Find the set of complex numbers such that $Arg(\frac{z}{z-2}) = \frac{\pi}{4}$ I have been stumped by the following question:
Find the set of complex numbers $z\in\Bbb C$ such that $$\operatorname{arg}\left(\frac{z}{z-2}\right) = \frac{\pi}{4}.$$
I think that complex numbers, $z = x + iy$, with principal argument $\frac{\pi}{4}$, all have the property $x = -y$. I then become a bit lost.

In general,
for $~w_1, w_2 \in \Bbb{C} ~: ~w_2 \neq 0,$ 
with $~\overline{w_2} = ~$ the complex conjugate of $w_2$, 
you have that 
$\displaystyle \frac{w_1}{w_2} = \frac{w_1 \times \overline{w_2}}{|w_2|^2}.$
This implies that
$\displaystyle ~\text{Arg}\left[\frac{w_1}{w_2}\right] 
~= ~~\text{Arg}\left[w_1 \times \overline{w_2}\right].$
Set $z = x+ iy$.
Then, you must have that
$\displaystyle ~\text{Arg}\left[(x + iy) \times (x - 2 - iy)\right] = \pi/4.$
As something of a shortcut, if you examine
Re$\left[(x + iy) \times (x - 2 - iy)\right]$ and 
Im$\left[(x + iy) \times (x - 2 - iy)\right]$
you must have that :

*

*The real component equals the imaginary component and

*Both components are positive.

The real component is $(x^2 - 2x + y^2),$ 
while the imaginary component is $-2y$.
So, you can guarantee the 2nd constraint above (i.e. both components positive), based on the 1st constraint, merely by requiring that $y < 0$.
So, the problem reduces to identifying all $(x,y) \in \Bbb{R^2}$ such that

*

*$y < 0$.

*$x^2 - 2x + y^2 = -2y.$
Edit
Originally, my work had one arithmetic mistake, which I corrected, and one (can't see the forest for the trees) simplification that I totally overlooked.
Once Charlotte left me a comment (following my answer), I proofread my answer and found both flaws.
The second constraint above may be re-expressed as $(x - 1)^2 + (y + 1)^2 = 2.$
