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Math Help - Cube roots of -8

  1. #1
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    Cube roots of -8

    I am totally stuck on this problem. I need to find the cube roots of -8 in polar coordinates. I know that -8 lies on the real axis, so the angle of reference is \pi. I also know that  cos \pi=-1 . I know that the formula I need to use is cube root of |z|= cis \frac{\theta}{3} and I know that cos \frac{\theta}{3}=Sqrt(\frac{1+cos\theta}{3}), sin\theta=Sqrt(\frac{1-cos\theta}{3}) and the cube root =2(cos\frac{\theta}{3}+i sin\frac{\theta}3}. I must be tired and doing something wrong with the signs or something. I keep getting a zero for the first term and I am pretty sure the answer is 1+i \sqrt{3}. Please help! Thanks!
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  2. #2
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    Re: Complex Analysis - Cube roots of -8

    First of all, you'll need a 2, somewhere.

    |z| = cis(x/3) really doesn't make quite enough sense. It is as if you ALMOST know where you are going.

    Start with z^{3} = -8 = 8\cdot cis(\pi)

    Just apply the theorem from DeMoivre.

    z = 8^{\frac{1}{3}}\cdot cis\left(\frac{\pi +2k\pi}{3}\right) for k = 0, 1, 2

    k = 0 leads to 2\cdot cis(\pi/3)

    k = 1 leads to 2\cdot cis(\pi)

    k = 2 leads to 2\cdot cis(5\pi/3)

    One for free. You show us the next one.
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  3. #3
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    Re: Complex Analysis - Cube roots of -8

    I think you're right; I ALMOST know where I am going! OK... I began with 2 cis \pi=0-i\frac {\sqrt{3}}{2} which is where I thought I was off track because of the zero in the real part. However, I should have known that was ok because -8 has nothing in the imaginary part... At any rate, moving on,  2 cis \frac{\pi}{3}=\frac{1}{2} + i \frac{\sqrt{3}}{2} and  2 cis \frac{5\pi}{3} = \frac{1}{2} - i \frac{\sqrt{3}}{2}. But you say one for free and I should show you the next one. Aren't these three roots all of the roots of -8? Thank you so much for your help and your kind attitude.
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  4. #4
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    Re: Complex Analysis - Cube roots of -8

    That was one problem for free, not one cube root of -8.

    2\cdot cis(\pi) = -2 + i\cdot 0

    How are you managing square roots and imaginary values?
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  5. #5
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    Re: Complex Analysis - Cube roots of -8

    I think you need to know that cos(\pi/3)= \frac{1}{2} and sin(\pi/3}= \frac{\sqrt{3}}{2}.
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  6. #6
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    Re: Complex Analysis - Cube roots of -8

    Thank you; I thought that was the end of it! I managed the square roots with imaginary values with a graph of a right triangle and de Moivre's. This one threw me because it was just a line and there was no imaginary value, but you helped me see that it was basically the same as the other; I just had trouble with the angles. Thanks!
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