Let u = 2 z + i z - k i , z = x + i y and k > 0 . If the curve represented by Re ( u ) + Im ( u ) = 1…

$u=\frac{2\left(x+iy\right)+i}{\left(x+iy\right)-ki}=\frac{2x+\left(2y+1\right)i}{x+\left(y-k\right)i}\times \frac{x-\left(y-k\right)i}{x-\left(y-k\right)i}$

Real part of $u\mathit{=}Re\left(u\right)\mathit{=}\frac{\mathit{2}{x}^{\mathit{2}}+\left(\mathit{2}y\mathit{+}\mathit{1}\right)\left(y\mathit{-}k\right)}{x^{\mathit{2}}\mathit{+}{\left(y\mathit{-}k\right)}^{\mathit{2}}}$

Imaginary part of $u=\mathrm{Im}\left(u\right)\frac{x\left(2y+1\right)-2x\left(y-k\right)}{x^2+{\left(y-k\right)}^2}$

Now $\mathrm{Re}\left(u\right)+\mathrm{Im}\left(u\right)=1$

$\frac{2x^2+\left(2y+1\right)\left(y-k\right)+x\left(2y+1\right)-2x\left(y-k\right)}{x^2+{\left(y-k\right)}^2}=1$

for $y$-axis put $x=0$

$\Rightarrow$ $\frac{\left(2y+1\right)\left(y-k\right)}{{\left(y-k\right)}^2}=1$

$\Rightarrow$ $\left(y-k\right)\left(y+1+k\right)=0$

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

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, $k=2 \left(k>0\right)$.