Argument of fraction of complex numbers is constant How can I find all  $z \in \Bbb C$ such that:
$$ \arg\left(\frac{z-a}{z-b}\right) = \text{constant value}  $$
Where $a,b \in \mathbb{C}  $ are constants.
 A: Write $\frac{z-a}{z-b} =  re^{\alpha i}$ ($r,\alpha$ a real numbers) and we have $\text{arg}\left(\frac{z-a}{z-b}\right) = \alpha =$ constant, hence $$z = \frac{b re^{\alpha i} - a}{re^{\alpha i} -1},$$ where $\alpha$ is constant and $r$ is variable.
A: Brute force method -
You can write:
$x = \frac{z-a}{z-b} = \frac{(z-a)(z^\star-b^\star)}{|z|^2+|b|^2}$.
With a little bit of algebra, you can see:
$(|z|^2+|b|^2)\Re x = (\Re z)^2 + (\Im z)^2 - \Re z(\Re a + \Re b) - \Im z(\Im a + \Im b) + \Re a\Re b + \Im a \Im b$, where $\Re$ and $\Im$ denote the real and imaginary parts, respectively.
$(|z|^2+|b|^2)\Im x = \Re z(\Im b - \Im a) + \Im z(\Re b- \Re a) + \Re a \Im b - \Im a \Re b$
Since $\arg(x) = C$, you have $\Re x/ \Im x = C'$. You can plug in $\Re x$ and $\Im  x$ from above and try to establish a relation between $\Re z$ and $\Im z$ to get the locus of $z$. It's hard to say whether you will get a neat solution in the most general case.  
A: Calling 
$$
z-a = \rho_a e^{i\phi_a}\\
z-b = \rho_b e^{i\phi_b}
$$
then
$$
\arg\left(\frac{z-a}{z-b}\right) = \phi_a-\phi_b = \arctan\left(\frac{y-a_y}{x-a_x}\right)-\arctan\left(\frac{y-b_y}{x-b_x}\right) = C_0
$$
but
$$
\tan\left(\arctan\left(\frac{y-a_y}{x-a_x}\right)-\arctan\left(\frac{y-b_y}{x-b_x}\right)\right) = \frac{a_x (y-b_y)+a_y (b_x-x)-b_x y+b_y x}{(a_x-x) (b_x-x)+(a_y-y) (b_y-y)} = \tan C_0
$$
so choosing $a = a_x+ i a_y = 1+ 2 i,\,\, b = b_x + i b_y = -1-i$ we get the curves for distinct values of $C_0$

