Question

If z is a point on the Argand plane such that $|z-1|=1$, then $\frac{z-2}{z}$ is equal to

• A. tan (arg (z-1))
• B. cot (arg (z-1))
• C. i tan (arg (z-1))
• D. none of these

Answer 1

Answer: (C)

i tan (arg (z-1))

Since $|z-1|=1$,
$\therefore$ Let $z-1=\cos \theta +i\sin \theta$
Then, $z-2=\cos \theta +i\sin \theta -1$
$=-2{\sin }^2\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}=2i\sin \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})$ ... (1)
and $z=1+\cos \theta +i\sin \theta$
$=2{\cos }^2\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}=2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})$ ... (2)
From (1) and (2), we get $\frac{z-2}{z}=i\tan \frac{\theta }{2}=i\tan \left(arg\right(z-1\left)\right)(\because arg=\frac{\theta }{2})$