# What is the locus of $\mathrm{arg}(\frac{z-a}{z-b})=\theta$?

This isn't quite homework, but it might as well be. I'm working through Needham's Visual Complex Analysis and I'm finding myself stuck on the following problem:

Explain geometrically why the the locus of z such that $\mathrm{arg}(\frac{z-a}{z-b})=\theta$ is an arc of a certain circle passing through the fixed points a and b.

My first thought was to think about what $\mathrm{arg}(z)=\theta$ means, and I think it should be a line starting at the origin and going in the direction of $\theta$. I also understand that, supposing that we were able to find one z that satisfies this condition, we could find another z by moving the vector in a way that increases the argument of $z-a$ while decreasing the argument if $z-b$.

I'm looking for a hint. Can anyone subtly nudge me in the right direction? I'm having trouble seeing why this has to be the arc of a circle. Also, how can the arc pass through b? Would the fraction $\frac{z-a}{z-b}$ be undefined for z = b? How can it pass through a? Wouldn't the fraction be the zero vector for z = a ?

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Note that $arg((z-a)/(z-b))$ gives the angle between $z-a$ and $z-b$. – Potato Aug 26 '12 at 3:19

The problem says that the circle passes through a and b, not necessarily the arc of the circle . I hope the image below jogs some memories from geometry class, or leads you to a new fun fact--as the case may be.

Good luck!

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I'm pretty familiar with this theorem, but am having trouble connecting it to the problem. – MBP Aug 26 '12 at 3:38
Jeeze, ok. Looking back, I think that there were three things that kept me from solving this problem. (1) I was trying to visualize $\frac{z-a}{z-b}$ all at once. (2) I was thinking of the angle between $z-a$ and $z$, and the angle between $z-b$ and $z$ as two separate angles instead of as one fixed angle between $z-a$ and $z-b$. (3) My drawings of z, a and b weren't done in a way that would remind me of inscribed angles. Should've tinkered more with the picture. – MBP Aug 26 '12 at 4:18

with $r$ free and $\theta$ fixed. Think of $z-a$ as the vector from $a$ to $z$, and $z-b$ similarly. Dividing subtracts their angles, giving the angle between them. So in plain english, this is the same thing as saying "all $z$ that form the same angle with $a$ and $b$". Do you see why this definition gives a circular arc?

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Thanks, but that isn't helping me. If the angle between a and z is the same as the angle between b and z, then z is the bisector of the angle between a and b. But the z that bisect the angle between a and b are all in one line. Help me out: I'm sure I'm misunderstanding you. – MBP Aug 26 '12 at 3:48
Ah, wait, OK, I think I'm seeing this. a, z and b are all points on the edge of the circle? – MBP Aug 26 '12 at 4:08
@MBP Yes! $a,b,z$ are all points on the circle that Justin Laimer posted. The key thing looking at that image is that the angle between $z-a$ (the line from $z$ to $a$) and $z-b$ (similar) is constant. – Robert Mastragostino Aug 26 '12 at 4:15