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

Given, z = x + iy<br><br>and $u = {{2z + i} \over {z - ki}}$<br><br>$= {{2(x + iy) + i} \over {(x + iy) - ki}}$<br><br>$$ = {{2x + i(2y + 1)} \over {x + i(y - k)}} \times {{x - i(y - k)} \over {x - i(y - k)}}$$<br><br>$$ = {{2{x^2} + (2y + 1)(y - k) + i(2xy + x - 2xy + 2kx)} \over {{x^2} + {{(y - k)}^2}}}$$<br><br>Given, <br><br>${\mathop{\rm Re}\nolimits} (u) + {\mathop{\rm Im}\nolimits} (u) = 1$<br><br>$$ \Rightarrow {{2{x^2} + (2y + 1)(y - k)} \over {{x^2} + {{(y - k)}^2}}} + {{x + 2kx} \over {{x^2} + {{(y - k)}^2}}} = 1$$<br><br>$\Rightarrow 2{x^2} + (2y + 1)(y - k) + x + 2kx = {x^2} + {(y - k)^2}$<br><br>This curve intersect the y-axis at point P and Q, so at point P and Q x = 0<br><br>Putting x = 0 at the above equation,<br><br>$\therefore$ $(2y + 1)(y - k) = {(y - k)^2}$<br><br>$\Rightarrow 2{y^2} + y - 2yk - k = {y^2} + {k^2} - 2ky$<br><br>$\Rightarrow {y^2} + y - (k + {k^2}) = 0$<br><br>Let roots of this quadratic equation y₁ and y₂<br><br>$\therefore$ Point P (0, y₁) and Q (0, y₂)<br><br>and ${y_{_1}} + {y_2} = 1$ , ${y_1}{y_2} = - k - {k^2}$<br><br>$\therefore$ ${({y_1} - {y_2})^2} = {({y_1} + {y_2})^2} - 4{y_1}{y_2}$<br><br>$= 1 + 4k + 4{k^2}$<br><br>$\Rightarrow |{y_1} - {y_2}| = \sqrt {1 + 4k + 4{k^2}}$<br><br>Given, PQ = 5<br><br>$\Rightarrow |{y_1} - {y_2}| = 5$<br><br>$\Rightarrow \sqrt {1 + 4k + 4{k^2}} = 5$<br><br>$\Rightarrow {k^2} + k - 6 = 0$<br><br>$\Rightarrow k = - 3,\,2$<br><br>So, k = 2 (Given k &gt; 0)