# Solve this

Question:

Let $u=\frac{2 z+i}{z-k i}, z=x+i y$ and $k>0$. If the curve

represented by $\operatorname{Re}(u)+\operatorname{Im}(u)=1$ intersects the $y$-axis at the points $P$ and $Q$ where $P Q=5$, then the value of $\mathrm{k}$ is :

1. $3 / 2$

2. 4

3. 2

4. $1 / 2$

Correct Option: , 3

Solution:

$\mathrm{u}=\frac{2 \mathrm{z}+\mathrm{i}}{\mathrm{z}-\mathrm{ki}}$

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

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

$\Rightarrow 2 x^{2}+(2 y+1)(y-k)+x(2 y+1)-2 x(y-k)$

$=x^{2}+(y-k)^{2}$

$\left.\begin{array}{l}\mathrm{P}\left(0, \mathrm{y}_{1}\right) \\ \mathrm{Q}\left(0, \mathrm{y}_{2}\right)\end{array}\right) \Rightarrow \mathrm{y}^{2}+\mathrm{y}-\mathrm{k}-\mathrm{k}^{2}=0\left\{\begin{array}{l}\mathrm{y}_{1}+\mathrm{y}_{2}=-1 \\ \mathrm{y}_{1} \mathrm{y}_{2}=-\mathrm{k}-\mathrm{k}^{2}\end{array}\right.$

$\because \mathrm{PQ}=5$

$\Rightarrow\left|y_{1}-y_{2}\right|=5 \Rightarrow k^{2}+k-6=0$

$\Rightarrow \mathrm{k}=-3,2$

So, $k=2(k>0)$