find $z$ that satisfies $z^2=3+4i$ Super basic question but some reason either I'm not doing this right or something is wrong.
The best route usually with these questions is to transform $3+4i$ to $re^{it}$ representation.
Ok, so $r^2=3^2+4^2 = 25$, so $r=5$. And $\frac{4}{3}=\tan(t)$ so that means $t \approx 0.3217$ and I'm not going to get an exact answer like that.
Another method would be to solve quadratic formula $z^2-3-4i=0$ that means $z_0=\frac{\sqrt{12+16i}}{2}$ and $z_1=\frac{-\sqrt{12+16i}}{2}$
But now I have the same problem, $12+16i$ doesn't have a "pretty" polar representation so its difficult to find $\sqrt{12+16i}$
I want to find an exact solution, not approximate, and it should be easy since the answers are $2+i$ and $-2-i$
Edit:
Also, something else is weird here. I know that if $z_0$ is some root of a polynomial then it's conjugate is also a root.but $2+i$ and $-2-i$ are not conjugates.

We can use the double angle formula to find the components of $z$:
$$ \cos \theta = \pm \sqrt{\frac{1 + \cos 2\theta}{2}}$$
$$ \sin \theta = \pm \sqrt{\frac{1 - \cos 2\theta}{2}}$$
Here $\theta$ is the argument of $z$ and $2\theta$ the argument of $z^2$. 
We know $ \cos 2\theta = \frac{3}{5}$ so $\cos \theta = \pm \sqrt{\frac{4}{5}}$ and $\sin \theta = \pm\sqrt{\frac{1}{5}}$. Since $z^2$ lies on the first quadrant (between $0$ and $\pi/2$), its two square roots must lie on the first and third quadrants, respectively, therefore the final answer is
$$ z_1 = \sqrt{5} \left( \sqrt{\frac{4}{5}} + i\sqrt{\frac{1}{5}} \right) = 2 + i $$
$$ z_2 = -z_1 = -2 - i $$
Alternatively: Since the numbers are relatively simple, we can let $z = x + yi$. Then
$$ z^2 = (x^2 - y^2) + 2xyi = 3 + 4i $$
Now equate the real and imaginary part to get this system of equations:
$$ x^2 - y^2 = 3 $$
$$ 2xy = 4 $$
