Problem with equation in complex numbers

Question

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?

Answer 1

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 $$20a^2=4a^4-1$$

Answer 2

$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\\ $