# Complex numbers

• September 2nd 2010, 01:18 PM
sinichko
Complex numbers

Prove
$Rez>0 \; \: and \; \; z\neq 1 \Rightarrow \left |\frac{z+1}{z-1} \right |>1$

Thanks a lot
• September 2nd 2010, 02:30 PM
Plato
Quote:

Originally Posted by sinichko
Prove
$Rez>0 \; \: and \; \; z\neq 1 \Rightarrow \left |\frac{z+1}{z-1} \right |>1$

$\left| z \right|^2 + 2\text{Re} (z) + 1 > \left| z \right|^2 - 2\text{Re} (z) + 1~~~[1]$
From the given it should be clear that [1] is true.

Recall that $z + \overline z = 2\text{Re} (z)$

You need to verify that $\left| z \right|^2 + 2\text{Re} (z) + 1 = \left| z \right|^2 + z + \overline z + 1 = \left( {z + 1} \right)\left( {\overline z + 1} \right)$
and $\left| z \right|^2 - 2\text{Re} (z) + 1 = \left| z \right|^2 - z - \overline z - 1 = \left( {z - 1} \right)\left( {\overline z - 1} \right)$

But $\left( {z + 1} \right)\left( {\overline z + 1} \right) = \left| {z + 1} \right|^2 \;\& \,\left( {z - 1} \right)\left( {\overline z - 1} \right) = \left| {z - 1} \right|^2$ which implies $\left| {z + 1} \right|^2 > \left| {z - 1} \right|^2$.

You can finish.