Considering the complex number $z = m+i$ for which values of $m$ do we have $ \left|\overline{z}+\frac{2}{z}\right| \ge 1 $ Good evening to everyone. I have the following problem that I tried to solve but my mathematical instinct tells me that I didn't solve it right:
Considering the complex number $z = m+i$ for which values of $m$ do we have: $$ \left|\overline{z}+\frac{2}{z}\right| \ge 1 $$
Here's what I've tried: $$ \left|\overline{z}+\frac{2}{z}\right| \ge 1 \rightarrow \left|m-i+\frac{2}{m+i}\right|\ge 1 \rightarrow \left|\frac{m^2-1}{m+i}\right|\ge 1 \rightarrow -1\le \frac{m^2-1}{m+i}\le 1 \rightarrow\begin {cases} -1\le m^2-1\le  1 \\ -1\le m+i\le 1 \end{cases}\rightarrow \begin{cases} m \in [-2,2] \\ m \in [-1-i,1-i] \end{cases} $$

you have mistakes ,it should be $$\left| \overline { z } +\frac { 2 }{ z }  \right| \ge 1\rightarrow \left| m-i+\frac { 2 }{ m+i }  \right| \ge 1\rightarrow \left| \frac { m^{ 2 }+3 }{ m+i }  \right| \ge 1\rightarrow \frac { \left| { m }^{ 2 }+3 \right|  }{ \sqrt { { m }^{ 2 }+1 }  } \ge 1\\ { m }^{ 2 }+3\ge \sqrt { { m }^{ 2 }+1 } \Rightarrow { m }^{ 4 }+5{ m }^{ 2 }+8\ge 0 $$ which true for  all real $m$
