Question

If $|z+\frac{2}{z}|=2,$ then prove that the maximum value of $\left|z\right|$ is $\sqrt{3}+1.$

Answer 1

$\left| z \right| =\left| z+\dfrac { 2 }{ z } -\dfrac { 2 }{
z } \right| =\left| z+\dfrac { 2 }{ z } +\left( -\dfrac { 2 }{ z
} \right) \right| \le \left| z+\dfrac { 2 }{ z } \right| +
\left| -\dfrac { 2 }{ z } \right| =2+\left| \dfrac { 2 }{ z }
\right|$
$\therefore \left| z \right| \le 2+\dfrac { 2 }{ \left| z
\right| }.$Put$\left| z \right|=t$
$t^2-2t-2\le 0$
$\therefore (t-a)(t-b)\le 0$ where $a$ and $b$ are roots of
$t^2-2t-2=0,a<b$
$\therefore a=1-\sqrt{3},b=1+\sqrt{3},a\le t\le b$
or $\left( 1-\sqrt { 3 } \right) \le \left| z \right| \le 1+\sqrt
{ 3 }$
Hence the max. value of $\left|z\right|$ is $1+\sqrt{3}$.