1. Given

   $u=\frac{2z+i}{z-ki},\qquad z=x+iy,\qquad k>0$

   and the curve is defined by

   $\Re(u)+\Im(u)=1.$

   We need the intersection of this curve with the y-axis, so set

   $x=0,\quad z=iy.$

2. Substitute $z=iy$ into $u$

   $u=\frac{2(iy)+i}{iy-ki}=\frac{i(2y+1)}{i(y-k)}=\frac{2y+1}{y-k}.$

   This is purely real (provided $y\neq k$), so

   $\Re(u)=\frac{2y+1}{y-k},\qquad \Im(u)=0.$

   Hence the condition becomes

   $\frac{2y+1}{y-k}=1.$

3. Solve for $y$

   $2y+1=y-k$ $y=-k-1.$

   This gives only one point on the y-axis, which contradicts the statement that the curve intersects the y-axis at two points $P$ and $Q$ with $PQ=5$.

4. Find the actual Cartesian equation of the curve

   Let

   $u=\frac{2z+i}{z-ki}=\frac{2x+i(2y+1)}{x+i(y-k)}.$

   Multiply numerator and denominator by the conjugate $x-i(y-k)$:

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

   Expanding the numerator:

   =2x^2+2(y-k)\,x+(2y+1)(y-k)+i\big(x(2y+1)-2x(y-k)\big). $$ So, $$\Re(u)=\frac{2x^2+2x(y-k)+(2y+1)(y-k)}{x^2+(y-k)^2},$$ $$\Im(u)=\frac{x(2y+1)-2x(y-k)}{x^2+(y-k)^2}=rac{x(2k+1)}{x^2+(y-k)^2}.$$ Therefore, $$\Re(u)+\Im(u)=\frac{2x^2+2x(y-k)+(2y+1)(y-k)+x(2k+1)}{x^2+(y-k)^2}=1.$$ Simplify the numerator part involving $x$: $$2x(y-k)+x(2k+1)=x(2y+1).$$ Hence, $$\frac{2x^2+x(2y+1)+(2y+1)(y-k)}{x^2+(y-k)^2}=1.$$ Cross-multiplying, $$2x^2+x(2y+1)+(2y+1)(y-k)=x^2+(y-k)^2.$$ Rearranging, $$x^2+x(2y+1)+(2y+1)(y-k)-(y-k)^2=0.$$ Now simplify the $y$-part: $$ (2y+1)(y-k)-(y-k)^2=(y-k)\big((2y+1)-(y-k)\big)=(y-k)(y+k+1). $$ Thus the curve is $$x^2+x(2y+1)+(y-k)(y+k+1)=0.$$

5. Intersect with the y-axis

   Put $x=0$:

   $(y-k)(y+k+1)=0.$

   So the two intersection points are at

   $y=k \quad \text{and} \quad y=-k-1.$

   Therefore,

   $P=(0,k),\qquad Q=(0,-k-1).$

6. Use the distance condition $PQ=5$

   Since both points lie on the y-axis,

   $PQ=|k-(-k-1)|=|2k+1|.$

   Given $k>0$, we have $2k+1>0$, so

   $2k+1=5.$

   Hence,

   $2k=4 \implies k=2.$

7. Check options

   • A: $2$ ✅
   • B: $4$ ❌
   • C: $\tfrac12$ ❌
   • D: $\tfrac32$ ❌

Therefore, the correct answer is A.