# If arg $\left ( z-1 \right )$ = $\large\frac{\pi}{6}$ and arg $\left ( z+1 \right )$ = 2$\large\frac{\pi}{3}$ then prove that $\left | z \right |=1$

Toolbox:
• If $z=x+iy$ is written in exponential form as $z=r(\cos \theta+i\sin \theta),r=\sqrt{x^2+y^2}$ and the argument $\theta$ is given by the following rule
• $\theta=\pi-\alpha\Rightarrow \theta=\alpha$
• $\theta=-\pi+\alpha\Rightarrow \theta=-\alpha$
• Where $\alpha=\tan^{-1}\mid\large\frac{y}{x}\mid$ and $(x,y)$ lies in one of the four quadrants (or the axes).
Step 1:
Let $z=x+iy\Rightarrow z-1=x+iy-1=(x-1)+iy$
$z+1=x+iy+1=(x+1)+iy$
$arg(z-1)=\tan^{-1}\large\frac{y}{x-1}=\frac{\pi}{6}$
$arg(z+1)=\tan^{-1}\large\frac{y}{x+1}=\frac{2\pi}{3}$
Therefore $\large\frac{y}{x-1}$$=\tan\large\frac{\pi}{6} (i.e)\large\frac{y}{x-1}=\large\frac{1}{\sqrt 3} \Rightarrow x-\sqrt 3y=1------(1) \large\frac{y}{x+1}$$=\tan\large\frac{2\pi}{3}$