozzi
    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 ?

    anyone could explain this please?
    (6 votes)
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  • duskpin ultimate style avatar for user samueltucker22
    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 :)
    (2 votes)
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  • winston baby style avatar for user AJ
    In Example 1, how is (cos(60∘))^3 the same as (cos(60∘+60∘+60∘))? I've seen almost all of Khan Academy's algebra module and I've never seen this concept till now.
    (3 votes)
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    • piceratops ultimate style avatar for user g.wallace.wyatt
      Each time we multiply the plane by z, every point in the plane rotates through 60 degrees; therefore, if we multiply the plane by z three times, everything will rotate through 180 degrees. If we multiply 1 by z, we get z = (modulus, argument) = (2, 60). If we multiply z by z, we get z^2 = (4, 120). And if we multiply z^2 by z, we get z^3 = (8, 180). The trigonometric form of a complex number with magnitude 8 and angle 180 is 8[cos(180) + isin(180)].

      The cos(60+60+60) follows from the fact that when we multiply complex numbers, we add the angles and multiply the magnitudes. This can be verified by using the angle addition formulas for sine and cosine. Try multiplying z = 2[cos(60) + isin(60)] (trigonometric form) by itself twice using the angle addition formulas for sine and cosine each time. You should arrive at the answer z^3 = 8[cos(180) + isin(180)].
      (2 votes)
  • blobby green style avatar for user mohamad
    i just didn't understand how we have figured out that n-1 is the highest possible amount for k such that we don't get redundant results for roots. in"Therefore one sees all the answers just by considering values of k ranging from 0 to n-1"
    any help is really appreciated
    (2 votes)
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