If \(z=x+i y, x, y \in R\) and if the point \(P\) in the argand plane represents \(z\), then the locus of…

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
  1. $\left\{z \in C \mid \left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}$
  2. \(\{z \in C /(3-i) z+(3+i) \bar{z}-6=0\}\)
  3. $\{z \in \mathbb{C} / (3-i) z+(3+i) \overline{z}-6 > 0\}$, $\left\{\left|z-\frac{1+3i}{2}\right|=\frac{\sqrt{10}}{2}\right\}$
  4. \(\{z \in C /(3-i) z+(3+i) \bar{z}-6 < 0\),

Solution

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

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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