Finding the loci of $\arg\dfrac{z-a}{z-b}=\theta$

Question

I am trying to figure out what the best method is to go about finding this locus.

$$ \arg\dfrac{z-a}{z-b}=\theta $$

I am aware that it must be part of an arc of a circle that passes through the points $a$ and $b$. The argument means that the vector $(z-a)$ leads the vector $(z-b)$ by $\theta$ and so by the argument must lie on an arc of a circle due to the converse of 'angles in the same segment theorem'.

My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don't need to know this, how will I know what the circle looks like.

Finally, if I changed the locus to $$ \arg\dfrac{z-a}{z-b}=-\theta $$

Would this just be the other part of the same circle as previously or a different circle?

Perhaps you could help by showing my how it would work with the question

$$ \arg\frac{z-2j}{z+3} = π/3 $$

Answer 1

Since you are aware that the locus would be a circle, let me just proceed to how you can get the radius of the circle. Let $A$, $B$ and $C$ be the points representing the complex numbers $a$, $b$ and $z$. Then, in $\Delta ABC$, $\angle C = \theta$. By applying the sine rule, i.e.

$$ {\sin A \over BC} = {\sin B \over AC} = {\sin C \over AB} = 2R $$

where $R$ is the circumradius of $\Delta ABC$, we get

$$ R = \frac{\sin C}{2AB} = \frac{\sin \theta}{2|a - b|}. $$

If $\arg\left( \frac{z-a}{z-b} \right) = -\theta$, then the resulting circle will be the reflection of the circle mentioned before about the line $AB$. This happens because a change from $+\theta$ to $-\theta$ only changes the sense of rotation. I leave it up to you to verify this.

EDIT: The locus will not be a circle, but only a part of the circle. Moreover, the curve will have poles at $z = a$ and $z = b$.