I have to find the roots of $(i)^{1/6}$ ...so I find $k= 0, 1, 2, 3, 4, 5$... the angle is zero degrees apparently...so the first root is $i^{1/6}\times [\cos (0+2\times 0\times \pi)/6 + i\times \sin(0+2\times 0\times \pi)/6]$ Now what?
Math question complex numbers?
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complex-numbers
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1Apparently, you want to find the sixth-roots of $i$. There are six of them. The "angle" is then $\pi/2$, not $0$; and $r=1$, not $i$. What you need to do is plug the values $k=0,1,2,3,4,5$ into the formula $1\cdot\text{cis}\,\Bigl({{2k\pi+{\pi\over 2}}\over 6}\Bigr) $ where $\text{cis}\, \theta=\cos\theta+i\sin\theta$ to obtain the six sixth-roots. – 2012-12-10