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 $2$

$u=\frac{2z+i}{z-ki}$
$\Rightarrow u=\frac{2(x+iy)+i}{(x+iy)-ki}$
$\Rightarrow u=\frac{2x+i(2y+1)}{x+i(y-k)}$
$\Rightarrow u=\frac{2x+i(2y+1)}{x+i(y-k)}\times \frac{x-i(y-k)}{x-i(y-k)}$
$\Rightarrow u=\frac{2x^2-2xi(y-k)+xi(2y+1)+(2y+1)(y-k)}{x^2+(y-k{)}^2}$
$\Rightarrow u=\frac{2x^2+(2y+1)(y-k)}{x^2+(y-k{)}^2}+i\frac{2xk+x}{x^2+(y-k{)}^2}$

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

At $y-$axis, $x=0$
$(2y+1)(y-k)=$$(y-k{)}^2$
$\Rightarrow (y-k)\left(\right(2y+1)-(y-k\left)\right)=0$
$\Rightarrow (y-k)(y+1+k)=0$
$\Rightarrow y=k,-1-k$
Let $P(0,k)$ and $(0,-1-k).$
Then, $\sqrt{(k+1+k{)}^2}=5$
$\Rightarrow \pm (2k+1)=5$
$\Rightarrow k=2$ and $k=-3$ (rejected)
$\therefore k=2$