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## Algebra (all content)

### Course: Algebra (all content)>Unit 16

Lesson 10: Multiplying & dividing complex numbers in polar form

# Complex number equations: x³=1

Sal finds all complex solutions to the equation x^3=1. Created by Sal Khan.

## Want to join the conversation?

• How does this work for z raised to a non-integer exponent?

For example: z^2.5 = 1

What could z be? • In general, for a^b = c, the complex and real solutions are exp() of any multiple of (2pi * 1/b)i. In this case, the answers are e^(2pi/2.5)i = e^(4pi/5)i, e^(8pi/5)i, e^(12pi/5), e^(16pi/5)i. The next one just simplifies into the first, so we are done. If you don't believe me, you can try making the 2.5th power of each of these and verify by yourself: They reduce to 1.
• So if the equation would be x^6=1, would there be six roots? •  Yes, there will be 6 roots, 2 reals and 4 imaginary. 1, (1/2) + (sqrt(3)/2)i, (-1/2) + (sqrt(3)/2)i, -1, (-1/2) - (sqrt(3)/2)i, (1/2) - (sqrt(3)/2)i. Just divide the degrees of a circle, 2pi, into 6 angles and plug into the form x = r (cos (theta) + i sin (theta)). (theta) = each angle you found.
Hope this helps!
• So I get the concept, and understand the math in the video. But what does it mean? What does taking the roots and drawing it on the argand diagram really prove? • If 0 = 2pi = 4pi (all in radians), then e^0i = e^2pi i = e^4pi i, then when you take the cube root of both sides, shouldn't they all equal each other? However, in the video, the cube roots are clearly different. Why is that so? •  As you point out, if you put e^(2*pi*i) in an advanced calculator it is = to 1, just like any e^((2n)*pi*i) where n is an integer. However, 0 != 2pi != 4pi. They are NOT equal. The reference angle measure at 0, 2pi, and 4pi is equal, not the measures themselves. So the difference is between the returned value of a reference angle and the angle itself.

Another way to think about it is this: Start facing North. Spin once, you are facing North. Spin a second time, you are facing North. All three times the result is the same, you are facing North. However, the number of times you have spun are 0, 1, and 2 and NOT equal. This allows a calculation that involves both the facing and the spin to use elements that are both equal and not.
• I know this is a really stupid question and against the point of this video, but why couldn't you just take the cube root of both sides at ""? • If you take the square root of both sides, you get x=1. But x=-1 is also valid. Because you're taking the principal square root to get x=1. Same in this case, you would be taking the principal cube root if you would be x=1. but if you think about the non-principal cube roots, either you use the method of this video or you use factorisation.
• Is deMoivre's Theorem needed to find roots if we use the method describe in the video to find roots? • z^5 =-243i
Solving for theta
1) 5⋅theta=270+k⋅360
theta=54+k⋅72
Remember that theta is strictly between 180-270
Therefore, we need to find the multiple of 90
​​ 90, degree that is strictly within the range of 180-54=126
and 270-54=216
multiple is simply 144
theta = 198
​​
My questions are as follows:
1) How do you know 5⋅theta=270+k⋅360?
Where does the 270 come from I wanted to say either 180 or 90.

2) How do you get 144 from 126, 216 and even after theta=198
is 144 found from 360-216. and you pick 216 as the right value because it alone is between 180 and 270?
• Why is the magnitude of z notated as |z|? Isn't that absolute value notation? • At , How can Khan say its √3/2 ? Plz HELP ME!   