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# Roots of complex numbers

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1. Using De moivre's theorem;

So the only solution I have is , but there are other solutions, but for any integer value of k other than 0, it will fall outside of the -pi to pi range. I think I used De moivre's wrong, but it is true for all rational numbers.
2. (Original post by NotNotBatman)

Using De moivre's theorem;

So the only solution I have is , but there are other solutions, but for any integer value of k other than 0, it will fall outside of the -pi to pi range. I think I used De moivre's wrong, but it is true for all rational numbers.
so subtract multiples of 2pi away from your angle till it's in the range you want.
3. (Original post by Zacken)
so subtract multiples of 2pi away from your angle till it's in the range you want.
But that would only give solutions outside of the range.
4. (Original post by NotNotBatman)
But that would only give solutions outside of the range.
What's the range you're considering? 0 to 2pi or -pi to pi?
5. (Original post by Ayman!)
What's the range you're considering? 0 to 2pi or -pi to pi?
-pi to pi
6. (Original post by NotNotBatman)
But that would only give solutions outside of the range.
say you take k = 1, that gets you the angle 2pi/9 + 8pi/3 = 26pi/9. Subtract 2pi, that gets you 8pi/9, which is most definitely in your range.
7. (Original post by Zacken)
say you take k = 1, that gets you the angle 2pi/9 + 8pi/3 = 26pi/9. Subtract 2pi, that gets you 8pi/9, which is most definitely in your range.
Got it, thank you.

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