Let u = frac{2z + i}{z - ki}, where z = x + i | vedclass

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(C) Given $u = \frac{2z + i}{z - ki} = \frac{2(x + iy) + i}{(x + iy) - ki} = \frac{2x + i(2y + 1)}{x + i(y - k)}$.Multiplying numerator and denominato

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$2$

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(C) Given $u = \frac{2z + i}{z - ki} = \frac{2(x + iy) + i}{(x + iy) - ki} = \frac{2x + i(2y + 1)}{x + i(y - k)}$.Multiplying numerator and denominator by the conjugate $x - i(y - k)$:$u = \frac{[2x + i(2y + 1)][x - i(y - k)]}{x^2 + (y - k)^2} = \frac{2x^2 + (2y + 1)(y - k) + i[x(2y + 1) - 2x(y - k)]}{x^2 + (y - k)^2}$.Given $\operatorname{Re}(u) + \operatorname{Im}(u) = 1$, we have:$2x^2 + (2y + 1)(y - k) + x(2y + 1) - 2x(y - k) = x^2 + (y - k)^2$.For intersection with the $y$-axis, set $x = 0$:$(2y + 1)(y - k) = (y - k)^2$.$(y - k)[(2y + 1) - (y - k)] = 0 \Rightarrow (y - k)(y + k + 1) = 0$.This gives $y_1 = k$ and $y_2 = -k - 1$.The distance $PQ = |y_1 - y_2| = |k - (-k - 1)| = |2k + 1| = 5$.Since $k > 0$, $2k + 1 = 5$ $\Rightarrow 2k = 4$ $\Rightarrow k = 2$.

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

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