# how to find the locus of a complex number that satisfies a given equation

describe the locus of the complex number $z$ that satisfies the equation

$\arg(z-a)-\arg(z-z_1)-\arg(z-\overline{z}_1)=k\pi$

where $a\in\mathbb{R}$, $z_1\in\mathbb{C},\Im(z_1)\ne0$ and $k\in\mathbb{Z}$

\begin{align} \Im\left[\frac{z-a}{(z-z_1)(z-\overline{z}_1)}\right]&=0\\ y[(u-a)^2+v^2-(x-a)^2-y^2]&=0\\ [(x-u)^2+v^2-y^2]^2+4y^2(x-u)^2&\ne0 \end{align}

i made $\arg(z-a)=\alpha,\arg(z-z_1)=\beta,\arg(z-\overline{z}_1)=\gamma$ and then i made a drawing

with some geometrical arguments i found that for $z\in\mathbb{R}$ we have $\beta+\gamma=2\pi$

for the angle $\alpha$, for $z\in\mathbb{R}$, $\alpha=180^\circ$ for $z<a$ and $\alpha=0^\circ$ for $z>a$, then $z\in\mathbb{R},z\ne a$ is one of the Solutions.

i found that the locus are a line paralel to $x$ that pass through $(a,0)$ and a sphere with center $(a,0)$ and radius $\sqrt{(u-a)^2+v^2}$ with exception of point $(a,0)$ and the points $(u,-v)$ and $(u,v)$, where $u=\Re(z_1)$ and $v=\Im(z_1)$

however i don't know how to give a geometric argument to this fact

• Forget all these equations. Think about the angles between $z$ and each of $a$, $z_1$, and $\bar z_1$. – Ted Shifrin Feb 22 '15 at 12:26

We may assume $a=0$ and write $z_1=x_1+iy_1$. Then your equation reads $$\arg z-\arg(z^2-2x_1 z+|z_1|^2)=k\pi\ .$$ This implies ${\displaystyle{z^2-2x_1 z+|z_1|^2\over z}}\in{\Bbb R}$ or $z+{\displaystyle{|z_1|^2\over z}}\in{\Bbb R}$. Writing $z=re^{i\phi}$ we see that necessarily $$\sin\phi\left(r-{|z_1|^2\over r}\right)=0\ .$$ It follows that $z$ has to lie on the real axis or on the circle with radius $|z_1|$ around $a=0$.
Conversely: It is easy to check that points $z\in\dot{\Bbb R}$ satisfy $$\arg z+\arg(z-z_1)+\arg(z-\bar z_1)=k\pi\ .$$ For the circle in question we may assume $z=e^{i\phi}$, $z_1=e^{i\beta}$, $\>\phi\ne\pm\beta$. Then \eqalign{\arg(z-z_1)+\arg(z-\bar z_1)&=\arg(e^{i\phi}-e^{i\beta})+\arg(e^{i\phi}-e^{-i\beta})\cr &={\phi+\beta\over2}+\arg(e^{i(\phi-\beta)/2}-e^{-i(\phi-\beta)/2}) \cr &\quad+{\phi-\beta\over2}+\arg(e^{i(\phi+\beta)/2}-e^{-i(\phi+\beta)/2}) \cr &=\arg z +k\pi\ .\cr}