Let u = (2z + i)/(z - ki), z = x + iy and k > 0 . If the curve...

u = \dfrac{2\left(x + iy\right) + i}{\left(x + iy\right) - ki} = \dfrac{2x + \left(2y + 1\right)i}{x + \left(y - k\right)i} \times \dfrac{x - \left(y - k\right)i}{x - \left(y - k\right)i}$</p><p>Real part of$u = Re\left(u\right) = \dfrac{2{x}^{2} + \left(2y + 1\right)\left(y - k\right)}{{x}^{2} + {\left(y - k\right)}^{2}}$</p><p>Imaginary part of$u = Im\left(u\right)\dfrac{x\left(2y + 1\right) - 2x\left(y - k\right)}{{x}^{2} + {\left(y - k\right)}^{2}}$</p><p>Now$Re\left(u\right) + Im\left(u\right) = 1$</p><p>$\dfrac{2{x}^{2} + \left(2y + 1\right)\left(y - k\right) + x\left(2y + 1\right) - 2x\left(y - k\right)}{{x}^{2} + {\left(y - k\right)}^{2}} = 1$</p><p>for$y$-axis put$x = 0$</p><p>$⇒ \dfrac{\left(2y + 1\right)\left(y - k\right)}{{\left(y - k\right)}^{2}} = 1$</p><p>$⇒$\left(y - k\right)\left(y + 1 + k\right) = 0$

$y = k, - \left(1 + k\right)$

Now point $P\left(0, k\right), Q\left(0, - \left(1 + k\right)\right)$

$PQ = \left|2K + 1\right| = 5$

$2k + 1 = \pm 5$

$2k = 4, - 6$

$k = 2, - 3$

Hence, $k = 2\left(k > 0\right)$.