If $z\in\mathbb{C}$ satisfying $|z-1|=1$ then prove $\arg(z-1)=2\arg z$ 
If $z\in\mathbb{C}$ satisfying $|z-1|=1$ then which of the following is correct 
  \begin{align}
&a)\quad \arg(z-1)=2\arg z\\
&b)\quad 2\arg z=2/3.\arg(z^2-z)\\
&c)\quad \arg(z-1)=\arg(z+1)\\
&d)\quad \arg z=2\arg(z+1)
\end{align}

My Attempt
$$
(x-1)^2+y^2=1\implies
$$
$$
\arg(z-1)=\alpha=\tan^{-1}\frac{y}{x-1}\\
2\arg z=2\tan^{-1}\frac{y}{x}=\tan^{-1}\frac{2y}{x}.\frac{x^2}{x^2-y^2}=\tan^{-1}\frac{2xy}{x^2-y^2}
$$
As it was asked as a multiple choice question, can I easily solve it using geometry ?

the geometric argument is neater, but the following manipulation using the double-angle formula for the cosine is standard, and worth being acquainted with.
$$
z = 1 + e^{i\theta}
$$
so
$$
\arg (z-1) = \arg e^{i\theta} = \theta
$$
and
$$
\arg z = \arg (1+\cos \theta + i\sin \theta) \\
= \arg \left( 2 \cos \frac{\theta}2 \left(\cos \frac{\theta}2+ i \sin \frac{\theta}2 \right)\right) \\
= \frac{\theta}2
$$
