答案：C
$\begin{array}{cl}
& z_1 \\
= & \dfrac{2+ki}{1+i} \\
= & \dfrac{2+ki}{1+i} \times \dfrac{1-i}{1-i}\\
= & \dfrac{2-2i + ki -ki^2}{1^2-i^2} \\
= & \dfrac{2-2i + ki -k(-1)}{1 – (-1)^2} \\
= & \dfrac{2+k}{2} + \dfrac{k-2}{2}i
\end{array}$
$\begin{array}{cl}
& z_1 \\
= & \dfrac{2+ki}{1+i} \\
= & \dfrac{2+ki}{1+i} \times \dfrac{1-i}{1-i}\\
= & \dfrac{2-2i + ki -ki^2}{1^2-i^2} \\
= & \dfrac{2-2i + ki -k(-1)}{1 – (-1)^2} \\
= & \dfrac{2+k}{2} + \dfrac{k-2}{2}i
\end{array}$
$\begin{array}{cl}
& z_2 \\
= & \dfrac{k+5i}{2-i} \\
= & \dfrac{k+5i}{2-i} \times \dfrac{2+i}{2+i} \\
= & \dfrac{2k +ki +10i +5i^2}{2^2 – i^2} \\
= & \dfrac{2k +ki +10i +5(-1)}{4 – (-1)} \\
= & \dfrac{2k -5}{5} + \dfrac{k+10}{5}i
\end{array}$
由於 $z_1$ 及 $z_2$ 的虛部相等，所以
$\begin{array}{rcl}
\dfrac{k-2}{2} & = & \dfrac{k+10}{5} \\
5k – 10 & = & 2k + 20 \\
3k & = & 30 \\
k & = & 10
\end{array}$
由此，$z_1 – z_2$
$\begin{array}{cl}
= & \left(\dfrac{2+10}{2} + \dfrac{10-2}{2} i\right) – \left( \dfrac{2(10)-5}{5} + \dfrac{10+10}{5} i \right) \\
= & 6+4i – 3 – 4i \\
= & 3
\end{array}$
