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

We have

$$u=\dfrac{2z+i}{\,z-ki\,},\qquad z=x+iy,\qquad k\gt 0.$$

The condition that generates the required locus in the $$z$$-plane is

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

First we write $$u$$ in the form $$a+ib$$ with real numerator and denominator.

Denominator:

$$z-ki=x+i(y-k).$$

Its modulus-squared is

$$|z-ki|^{2}=x^{2}+(y-k)^{2}.$$

Numerator:

$$2z+i=2(x+iy)+i=2x+i(2y+1).$$

Multiplying numerator and denominator by the conjugate of the denominator,

$$u=\dfrac{\bigl[2x+i(2y+1)\bigr]\bigl[x-i(y-k)\bigr]} {\,x^{2}+(y-k)^{2}\,}.$$ Expanding the product in the numerator step by step,

$$\begin{aligned} [2x]\,[x] &=2x^{2},\\[2pt] [2x]\,[ -i(y-k)] &=-2xi(y-k),\\[2pt] [i(2y+1)]\,[x] &=xi(2y+1),\\[2pt] [i(2y+1)]\,[ -i(y-k)] &=-i^{2}(2y+1)(y-k) =(2y+1)(y-k). \end{aligned}$$

Collecting real and imaginary parts,

$$u=\dfrac{2x^{2}+(2y+1)(y-k)\;+\;i\,x(2k+1)} {\,x^{2}+(y-k)^{2}\,}.$$

Therefore

$$\operatorname{Re}(u)=\dfrac{2x^{2}+(2y+1)(y-k)}{x^{2}+(y-k)^{2}},$$ $$\operatorname{Im}(u)=\dfrac{x(2k+1)}{x^{2}+(y-k)^{2}}.$$

Adding them and equating to $$1$$,

$$ \dfrac{2x^{2}+(2y+1)(y-k)+x(2k+1)}{x^{2}+(y-k)^{2}}=1. $$

Multiplying both sides by the positive denominator,

$$2x^{2}+(2y+1)(y-k)+x(2k+1)=x^{2}+(y-k)^{2}.$$

Transposing the right-hand terms to the left,

$$x^{2}+(2y+1)(y-k)+x(2k+1)-(y-k)^{2}=0.$$

Now expand the $$y$$-terms explicitly:

$$\begin{aligned} (2y+1)(y-k) &=2y^{2}-2ky+y-k,\\ (y-k)^{2} &=y^{2}-2ky+k^{2}. \end{aligned}$$

Substituting these expansions,

$$ x^{2}+[2y^{2}-2ky+y-k]+\;x(2k+1)\;-\;[y^{2}-2ky+k^{2}]=0. $$

Combining like terms carefully,

$$ x^{2}+y^{2}+y-k-k^{2}+x(2k+1)=0. $$

The curve is required to meet the $$y$$-axis, so we put $$x=0$$:

$$y^{2}+y-k-k^{2}=0.$$

This quadratic in $$y$$ has its two roots $$y_{1},y_{2}$$ corresponding to the intersection points $$P$$ and $$Q$$ on the $$y$$-axis. For a quadratic $$y^{2}-Sy+P=0,$$ the distance between its roots is $$|y_{1}-y_{2}|=\sqrt{S^{2}-4P}.$$

In the present quadratic the sum and product of the roots are

$$S=-1,\qquad P=-(k+k^{2}).$$

Hence

$$|y_{1}-y_{2}|=\sqrt{(-1)^{2}-4\bigl[-(k+k^{2})\bigr]} =\sqrt{1+4k+4k^{2}} =\sqrt{(2k+1)^{2}}=2k+1.$$

The length $$PQ$$ is given to be $$5$$, so

$$2k+1=5\quad\Longrightarrow\quad k=2.$$

$$k$$ is required to be positive and $$2$$ satisfies this. No other positive value fulfils the condition.

Hence, the correct answer is Option D.