曲线 C 为 1+i 到 3-4i 的直线段，int_(C) 2z , dz=A. -7-26iB. -7+22iC. 23-26iD. 25-26i

本题考查复变函数沿曲线的积分计算，解题思路是先求出曲线$C$的参数方程，然后将$z$、$dz$用参数表示出来，最后将其代入曲线积分表达式进行计算。

步骤一：求曲线$C$的参数方程

已知曲线$C$为从$1 + i$到$3 - 4i$的直线段，设$z(t)=(1 + i)+t[(3 - 4i)-(1 + i)]$，其中$t\in[0,1]$。
对$z(t)$进行化简：
$\begin{align*} z(t)&=(1 + i)+t(3 - 4i - 1 - i)\\ &=(1 + i)+t(2 - 5i)\\ &=1 + 2t+(1 - 5t)i \end{align*}$

步骤二：求$dz$

对$z(t)$求导，根据求导公式$(u+v)^\prime=u^\prime+v^\prime$，可得：
$z^\prime(t)=(1 + 2t+(1 - 5t)i)^\prime=(1 + 2t)^\prime+(1 - 5t)^\prime i=2 - 5i$
所以$dz = z^\prime(t)dt=(2 - 5i)dt$。

步骤三：将$z$和$dz$代入曲线积分表达式

已知$\int_{C} 2z \, dz$，将$z = 1 + 2t+(1 - 5t)i$和$dz = (2 - 5i)dt$代入可得：
$\begin{align*} \int_{C} 2z \, dz&=\int_{0}^{1} 2[1 + 2t+(1 - 5t)i](2 - 5i)dt\\ &=2\int_{0}^{1}[(1 + 2t)+(1 - 5t)i](2 - 5i)dt\\ &=2\int_{0}^{1}[(1 + 2t)\times2-(1 + 2t)\times5i+(1 - 5t)i\times2-(1 - 5t)i\times5i]dt\\ &=2\int_{0}^{1}[2 + 4t - 5i - 10ti + 2i - 10ti - 5i^2 + 25ti^2]dt\\ &=2\int_{0}^{1}[2 + 4t - 5i - 20ti + 5 - 25t]dt\\ &=2\int_{0}^{1}[(2 + 5)+(4t - 25t)- 20ti - 5i]dt\\ &=2\int_{0}^{1}(7 - 21t - 20ti - 5i)dt \end{align*}$

步骤四：计算定积分

根据定积分的性质$\int_{a}^{b}(u+v)dt=\int_{a}^{b}u dt+\int_{a}^{b}v dt$，可得：
$\begin{align*} &2\int_{0}^{1}(7 - 21t - 20ti - 5i)dt\\ =&2\left(\int_{0}^{1}7dt - \int_{0}^{1}21tdt - \int_{0}^{1}20ti dt - \int_{0}^{1}5i dt\right)\\ =&2\left(7t\big|_{0}^{1}-\frac{21}{2}t^2\big|_{0}^{1}-10t^2i\big|_{0}^{1}-5it\big|_{0}^{1}\right)\\ =&2\left(7\times(1 - 0)-\frac{21}{2}\times(1^2 - 0^2)-10\times(1^2 - 0^2)i-5i\times(1 - 0)\right)\\ =&2\left(7 - \frac{21}{2}-10i - 5i\right)\\ =&2\left(\frac{14}{2} - \frac{21}{2}-15i\right)\\ =&2\left(-\frac{7}{2}-15i\right)\\ =&-7 - 30i \end{align*}$
发现计算结果与答案不一致，重新检查步骤三的展开式：
$\begin{align*} &2\int_{0}^{1}[(1 + 2t)+(1 - 5t)i](2 - 5i)dt\\ =&2\int_{0}^{1}[(1 + 2t)\times2-(1 + 2t)\times5i+(1 - 5t)i\times2-(1 - 5t)i\times5i]dt\\ =&2\int_{0}^{1}[2 + 4t - 5i - 10ti + 2i - 10ti - 5i^2 + 25ti^2]dt\\ =&2\int_{0}^{1}[2 + 4t - 5i - 20ti + 5 - 25t]dt\\ =&2\int_{0}^{1}[(2 + 5)+(4t - 25t)- 20ti - 5i]dt\\ =&2\int_{0}^{1}(7 - 21t - 20ti - 5i)dt \end{align*}$
发现错误，正确展开式为：
$\begin{align*} &2\int_{0}^{1}[(1 + 2t)+(1 - 5t)i](2 - 5i)dt\\ =&2\int_{0}^{1}[(1 + 2t)\times2-(1 + 2t)\times5i+(1 - 5t)i\times2-(1 - 5t)i\times5i]dt\\ =&2\int_{0}^{1}[2 + 4t - 5i - 10ti + 2i - 10ti + 5 - 25t]dt\\ =&2\int_{0}^{1}[(2 + 5)+(4t - 25t)+(- 10t - 10t)i + (-5 + 2)i]dt\\ =&2\int_{0}^{1}(7 - 21t - 20ti - 3i)dt \end{align*}$
重新计算定积分：
$\begin{align*} &2\int_{0}^{1}(7 - 21t - 20ti - 3i)dt\\ =&2\left(\int_{0}^{1}7dt - \int_{0}^{1}21tdt - \int_{0}^{1}20ti dt - \int_{0}^{1}3i dt\right)\\ =&2\left(7t\big|_{0}^{1}-\frac{21}{2}t^2\big|_{0}^{1}-10t^2i\big|_{0}^{1}-3it\big|_{0}^{1}\right)\\ =&2\left(7\times(1 - 0)-\frac{21}{2}\times(1^2 - 0^2)-10\times(1^2 - 0^2)i-3i\times(1 - 0)\right)\\ =&2\left(7 - \frac{21}{2}-10i - 3i\right)\\ =&2\left(\frac{14}{2} - \frac{21}{2}-13i\right)\\ =&2\left(-\frac{7}{2}-13i\right)\\ =&-7 - 26i \end{align*}$