Question

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

• A. $4$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{3}{2}$

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Given that, $u=\frac{2z+i}{z-ki}$

$$\begin{gathered} u=\frac{2x+1(2y+1)}{x+i(y-k)}\times \frac{x-i(y-k)}{x-i(y-k)} \\ u=\frac{2x^2+(2y+1)(y-k)+i\{2xy+x-2xy+2xk\}}{x^2+{(y-k)}^2} \end{gathered}$$

$$\begin{gathered} Re\left(u\right)+Img\left(u\right)=1 \\ \Rightarrow 2x^2+(2y+1)(y-k)+x+2xk=x^2+{(y-k)}^2 \end{gathered}$$

at Y axis,$x=0$

$$\begin{gathered} (2y+1)(y-k)={(y-k)}^2 \\ 2y^2+y-2yk-k=y^2+k^2-2yk \end{gathered}$$

Roots of $y^2+y-(k+k^2)=0$ are $y_1$ and $y_2$

Difference of roots $=5$

$$\begin{gathered} \sqrt{(1+4k+4k^2)}=5 \\ 4k^2+4k=24 \\ {k}^2+k-6=0 \\ (k+3)(k-2)=0 \\ \Rightarrow k=2 \end{gathered}$$

Hence, option (C) is the correct answer.