Mentallic said:
Well then that makes two of us

ok I will try that:
Y=\pm X (1)
2XY=\frac{sin\theta}{1-cos\theta} (2)
Substituting (1) into (2) \pm 2X^2=\frac{sin\theta}{1-cos\theta}
Therefore, X=\pm \sqrt{\pm \frac{sin\theta}{2-2cos\theta}} (3)
Substituting (3) back into (1) Y=\pm \sqrt{\pm \frac{sin\theta}{2-2cos\theta}} (4)
I'm unsure if these equations are completely correct so please correct me if any of my steps seem to be wrong. But now what do I do? This doesn't seem to be the locus yet
Okay good, so your possible solution pairs are
(X,Y)=\{(\sqrt{\frac{sin\theta}{2-2cos\theta}},\sqrt{\frac{sin\theta}{2-2cos\theta}}),(-\sqrt{\frac{sin\theta}{2-2cos\theta}},-\sqrt{\frac{sin\theta}{2-2cos\theta}}),(\sqrt{\frac{-sin\theta}{2-2cos\theta}},-\sqrt{\frac{-sin\theta}{2-2cos\theta}}),(-\sqrt{\frac{-sin\theta}{2-2cos\theta}},\sqrt{\frac{-sin\theta}{2-2cos\theta}})\}
Now, remember your definition Z=X+iY
where X and Y are real...That tells you that for a given value of theta, only solutions that have real values for X and Y are valid.
So, let's see what happens when you slowly go counterclockwise around the circle |z|=1; by starting with \theta=0 and slowly increasing it to \theta=2\pi...What are your real solutions for \theta=0? How about \theta=\frac{\pi}{4}? How about \theta=\frac{\pi}{2}, \theta=\frac{3\pi}{4}, \theta=\pi, \theta=\frac{5\pi}{4}, \theta=\frac{3\pi}{2} and \theta=\frac{7\pi}{4}?...what do these solutions look like?
Apparently your locus is the two lines Y=\pm X, and as you go counterclockwise around the circle |z|=1, You start at the points (X,Y)=(\infty,\infty)
and (X,Y)=(-\infty,-\infty) and move inwards along the line Y=X from both directions until you reach the origin at \theta=\pi. Then as you continue along the unit circle, you find that you move from the origin outwards in both directions along the line Y=-X.
Make sense?