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 Vi Nguyen Posts: 48, Reputation: 2 Junior Member #1 Jul 2, 2009, 09:11 AM
complex no.s
Can anyone help me with finding the solutions to:

(z^8)-256=0

I can find z= 2, -2, 2i, -2i but don't know how to find the other 4 solutions, apparently I can let z^2=(a+ib)^2 and solve for a and b, but don't know how to do this. Thanks in advance.
 galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Jul 2, 2009, 10:34 AM
There are n different nth roots of $z=r(cos{\theta}+i\cdot sin{\theta})$

Therefore, these are given by:

$\sqrt[n]{r}\left[cos(\frac{\theta}{n}+\frac{2k{\pi}}{n})+i\cdot sin(\frac{\theta}{n}+\frac{2k{\pi}}{n})\right]$

Where $\theta=2{\pi}$ and n=8, so we have $\frac{2\pi}{8}=\frac{\pi}{4}$

$\sqrt[8]{256}\left[cos(\frac{\pi}{4}+\frac{2k{\pi}}{8})+i\cdot sin(\frac{\pi}{4}+\frac{2k{\pi}}{8})\right]$

and let k=1,2,3...

If k=1, we get 2i

If k=2, we get $-\sqrt{2}+\sqrt{2}i$

and so on.
 Vi Nguyen Posts: 48, Reputation: 2 Junior Member #3 Jul 5, 2009, 05:02 AM
Hey thanks I get confused about which form to express it in to work out the solution, I had tried to put it in polar exponential form and didn't think on expanding etc.

Originally Posted by galactus
There are n different nth roots of $z=r(cos{\theta}+i\cdot sin{\theta})$

Therefore, these are given by:

$\sqrt[n]{r}\left[cos(\frac{\theta}{n}+\frac{2k{\pi}}{n})+i\cdot sin(\frac{\theta}{n}+\frac{2k{\pi}}{n})\right]$

Where $\theta=2{\pi}$ and n=8, so we have $\frac{2\pi}{8}=\frac{\pi}{4}$

$\sqrt[8]{256}\left[cos(\frac{\pi}{4}+\frac{2k{\pi}}{8})+i\cdot sin(\frac{\pi}{4}+\frac{2k{\pi}}{8})\right]$

and let k=1,2,3.....

If k=1, we get 2i

If k=2, we get $-\sqrt{2}+\sqrt{2}i$

and so on.

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