# How to describe the locus of $z$ where $\arg \left(\frac{z-z_1}{z-z_2}\right)=\theta$.

I need to describe the locus of $z$ which stisfies $\displaystyle\arg \left(\frac{z-z_1}{z-z_2}\right)=\theta$.

$\displaystyle\arg \left(\frac{z-z_1}{z-z_2}\right)=\theta \Rightarrow \arg(z-z_1)-\arg(z-z_2)=\theta$

So I got that ,

$\arg(z-z_1)=\arg(z-z_2)+\theta$

So $z$ lies on an arc.

If $0<\theta \leq\frac{\pi}{2}$ , $z$ lies on the major arc of a circle which passes through $z_1$ and $z_2.$ But there are two such circles ! How can we identify the correct circle ?

If $\frac{\pi}{2}<\theta < \pi$ , $z$ lies on the major arc of a circle which passes through $z_1$ and $z_2.$ But there are two such circles ! How can we identify the correct circle ?

• The positive orientation ($\theta >0$) is counterclockwise. It is correct at your drawings. – user376343 Nov 26 '18 at 20:26

Since $\theta$ is a given real number, the arc is oriented. Only one of two possible circles has the right orientation of the angle from $z_1$ via $z$ to $z_2.$

• Yes! I am asking how to choose the correct one ? – Angelo Mark Dec 13 '15 at 15:51
• The one where points on the arc satisfy $\arg(z-z_1)-\arg(z-z_2)=\theta,$ not $-\theta.$ – Justpassingby Dec 13 '15 at 15:53
• Well how can I find a point on the arc ? – Angelo Mark Dec 13 '15 at 15:56
• If $\theta$ is sharp, then the angle of the segment $[z_1z_2]$ viewed from the centre of the circle is $2\theta.$ This helps you to find the centre of the circle on the bisector line of $z_1$ and $z_2.$ Again, be careful to select the correct orientation of $2\theta$ or you end up with 2 possible centres. If $\theta$ is obtuse then the centre angle is $-2(\pi-\theta).$ – Justpassingby Dec 13 '15 at 16:01