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 $$x_1=\frac{4\sqrt3-3}{3}$$ $$y_1=\frac{6-2\sqrt3}{3}$$ $$x_2=\frac{-3-4\sqrt3}{3}$$ $$y_2=\frac{6+2\sqrt3}{3}$$


$$z_1=\frac{4\sqrt3-3}{3}+i\frac{6-2\sqrt3}{3}$$ $$z_2=\frac{-3-4\sqrt3}{3}+i\frac{6+2\sqrt3}{3}$$

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.

  • $\begingroup$ I think you made a mistake in your 5th line down (see @mathlove's answer) $\endgroup$
    – pshmath0
    Jun 25, 2015 at 19:41
  • $\begingroup$ $\operatorname{Re}$ is additive, so $\operatorname{Re}(z^2+1)=1$ is equivalent to $\operatorname{Re}(z^2)=0$. $\endgroup$
    – egreg
    Jun 25, 2015 at 20:03

1 Answer 1


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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