# Let z=a+ib, where a and b are real. If z/(z-i) is real, show that z is imaginary or 0. Help?

## Let $z = a + i b$, where $a$ and $b$ are real. If $\frac{z}{z - i}$ is real, show that $z$ is imaginary or $0$. Thanks!

Dec 31, 2017

Here's one method...

#### Explanation:

Note that:

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

If this is real then so is $\frac{1}{\frac{z}{i} - 1}$ and therefore $\frac{z}{i} - 1$ and therefore $\frac{z}{i}$.

So if $\frac{z}{i} = c$ for some real number $c$, then $z = c i$, which means that $z$ is either pure imaginary or $0$.