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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?

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Apparently, 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. –  David Mitra Dec 10 '12 at 16:37

1 Answer 1

Let $i=R(\cos \theta+i\sin\theta)$ where $R>0$

Equating the real & the imaginary parts, $R\sin\theta=1$ and $R\cos \theta=0\implies \cos \theta=0$

Squaring & adding we get $R^2=1\implies R=1,\sin\theta=1\implies \theta=\frac\pi2$

So, $i=\cos\frac\pi2+i\sin\frac\pi2=e^{i\frac\pi2}$ (using Euler's identity)

So,$i=e^{(2n\pi+\frac\pi2)i}=e^{\frac{(4n+1)\pi i}2}$ where $n$ is any integer.

$$i^{\frac16}=e^{\frac{(4n+1)\pi i}{12}}$$ where $0\le n\le 5$ or more generally $n$ can assume any $6$ in-congruent values $\pmod 6$

Also, if $$x_r=e^{\frac{(4r+1)\pi i}{12}}, x_{r+\frac62}=e^{\frac{\{4(r+3+1)\pi i}{12}}=-x_r$$

So, $-x_3=x_0=e^{\frac{i\pi }{12}},$

$-x_4=x_1=e^{\frac{5i\pi }{12}}$

and $-x_5=x_2=e^{\frac{9i\pi }{12}}=e^{\frac{3i\pi}4}$

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@DavidMitra, sorry for the typo, thanks for you observation. –  lab bhattacharjee Dec 10 '12 at 16:42

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