How to find the locus of points $z\in \mathbb{C}$ that satisfies $$\operatorname{arg}\left(\frac{z-z_1}{\overline{z}-z_2}\right)=\frac{\pi}{4}?$$ With $z_1,z_2 \in \mathbb{C}$.
My attempt:
Let $z = x+iy, z_1=a+ib,z_2=c+id$. Substituting this into the equation, things got very messy, but I think the locus is a hyperbolic curve rotated by an angle of $\frac{\pi}{8}$, but that's not very accurate. Is there another way to do this?
Thank you.
Simply consider $$\operatorname{arg}(\frac{z-z_1}{\overline{z}-z_2})=\operatorname{arg}\{(\frac{z-z_1}{\overline{z}-z_2})\cdot (\frac {\overline{\overline{z}-z_2}}{\overline{\overline{z}-z_2}})\}$$
$$=\operatorname{arg}(\frac{z^2-(\overline{z_2}+z_1)z+\overline{z_2}z_1}{\alpha}) = \operatorname{arg}(z^2-(\overline{z_2}+z_1)z+\overline{z_2}z_1)$$
where $\alpha$ is a real number.
We can do completing the square for the numerator, so that it becomes ${z^{'}}^2 + \beta$, where $\beta$ is a complex number.
$$=\operatorname{arg}({z^{'}}^2 + \beta)$$
By this diagram, we know $$ \operatorname{Re}(({z^{'}}^2 + \beta)) =\operatorname{Im}(({z^{'}}^2 + \beta))$$
Let ${z^{'}}=x+yi, \beta=a+bi$
so it satisfies $(x^2-y^2+a)=(2xy+b)$
$a,$$b$ can be known very easily, so is left for you to compute.
From the equation, it is hyperbola rotated by a certain degree.