Question

Prove that $\left|\frac{z}{\left|z\right|-1}\right|\le \left|argz\right|$

Answer 1

Clearly $|z|\ne 1$
Now
$|\frac{z}{|z|-1}|=\frac{|z|}{||z|-1|}$
Let
$z=r.e^{i\theta }$
Hence $\frac{|z|}{||z|-1|}$ $=\frac{r}{||r|-1|}=\frac{r}{1-r}$ For $0<r<1$
and for $r>1$ we get $\frac{r}{r-1}$
Now as $r$ increases this value decreases, and maximum is when $|r|=2$
Now considering $arg\left(z\right)$
$|arg\left(z\right)|ϵ[0,2\pi ]$
$=[0,6.28]$
Which is generally greater than $|\frac{|z|}{|z|-1}|$.
Hence the following inequality is true under certain condtions.