Sqaure roots of complex number

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

Turns out like this for me

:$\displaystyle sqrt(3) +i$

$\displaystyle abs:z = 2$

$\displaystyle argz = atan:(1/sqrt3) = 30deg$

$\displaystyle => 2(cos(pi/6)+isin(pi/6)$

Set that to be $\displaystyle z^2$

$\displaystyle z = sqrt(2)(cos(pi/12) +isin(pi/12))$ + the other root that gave the +/- sign to the equation...