# 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: $$\left|\overline{z}+\frac{2}{z}\right| \ge 1 \rightarrow \left|m-i+\frac{2}{m+i}\right|\ge 1 \rightarrow \left|\frac{m^2-1}{m+i}\right|\ge 1 \rightarrow -1\le \frac{m^2-1}{m+i}\le 1 \rightarrow\begin {cases} -1\le m^2-1\le 1 \\ -1\le m+i\le 1 \end{cases}\rightarrow \begin{cases} m \in [-2,2] \\ m \in [-1-i,1-i] \end{cases}$$

• The statement $m \in [-1-i,1-i]$ doesn't make sense.Is m real? Moreover,shouldn't it be $m^2+3$ Commented Jul 24, 2016 at 18:42
• @TushantMittal thanks for correcting me, I did some mistakes. Commented Jul 24, 2016 at 18:47

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 $m$

• Sorry can I ask you how did you arrive from $\left|m+i\right|$ to $\sqrt{m^2+1}$ in the denominator? Commented Jul 24, 2016 at 18:46
• Assuming m is real Commented Jul 24, 2016 at 18:47
• @T4yl0r yes,indeed Commented Jul 24, 2016 at 18:47
• I still can't find a way to arrive from $\left|m+i\right|$ to $\sqrt{m^2+1}$.. You didn;t amplify so I don't find a way to transform the first expression in the second one. Commented Jul 24, 2016 at 18:55
• @T4yl0r,it is a absolute value of complex number ,you know it ,remember $z=a+ib,\left| z \right| =\left| a+ib \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } }$ Commented Jul 24, 2016 at 18:57

$$\left|\bar{z}+\frac{2}{z}\right| = \frac{|z|^2 +2}{|z|} = |z|+\frac{2}{|z|}$$ By, AM-GM, $$|z|+\frac{2}{|z|} \geq 2\sqrt{2} \geq 1$$ Hence, the inequality is true for all z and hence all m.

• (+1) don't know why I didn't see yours before I posted. I hope you don't mind my edit.
– robjohn
Commented Jul 24, 2016 at 19:28
• Thanks.I definitely don't mind it. Commented Jul 24, 2016 at 19:42