If $z\in\mathbb{C}$ satisfying $|z-1|=1$ then prove $\arg(z-1)=2\arg z$

Question

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 ?

Answer 1

Consider the points $O = 0 + 0i$, $A = -1 + 0i$, $B = 1 + 0i$, and $C = z - 1$. Note that $\arg z$ is the angle $BAC$, and $\arg(z - 1)$ is the angle $BOC$. Now apply the theorem from circle geometry about angles subtended from the centre being double the angle subtended from the circumference.

Answer 2

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

$$ \arg z = \arg (1+\cos \theta + i\sin \theta) \\ = \arg \left( 2 \cos \frac{\theta}2 \left(\cos \frac{\theta}2+ i \sin \frac{\theta}2 \right)\right) \\ = \frac{\theta}2 $$