Question

State True and False

It is known that $|z+\frac{1}{z}|=a,$ where z is a complex number. The greatest and the least possible values of $\left|z\right|$ are $\frac{1}{2}[a+\sqrt{(a^2+4)}]$,$\frac{1}{2}[a-\sqrt{(a^2+4)}]$. Type 1 for true and 0 for false

Answer 1

Applying triangular inequality, we get
$|z+\frac{1}{z}|\le |z|+|\frac{1}{z}|$
Hence
$a\le |z|+|\frac{1}{z}|$
$|z{|}^2-a|z|+|1|>0$
Hence
$|z{|}^2-a|z|-1>0$ and $|z{|}^2-a|z|+1>0$
From the first equation we get
$z<\frac{a-\sqrt{a^2+4}}{2}$ and $z>\frac{a+\sqrt{a^2+4}}{2}$
And from the second equation we get
$z<\frac{a-\sqrt{a^2-4}}{2}$ and $z>\frac{a+\sqrt{a^2-4}}{2}$