<s>[INST] Find all complex numbers so that $Im(\frac{z+2}{2-i})=1$ and $Re(z^2+1)=1$ and for $z$ which is in the first quadrant find $\sqrt{z}$ 
Find all complex numbers so that $Im(\frac{z+2}{2-i})=1$ and $Re(z^2+1)=1$ and for $z$ which is in the first quadrant find $\sqrt{z}$.

If $z=x+iy$ then $$\frac{z+2}{2-i}=\frac{x+2+iy}{2-i}\times \frac{2+i}{2+i}=\frac{2x-y+4+i(x+2+2y)}{5}=\frac{2x-y+4}{5}+i\frac{x+2+2y}{5}$$
$Im(\frac{z+2}{2-i})=\frac{x+2+2y}{5}=1$ 
This gives the equation $$x+2y=3$$
$$z^2+1=(x+iy)^2+1=x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
$Re(z^2+1)=x^2-y^2=1$
Solving linear system of equations 
$$
\left\{ 
\begin{array}{c}
x+2y=3 \\
x^2-y^2=1
\end{array}
\right.
$$
I get $$x_1=\frac{4\sqrt3-3}{3}$$ $$y_1=\frac{6-2\sqrt3}{3}$$ $$x_2=\frac{-3-4\sqrt3}{3}$$ $$y_2=\frac{6+2\sqrt3}{3}$$
Now,
$$z_1=\frac{4\sqrt3-3}{3}+i\frac{6-2\sqrt3}{3}$$
$$z_2=\frac{-3-4\sqrt3}{3}+i\frac{6+2\sqrt3}{3}$$
Is this correct, because I don't get good values for $arg(z)$ and $module(z)$?
Complex number $z_1$ is in the first quadrant, so $\sqrt{z_1}=?$ 
I found the module of $z_1$ as $$\rho=\sqrt{\frac{35-16\sqrt3}{3}}$$ and argument of $z_1$ as $$cos\phi=0.84$$ $$sin\phi=0.54$$ which I think is wrong. Approximately, this would be $$\phi=\frac{\pi}{6}$$
Now, using Moivre's formula, 
$$\sqrt[n]{z_k}=\sqrt[n]{\rho}(cos\frac{\phi+2k\pi}{n}+isin\frac{\phi+2k\pi}{n})$$ there are two roots for $\sqrt{z_1}$ for $k=0$ and $k=1$
$$\sqrt{z_0}=\sqrt{\rho}(cos\frac{\pi}{12}+isin\frac{\pi}{12})$$ and
$$\sqrt{z_1}=\sqrt{\rho}(cos\frac{13\pi}{12}+isin\frac{13\pi}{12})$$
Could someone check this?
Thanks for replies.
 [/INST] Note that 
$$z^2+1=(x+iy)^2+1=x^2+2xyi-y^2+1=x^2-y^2\color{red}{+1}+2xyi.$$
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