Argument of fraction of complex numbers is constant

Question

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.

Answer 1

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 $C_0$

Answer 2

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.

Answer 3

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.