If $|z-1|=1$, find $\arg(z)$ 
If $|z-1|=1$, find $\arg(z)$

Putting $z=r(\cos\theta+i\sin\theta)$ in $|z-1|=1$,
$$|r\cos\theta-1+ir\sin\theta|=1\\\implies r^2\cos^2\theta+1-2r\cos\theta+r^2\sin^2\theta=1\\\implies r=2\cos\theta$$
Putting $r$ in $z$, $$z=2\cos^2\theta+i2\cos\theta\sin\theta\\\implies|z-1|=\cos2\theta+i\sin2\theta$$
So, $\arg(z)=\frac12\arg(z-1)$, which matches with the answer but I am looking for any other method to do it. Maybe a more intuitive method or one involving a diagram?
 A: The set of the points with $|z-1|=1$ is a circle around $1+0i$ with radius $1$. The argument of $z$ is the angle the vector has with the real axis. The answer $\arg(z)=\frac12\arg(z-1)$ corresponds to the familiar theorem regarding the inscribed angle.
In the picture here 
the vector $z_2 = z_1 - (1,0)$. To find the argument of $z_2$, you should translate it to the origin and calculate its angle with the $x$ axis. But this is the same as calculating the angle $\angle ABC$, which is why we can think of $\arg(z-1)$ as the argument of the point $z$ calculated from the center $1$.
A: Yes, there is. Note that $|z-1|=1$ is equivalent to the circle $(x-1)^2+y^2=1$ in the Cartesian plane. This represents a circle with centre $C(1,0)$ and radius $1$. Now, take a random point $P$ on this circle. Let $O$ be the origin. Note that $\Delta OPC$ is an isosceles triangle with $OC=PC$. This means that $\angle POC=\angle OPC=\frac 12 \angle PCX$ from the exterior angle property of of triangle, where $\angle PCX$ represents angle of $PC$ with positive $x$-axis. This is equivalent to $arg(z)=\frac 12 arg(z-1)$ in the Argand plane, since $\arg(z-1)=\angle PCX$ and $arg(z)=\angle POC$.
