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?

You can also write $$5+i=a^2-b^2+2abi$$ so we will get $$5=a^2-b^2$$ and $$1=2ab$$
Plugging $$b=\frac{1}{2a}$$ in the first equation you will get
$$20a^2=4a^4-1$$
