complex numbers - 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.

Answer

Hint:

$$z^2=3+4i=5e^{it+2k\pi i}\;,\;\;k\in\Bbb Z\;,\;\;t=\arctan\frac43\implies$$

$$z=\sqrt[2]5e^{\frac{it+2k\pi i}2}\;,\;\;k=0,1\;\;\text{(Why it is enough to take only these vales of}\;\;k\;?)$$

A more basic approach: put$\;z=a+bi\;,\;\;a,b\in\Bbb R\;$ , so that

$$3+4i=(a+bi)^2=(a^2-b^2)+2abi\implies\begin{cases}a^2-b^2=3\\{}\\2ab=4\implies b=\frac2a\end{cases}\;\;\implies$$

$$a^2-\frac4{a^2}=3\implies 0=a^4-3a^2-4=(a^2-4)(a^2+1)\implies a=\pm2$$

and thus

$$\;b=\pm\frac22=\pm1\;\implies a+bi=\begin{cases}\;\;\;2+i\\{}\\-2-i\end{cases}$$