1. ## Complex Proof

$z\in C$ such that $\frac{z}{z-i}$ is real. Show that z is imaginary.

just at the first stage of intro to complex numbers and beginning to struggle
so far i have this but don't know how to proceed any further

$\frac{z}{z-i}=\frac{z-i+i}{z-i}$
$=\frac{z-i}{z-i}+\frac{i}{z-i}$
$=1+\frac{i}{z-i}$
$=1+\frac{i\times i}{i(z-i)}$
$=1-\frac{1}{iz+1}$
hence $\frac{1}{iz+1}$ is real

multiply by conjugate
$=\frac{1}{iz+1}\times\frac{1-i\overline{z}}{1-i\overline{z}}$
$=\frac{1-i\overline{z}}{(iz+1)(-i\overline{z}+1)}$ which is real
hence $i\overline{z}$ must also be real

not quite sure what to do to proceed now.

any help is enormously appreciated

2. ## Re: Complex Proof

Originally Posted by jacs
$z\in C$ such that $\frac{z}{z-i}$ is real. Show that z is imaginary.
just at the first stage of intro to complex numbers and beginning to struggle
so far i have this but don't know how to proceed any further.
We can show that $\frac{z}{{z - i}} = \frac{{z(\overline z + i)}}{{{{\left| {z - i} \right|}^2}}} = \frac{{z\overline z + iz}}{{{{\left| {z - i} \right|}^2}}}$.
Because $\frac{{z\overline z + iz}}{{{{\left| {z - i} \right|}^2}}}$ is real conclude that $z$ must be imaginary.

3. ## Re: Complex Proof

let $z = a+bi$

$\dfrac{z}{z-i} = \dfrac{a+bi}{a+(b-1)i} \cdot \dfrac{a-(b-1)i}{a-(b-1)i} = \dfrac{a^2+abi -a(b-1)i + b(b-1)}{a^2+(b-1)^2} \implies ab = a(b-1) \implies a = 0 \implies z = bi \, , \, b \ne 0$

4. ## Re: Complex Proof

I would simply have written $\frac{z}{z- i}= a$ where a is real. Then $z= a(z- i)= az- ai$, $(a- 1)z= ai$ so $z= \frac{ai}{a- 1}= \frac{a}{a- 1}i$. Since a is real, so is $\frac{a}{a- 1}$.