24. Integration on the Complex Plane
本章主题：复平面上的积分。 定理 24.1：设 \(z=x+yi\in\mathbb C\)、\(u,v:\mathbb R^2\to\mathbb R\)、\(f:\mathbb C\to\mathbb C\)，若 \(f(z)=u(x,y)+v(x,y)i\) 沿曲线 \(C\) 连续，则 \(f(z)\) 沿 \(C\) 可积，且 $$\int_C f(z)\,dz=\int_C u\,dx-v\,dy+i\int_C v\,dx+u\,dy$$ 证明思路：在 \(C\) 上取 \(n\) 个点 \(z_k=x_k+y_k i\)，把黎曼和 \(\sum f(z_k)(z_{k+1}-z_k)\) 展成实部与虚部；由 \(f\) 沿（紧致的）\(C\) 连续，四个相应的极限都存在，故和的极限存在，从而以该极限定义 \(\int_C f(z)\,dz\)。积分依赖于具体的曲线 \(C\)。
Chapter theme: integration on the complex plane. Theorem 24.1: let \(z=x+yi\in\mathbb C\), \(u,v:\mathbb R^2\to\mathbb R\), \(f:\mathbb C\to\mathbb C\); if \(f(z)=u(x,y)+v(x,y)i\) is continuous along a curved line \(C\), then \(f(z)\) is integrable along \(C\) and $$\int_C f(z)\,dz=\int_C u\,dx-v\,dy+i\int_C v\,dx+u\,dy$$ Proof idea: pick \(n\) points \(z_k=x_k+y_k i\) on \(C\) and expand the Riemann sum \(\sum f(z_k)(z_{k+1}-z_k)\) into its real and imaginary parts; since \(f\) is continuous along the (compact) \(C\), the four corresponding limits all exist, so the limit of the sum exists, and we define \(\int_C f(z)\,dz\) by that limit. The integration depends on the specific curved line \(C\).
定理 24.1 设 \(x,y\in\mathbb R\) 且 \(z=x+yi\in\mathbb C\)，\(u,v:\mathbb R^2\to\mathbb R\)，\(f:\mathbb C\to\mathbb C\)。若 \(f(z)=u(x,y)+v(x,y)i\) 沿曲线 \(C\) 连续，则 \(f(z)\) 沿 \(C\) 可积，且 $$\int_C f(z)\,dz=\int_C u\,dx-v\,dy+i\int_C v\,dx+u\,dy$$
证明 对复平面上的一条曲线 \(C\)，从中取 \(n\) 个点，记为 \(z_1,z_2,\dots,z_n\)，其中 \(z_k=x_k+y_k i\)。则 $$\begin{aligned}\sum_{k=1}^n f(z_k)(z_{k+1}-z_k)&=\sum_{k=1}^n\left(u(x_k,y_k)+v(x_k,y_k)i\right)\left((x_{k+1}-x_k)+(y_{k+1}-y_k)i\right)\\&=\sum_{k=1}^n\left(u(x_k,y_k)(x_{k+1}-x_k)-v(x_k,y_k)(y_{k+1}-y_k)\right)\\&\quad+i\sum_{k=1}^n\left(v(x_k,y_k)(x_{k+1}-x_k)+u(x_k,y_k)(y_{k+1}-y_k)\right)\end{aligned}$$ 由于 \(f(z)=u(x,y)+v(x,y)i\) 沿曲线 \(C\) 连续、且曲线 \(C\) 紧致，下面四个极限都存在，即 $$\lim_{n\to\infty}\sum_{k=1}^n u(x_k,y_k)(x_{k+1}-x_k)$$ $$\lim_{n\to\infty}\sum_{k=1}^n u(x_k,y_k)(y_{k+1}-y_k)$$ $$\lim_{n\to\infty}\sum_{k=1}^n v(x_k,y_k)(x_{k+1}-x_k)$$ $$\lim_{n\to\infty}\sum_{k=1}^n v(x_k,y_k)(y_{k+1}-y_k)$$ 都存在。故极限 $$\lim_{n\to\infty}\sum_{k=1}^n f(z_k)(z_{k+1}-z_k)$$ 也存在，从而可定义 $$\begin{aligned}\int_C f(z)\,dz&\equiv\lim_{n\to\infty}\sum_{k=1}^n f(z_k)(z_{k+1}-z_k)\\&=\lim_{n\to\infty}\sum_{k=1}^n\left(u(x_k,y_k)(x_{k+1}-x_k)-v(x_k,y_k)(y_{k+1}-y_k)\right)\\&\quad+i\lim_{n\to\infty}\sum_{k=1}^n\left(v(x_k,y_k)(x_{k+1}-x_k)+u(x_k,y_k)(y_{k+1}-y_k)\right)\\&=\int_C u(x,y)\,dx-v(x,y)\,dy+i\int_C v(x,y)\,dx+u(x,y)\,dy\end{aligned}$$ 显然，该积分依赖于具体的曲线 \(C\)。\(\blacksquare\)
Theorem 24.1 Suppose \(x,y\in\mathbb R\) and \(z=x+yi\in\mathbb C\), and \(u,v:\mathbb R^2\to\mathbb R\) and \(f:\mathbb C\to\mathbb C\). If \(f(z)=u(x,y)+v(x,y)i\) is continuous along a curved line \(C\), then \(f(z)\) is integrable along \(C\) and $$\int_C f(z)\,dz=\int_C u\,dx-v\,dy+i\int_C v\,dx+u\,dy$$
Proof For a curved line \(C\) in the complex plane, pick \(n\) points from it and call it \(z_1,z_2,\dots,z_n\) where \(z_k=x_k+y_k i\). Then $$\begin{aligned}\sum_{k=1}^n f(z_k)(z_{k+1}-z_k)&=\sum_{k=1}^n\left(u(x_k,y_k)+v(x_k,y_k)i\right)\left((x_{k+1}-x_k)+(y_{k+1}-y_k)i\right)\\&=\sum_{k=1}^n\left(u(x_k,y_k)(x_{k+1}-x_k)-v(x_k,y_k)(y_{k+1}-y_k)\right)\\&\quad+i\sum_{k=1}^n\left(v(x_k,y_k)(x_{k+1}-x_k)+u(x_k,y_k)(y_{k+1}-y_k)\right)\end{aligned}$$ Since \(f(z)=u(x,y)+v(x,y)i\) is continuous along the curved line \(C\), and the line \(C\) is compact, the following four limits all exist, i.e. $$\lim_{n\to\infty}\sum_{k=1}^n u(x_k,y_k)(x_{k+1}-x_k)$$ $$\lim_{n\to\infty}\sum_{k=1}^n u(x_k,y_k)(y_{k+1}-y_k)$$ $$\lim_{n\to\infty}\sum_{k=1}^n v(x_k,y_k)(x_{k+1}-x_k)$$ $$\lim_{n\to\infty}\sum_{k=1}^n v(x_k,y_k)(y_{k+1}-y_k)$$ all exists. So, the limit $$\lim_{n\to\infty}\sum_{k=1}^n f(z_k)(z_{k+1}-z_k)$$ also exists, and we can define $$\begin{aligned}\int_C f(z)\,dz&\equiv\lim_{n\to\infty}\sum_{k=1}^n f(z_k)(z_{k+1}-z_k)\\&=\lim_{n\to\infty}\sum_{k=1}^n\left(u(x_k,y_k)(x_{k+1}-x_k)-v(x_k,y_k)(y_{k+1}-y_k)\right)\\&\quad+i\lim_{n\to\infty}\sum_{k=1}^n\left(v(x_k,y_k)(x_{k+1}-x_k)+u(x_k,y_k)(y_{k+1}-y_k)\right)\\&=\int_C u(x,y)\,dx-v(x,y)\,dy+i\int_C v(x,y)\,dx+u(x,y)\,dy\end{aligned}$$ Clearly, the integration depends on the specific curved line \(C\). \(\blacksquare\)