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

$$u=\frac{2(x+iy)+i}{(x+iy)-k i}=\frac{2x+(2y+1)i}{x+(y-k)i}\times \frac{x-(y-k)i}{x-(y-k)i}$$u=(x+iy)−ki2(x+iy)+i​=x+(y−k)i2x+(2y+1)i​×x−(y−k)ix−(y−k)i​

Real part of $u=Re(u)=\frac{2{x}^{2}+(2y+1)(y-k)}{{x}^{2}+{(y-k)}^{2}}$u=Re(u)=x2+(y−k)22x2+(2y+1)(y−k)​

Imaginary part of $u=\mathrm{Im}(u)\frac{x(2y+1)-2x(y-k)}{{x}^{2}+{(y-k)}^{2}}$u=Im(u)x2+(y−k)2x(2y+1)−2x(y−k)​

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

$$\frac{2{x}^{2}+(2y+1)(y-k)+x(2y+1)-2x(y-k)}{{x}^{2}+{(y-k)}^{2}}=1$$x2+(y−k)22x2+(2y+1)(y−k)+x(2y+1)−2x(y−k)​=1

for $y$y-axis put $x=0$x=0

$\Rightarrow$⇒ $\frac{(2y+1)(y-k)}{{(y-k)}^{2}}=1$(y−k)2(2y+1)(y−k)​=1

$\Rightarrow$⇒ $(y-k)(y+1+k)=0$(y−k)(y+1+k)=0

$$y=k, -(1+k)$$y=k,−(1+k)

Now point $P(0, k), Q(0, -(1+k))$P(0,k),Q(0,−(1+k))

$$PQ=|2K+1|=5$$PQ=∣2K+1∣=5

$$2k+1=\pm 5$$2k+1=±5

$$2k=4,-6$$2k=4,−6

$$k=2,-3$$k=2,−3

Hence, $k=2(k>0)$k=2(k>0).