（线性性）

$$\begin{aligned} \int_a^b [\alpha f(t) + \beta g(t)] \,dt &= \int_a^b [\alpha \mathrm{Re} f(t) + \beta \mathrm{Re} g(t)] \, dt +i \int_a^b [\alpha \mathrm{Im} f(t) + \beta \mathrm{Im} g(t)] \, dt \\ &= \alpha \int_a^b \mathrm{Re} f(t) \, dt + \beta \int_a^b \mathrm{Re} g(t) \, dt +i \left( \alpha \int_a^b \mathrm{Im} f(t) \, dt + \beta \int_a^b \mathrm{Im} g(t) \, dt \right) \\ &= \alpha \int_a^b f(t) \, dt + \beta \int_a^b g(t) \, dt \end{aligned}$$

（积分不等式）
取极坐标表示 $\int_a^b f(t) \, dt = R e^{i\theta}$，其中 $R = |\int_a^b f(t) \, dt|$，则

$$\begin{aligned} \left| \int_a^b f(t) \, dt \right| &= e^{-i\theta} \int_a^b f(t) \, dt\\ &= \int_a^b e^{-i\theta} f(t) \, dt \in \mathbb R\\ &= \int_a^b \mathrm{Re} \left( e^{-i\theta} f(t) \right) \, dt \\ &\leq \int_a^b \left| e^{-i\theta} f(t) \right| \, dt \\ &= \int_a^b |f(t)| \, dt \end{aligned}$$

$\square$

（反向路径）

$$\begin{aligned} \int_{-\gamma} f(z) \, dz &= \int_a^b f(\gamma(a+b-t)) \gamma'(a+b-t) \, dt \\ &= \int_b^a f(\gamma(s)) \gamma'(s) \, (-ds) \quad (s = a+b-t) \\ &= - \int_a^b f(\gamma(s)) \gamma'(s) \, ds \\ &= - \int_\gamma f(z) \, dz \end{aligned}$$

（曲线拼接）

$$\begin{aligned} \int_{\gamma_1 \vee \gamma_2} f(z) \, dz &= \int_a^b f((\gamma_1 \vee \gamma_2)(t)) (\gamma_1 \vee \gamma_2)'(t) \, dt \\ &= \int_a^{(a+b)/2} f(\gamma_1(2t - a)) \cdot 2 \gamma_1'(2t - a) \, dt +\int_{(a+b)/2}^b f(\gamma_2(2t - b)) \cdot 2 \gamma_2'(2t - b) \, dt \\ &= \int_a^b f(\gamma_1(s)) \gamma_1'(s) \, ds + \int_a^b f(\gamma_2(s)) \gamma_2'(s) \, ds \quad (s = 2t - a \text{ or } 2t - b) \\ &= \int_{\gamma_1} f(z) \, dz + \int_{\gamma_2} f(z) \, dz \end{aligned}$$

（线性性）

$$\begin{aligned} \int_\gamma [\lambda f(z) + g(z)] \, dz &= \int_a^b [\lambda f(\gamma(t)) + g(\gamma(t))] \gamma'(t) \, dt \\ &= \lambda \int_a^b f(\gamma(t)) \gamma'(t) \, dt + \int_a^b g(\gamma(t)) \gamma'(t) \, dt \\ &= \lambda \int_\gamma f(z) \, dz + \int_\gamma g(z) \, dz \end{aligned}$$

（积分不等式）

$$\begin{aligned} \left| \int_\gamma f(z) \, dz \right| &= \left| \int_a^b f(\gamma(t)) \gamma'(t) \, dt \right| \\ &\leq \int_a^b |f(\gamma(t))| \cdot |\gamma'(t)| \, dt \\ &= \int_\gamma |f(z)| \, |dz| \\ &\leq M \int_a^b |\gamma'(t)| \, dt = M L(\gamma) \end{aligned}$$

$\square$

与分段光滑的定义类似，均可通过对每一段 $C^1$ 曲线应用相应命题，然后将结果相加得到整体结论，证明过程与之前完全相同，此处省略
$\square$

若 $\gamma$ 本身为 $C^1$ 曲线，则由于

$$\frac{d}{dt} F(\gamma(t)) = F'(\gamma(t)) \cdot \gamma'(t) = f(\gamma(t)) \cdot \gamma'(t)$$

所以通过实数上的微积分基本定理，

$$\begin{aligned} \int_\gamma f(z) \, dz &= \int_a^b f(\gamma(t)) \gamma'(t) \, dt \\ &= \int_a^b \frac{d}{dt} \mathrm{Re}(F(\gamma(t))) \, dt + i \int_a^b \frac{d}{dt} \mathrm{Im}(F(\gamma(t))) \, dt \\ &= \bigg[\mathrm{Re}(F(\gamma(t)))\bigg]_a^b + i \bigg[\mathrm{Im}(F(\gamma(t)))\bigg]_a^b \\ &= F(E(\gamma)) - F(S(\gamma)) \end{aligned}$$

对于一般的路径 $\gamma$，将其分割为 $n$ 段 $C^1$ 曲线 $\gamma_i$，则

$$\begin{aligned} \int_\gamma f(z) \, dz &= \sum_{i=1}^n \int_{\gamma_i} f(z) \, dz \\ &= \sum_{i=1}^n [F(E(\gamma_i)) - F(S(\gamma_i))] \\ &= F(E(\gamma)) - F(S(\gamma)) \end{aligned}$$

$\square$