Problem with equation in complex numbers 
I am supposed to calculate:
$x^{2}=5+i$

I used formula:
$\left | \cos \frac{\alpha }{2} \right |=\sqrt{\frac{1+\cos \alpha }{2}}$ 
$\left | \sin \frac{\alpha }{2} \right |=\sqrt{\frac{1-\cos  \alpha }{2}}$
and I came to the point where:
x= $\sqrt[4]{26}\left ( \frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}\sqrt[4]{26}}  + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}\sqrt[4]{26}} \right)$= $\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}}  + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
I know that k=0,1 and also that  $\sin x, \cos x $ are positive in the first quadrant, so my solution will simple be :
$x_{0}=\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}}  + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
$x_{1}=-\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}}  + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
It does not seem like the correct solution. Can anyone please tell me, where I made the mistake?

$5 + i = \rho (\cos \theta + i\sin theta)\\
\sqrt{\rho (\cos \theta + i\sin \theta)} = \sqrt {\rho}(\cos \frac \theta2 + i\sin \frac \theta2)\\
\rho = \sqrt {5^2 + 1} = \sqrt {26}\\
\theta = \arctan \frac {1}{5}\\
\cos \theta = \frac {5}{\sqrt {26}}\\
\sin\theta = \frac {1}{\sqrt {26}}\\
\tan \frac{\theta}{2} = \frac {1-\cos\theta}{\sin\theta}= \frac {1 - \frac {5}{\sqrt{26}}}{\frac {1}{\sqrt {26}}} = \sqrt {26} - 5\\
$
