View Full Version : Complex no.s
Vi Nguyen
Jul 2, 2009, 09:11 AM
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
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
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.
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.