1. ## complex number question

if z = a + bi and|z| = 1, let w = (z+1)/(z-1). prove that w is a pure imaginary number of the form bi.

so from the given i have a^2 + b^2 = 1 since sqrt(a^2 + b^2) = 1 = |z|.

so i plugged in a + bi for z into w = ((a + bi) + 1) / ((a + bi) - 1) i multiplied the top and bottom by the conjugate of the bottom witch is ((a + bi) + 1) and i got something complicated: (2a^2 + 2abi + 2a + 2bi) / (2abi - 2b^2) when i split this up into separate fractions i don't get a pure imaginary number. how do i do this?

2. Originally Posted by oblixps
if z = a + bi and|z| = 1, let w = (z+1)/(z-1). prove that w is a pure imaginary number of the form bi.?
You should know that $\left| z \right| = 1\; \Rightarrow \;z\overline z = 1$ and that $\left| {z - 1} \right|^2$ is a real number.
Also that $\overline z - z = - 2\text{Im} (z)$.

$\frac{{z + 1}}{{z - 1}} = \frac{{\left( {z + 1} \right)\left( {\overline z - 1} \right)}}{{\left| {z - 1} \right|^2 }} = \frac{{z\overline z + \left( {\overline z - z}\right) - 1}}{{\left| {z - 1} \right|^2 }}$

Now put it all together.

3. Originally Posted by Plato
You should know that $\left| z \right| = 1\; \Rightarrow \;z\overline z = 1$ and that $\left| {z - 1} \right|^2$ is a real number.
Also that $\overline z - z = - 2\text{Im} (z)$.

$\frac{{z + 1}}{{z - 1}} = \frac{{\left( {z + 1} \right)\left( {\overline z - 1} \right)}}{{\left| {z - 1} \right|^2 }} = \frac{{z\overline z + \left( {\overline z - z}\right) - 1}}{{\left| {z - 1} \right|^2 }}$

Now put it all together.
what does the notation $\overline z$ and $\ - 2\text{Im} (z)$ mean? i haven't encountered those before in my class.

4. We have, as Plato said, $z\cdot\overline{z}=1\Rightarrow \overline{z}=\frac{1}{z}$

A complex number w is pure imaginary if and only if $\overline{w}=-w$.

$\overline{w}=\overline{\left(\frac{z+1}{z-1}\right)}=\frac{\overline{z+1}}{\overline{z-1}}=\frac{\overline{z}+1}{\overline{z}-1}=$

$=\frac{\displaystyle\frac{1}{z}+1}{\displaystyle\f rac{1}{z}-1}=\frac{1+z}{1-z}=-\frac{z+1}{z-1}=-w$