Hi,

The question i have is "Find the sqaure roots of $\displaystyle \sqrt{3} + i$ "

The book gives the answer:

$\displaystyle \sqrt{2}e^{(\frac{1}{2} + k)\pi i}$

$\displaystyle = \pm \sqrt{2}(cos(\frac{\pi}{12}) + isin(\frac{\pi}{12}))$

I can see how they get that but I thought you have to plug in k = 0, 1(n-1) and then use the the new arguements given from that to find them.

I get $\displaystyle arg = \frac{\pi}{6}, \frac{7\pi}{6}$

Which leads my answer to be:

$\displaystyle \pm(\frac{\sqrt{6}}{2} + i\frac{\sqrt{2}}{2})$

Could anyone clear up my grave error?

Many thanks