Let

Question:

Let $u=\frac{2 z+i}{z-k i}, z=x+i y$ and $k>0 .$ If the curve represented by $\operatorname{Re}(\mathrm{u})+\operatorname{Im}(\mathrm{u})=1$ intersects the $y$-axis at the points $P$ and $Q$ where $P Q=5$, then the value of $k$ is :

1. (1) $3 / 2$

2. (2) $1 / 2$

3. (3) 4

4. (4) 2

Correct Option: 2, 4,

Solution:

$u=\frac{2(x+i y)+i}{(x+i y)-k i}=\frac{2 x+i(2 y+1)}{x+i(y-k)}$

Real part of $u=\operatorname{Re}(u)=\frac{2 x^{2}+(y-K)(2 y+1)}{x^{2}+(y-K)^{2}}$

Imaginary part of $u$

$=\operatorname{Im}(u)=\frac{-2 x(y-K)+x(2 y+1)}{x^{2}+(y-K)^{2}}$

$\because \operatorname{Re}(u)+\operatorname{Im}(u)=1$

$\Rightarrow 2 x^{2}+2 y^{2}-2 K y+y-K-2 x y+2 K x+2 x y+x$

$=x^{2}+y^{2}+K^{2}-2 K y$

Since, the curve intersect at $y$-axis

$\therefore x=0$

$\Rightarrow y^{2}+y-K(K+1)=0$

Let $y_{1}$ and $y_{2}$ are roots of equations if $x=0$

$\because y_{1}+y_{2}=-1$

$y_{1} y_{2}=-\left(K^{2}+K\right)$

$\therefore\left(y_{1}-y_{2}\right)^{2}=\left(1+4 K^{2}+4 K\right)$

Given $P Q=5 \Rightarrow\left|y_{1}-y_{2}\right|=5$

$\Rightarrow 4 K^{2}+4 K-24=0 \Rightarrow K=2$ or $-3$

as $K>0, \therefore K=2$