Question

If $z=x+i y, x, y \in R$ and if the point $P$ in the argand plane represents $z$, then the locus of $P$ satisfying the condition arg $\left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}$, is

• A. $\left\{z \in C /\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}$
• B. $\{z \in C /(3-i) z+(3+i) \bar{z}-6=0\}$
• C. $\{z \in C /(3-i) z+(3+i) \bar{z}-6>0$ $\left.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}$
• D. $\{z \in C /(3-i) z+(3+i) \bar{z}-6<0$ $\left.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}$

Answer 1

Answer: (C)

We have,
$\arg \left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}$
$\Rightarrow \arg (z-1)-\arg (z-3 i)=\frac{\pi}{2}$
$\Rightarrow \arg [(x-1)+i y]-\arg [x+(y-3) i]=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y-3}{x}=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}\left[\frac{\frac{y}{x-1}-\frac{y-3}{x}}{1+\frac{y}{x-1} \cdot \frac{y-3}{x}}\right]=\frac{\pi}{2}$
$\Rightarrow \frac{x y-(x-1)(x-3)}{x(x-1)+y(x-3)}=\tan \frac{\pi}{2}$
$\Rightarrow \frac{x y-(x-1)(y-3)}{x(x-1)+y(y-3)}=\frac{1}{0}$
$\Rightarrow x(x-1)+y(y-3)=0 $
$ \Rightarrow x^{2}+y^{2}-x-3 \,y=0$
$\Rightarrow \left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{1}{4}+\frac{9}{4}$
$\Rightarrow \left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\left(\frac{\sqrt{10}}{2}\right)^{2}$
Which is a circle with centre
$\left(\frac{1}{2}, \frac{3}{2}\right)$ and radius
$\frac{\sqrt{10}}{2}$.
$\therefore z \in C:(3-i) z+(3+i) \bar{z}-6>0,\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}$