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.
 A: 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 
$$\;b=\pm\frac22=\pm1\;\implies a+bi=\begin{cases}\;\;\;2+i\\{}\\-2-i\end{cases}$$
A: This can be computed by a Simple Square Root Denesting Rule:
$\ 3+4\sqrt{-1}\ $ has norm $= 25.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = -5\,\ $ yields $\,\  8+4\sqrt{-1}\:$
which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{16}\, =\, 4.\ \  \rm \color{brown}{Dividing\ it\ out}\ $ of the above  yields $\ \ 2+\sqrt{-1}$ 
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
A: $$
z^2=3+4i\\
(a+bi)^2=3+4i,\;\;\;\;a,b\in\mathbb{R}\\
a^2+2abi-b^2=3+4i
$$
Now you have a system of equations:
$$
a^2-b^2=3\tag{1}
$$
$$
2ab=4\tag{2}
$$
$$
b=\dfrac2a
$$
Substitute $b$ in $(1)$:
$$a^2-\dfrac4{a^2}=3$$
Now let $t=a^2$, then
$$
t^2-3a-4=0\\
t=\dfrac{3\pm\sqrt{9+16}}{2}
$$
which have solutions $t_1=4$ and $t_2=-1$. Notice that $t_2$ cannot be solution, so only solution is
$$
t=4\\
a^2=4\\
a=\pm2
$$
So, we have solutions $a=2,b=1$ and $a=-2,b=-1$. Solutions of $z$ are
$$z=2+i\;\;\;\;\text{or}\;\;\;\;z=-2-i$$
A: 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:
$$ x^2 - y^2 = 3 $$
$$ 2xy = 4 $$
