Question

Let $z$ be a unit complex number having the argument $\theta$, $0<\theta <\frac{\pi }{2}$ and satisfying the relation $|z-3i|=3,$ then $\arg (\cot \theta -\frac{6}{z})$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{2}$

From polar form of complex numbers,
$z=|z|(\cos \theta +i\sin \theta )$
As $z$ is uni modular, $z=\cos \theta +i\sin \theta$
We have $|z-3i|=3$
$\Rightarrow |\cos \theta +i(\sin \theta -3)|=3$
$\Rightarrow \sqrt{{\cos }^2\theta +(\sin \theta -3{)}^2}=3$
$\Rightarrow {\cos }^2\theta +(\sin \theta -3{)}^2=9$
$\Rightarrow {\cos }^2\theta +{\sin }^2\theta -6\sin \theta +9=9$
$\Rightarrow 1-6\sin \theta =0$
$\Rightarrow \sin \theta =\frac{1}{6}\dots \left(1\right)$

Now, $\cot \theta -\frac{6}{z}=\cot \theta -\frac{6}{(\cos \theta +i\sin \theta )}$
$=\cot \theta -6(\cos \theta -i\sin \theta )$
$=\cot \theta -6\sin \theta (\cot \theta -i)$
By equation $\left(1\right)$
$=\cot \theta -\cot \theta +i=i$
Therefore, $\arg (\cot \theta -\frac{6}{z})=\arg \left(i\right)=\frac{\pi }{2}$