Search

7.1: Теорема Коші
https://ukrayinska.libretexts.org/%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B0/%D0%9B%D1%96%D0%BD%D1%96%D0%B9%D0%BD%D0%B0_%D0%B0%D0%BB%D0%B3%D0%B5%D0%B1%D1%80%D0%B0/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%9C%D0%B0%D1%82%D1%80%D0%B8%D1%87%D0%BD%D0%B8%D0%B9_%D0%B0%D0%BD%D0%B0%D0%BB%D1%96%D0%B7_(Cox)/07%3A_%D0%9A%D0%BE%D0%BC%D0%BF%D0%BB%D0%B5%D0%BA%D1%81%D0%BD%D0%B8%D0%B9_%D0%B0%D0%BD%D0%B0%D0%BB%D1%96%D0%B7_II/7.01%3A_%D0%A2%D0%B5%D0%BE%D1%80%D0%B5%D0%BC%D0%B0_%D0%9A%D0%BE%D1%88%D1%96
$\int z dz =\in_{0}^{2\pi} e^{it}ie^{it} dt = i \int_{0}^{2\pi} e^{i2t} dt = i \int_{0}^{2\pi} \cos(2t)+i \sin(2t) dt = 0 \nonumber$ \[\begin{align*} \int f(z) dz &=\int u(x,y)+iv(x,y) dx+i \int u(x... $\int z dz =\in_{0}^{2\pi} e^{it}ie^{it} dt = i \int_{0}^{2\pi} e^{i2t} dt = i \int_{0}^{2\pi} \cos(2t)+i \sin(2t) dt = 0 \nonumber$ $\begin{align*} \int f(z) dz &=\int u(x,y)+iv(x,y) dx+i \int u(x,y)+iv(x,y) dy \\[4pt] &=\int u(x,y)dx - \int v(x,y) dy+i \int v(x,y)dx+i \int u(x,y) dy \\[4pt] &=\int_{a}^{b} u(x(t), y(t))x'(t)-v(x(t), y(t))y'(t) dt+i \int_{a}^{b} u(x(t), y(t))y'(t)+v(x(t), y(t))x'(t) dt \end{align*}$
2.4: Комплексні змінні, стаціонарні інтеграли шляху
https://ukrayinska.libretexts.org/%D1%84%D1%96%D0%B7%D0%B8%D0%BA%D0%B8/%D0%9A%D0%B2%D0%B0%D0%BD%D1%82%D0%BE%D0%B2%D0%B0_%D0%BC%D0%B5%D1%85%D0%B0%D0%BD%D1%96%D0%BA%D0%B0/%D0%9A%D0%B2%D0%B0%D0%BD%D1%82%D0%BE%D0%B2%D0%B0_%D0%BC%D0%B5%D1%85%D0%B0%D0%BD%D1%96%D0%BA%D0%B0_(Fowler)/02%3A_%D0%94%D0%B5%D1%8F%D0%BA%D1%96_%D0%BE%D1%81%D0%BD%D0%BE%D0%B2%D0%BD%D1%96_%D0%BC%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B8/2.04%3A_%D0%9A%D0%BE%D0%BC%D0%BF%D0%BB%D0%B5%D0%BA%D1%81%D0%BD%D1%96_%D0%B7%D0%BC%D1%96%D0%BD%D0%BD%D1%96%2C_%D1%81%D1%82%D0%B0%D1%86%D1%96%D0%BE%D0%BD%D0%B0%D1%80%D0%BD%D1%96_%D1%96%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D0%B8_%D1%88%D0%BB%D1%8F%D1%85%D1%83
Прирівнюючи дійсну і уявну частини цього рівняння, знаходимо:\[ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \;\; \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}. \label{2...Прирівнюючи дійсну і уявну частини цього рівняння, знаходимо: $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \;\; \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}. \label{2.4.5}$ $\iint \left( -\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right) dxdy+i\int \int \left( \frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right) dxdy \label{2.4.14}$
4.5: Приклади
https://ukrayinska.libretexts.org/%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B0/%D0%90%D0%BD%D0%B0%D0%BB%D1%96%D0%B7/%D0%A1%D0%BA%D0%BB%D0%B0%D0%B4%D0%BD%D1%96_%D0%B7%D0%BC%D1%96%D0%BD%D0%BD%D1%96_%D0%B7_%D0%B4%D0%BE%D0%B4%D0%B0%D1%82%D0%BA%D0%B0%D0%BC%D0%B8_(Orloff)/04%3A_%D0%9B%D1%96%D0%BD%D1%96%D0%B9%D0%BD%D1%96_%D1%96%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D0%B8_%D1%82%D0%B0_%D1%82%D0%B5%D0%BE%D1%80%D0%B5%D0%BC%D0%B0_%D0%9A%D0%BE%D1%88%D1%96/4.05%3A_%D0%9F%D1%80%D0%B8%D0%BA%D0%BB%D0%B0%D0%B4%D0%B8
$\int_{\gamma} \dfrac{1}{z} \ dz = \int_{0}^{2\pi} \dfrac{1}{e^{i \theta}} ie^{i \theta} \ dt = \int_{0}^{2\pi} i \ dt = 2\pi i.$ (ii) Як завжди, ми параметризуємо одиничну окружність як\(\gamma (\t... $\int_{\gamma} \dfrac{1}{z} \ dz = \int_{0}^{2\pi} \dfrac{1}{e^{i \theta}} ie^{i \theta} \ dt = \int_{0}^{2\pi} i \ dt = 2\pi i.$ (ii) Як завжди, ми параметризуємо одиничну окружність як $\gamma (\theta = e^{i \theta}$ з $0 \le \theta \le 2\pi$ . $\int_{\gamma} \dfrac{1}{z^2} \ dz = \int_{0}^{2 \pi} \dfrac{1}{e^{2i \theta}} ie^{i \theta}\ d \theta = \int_{0}^{2\pi} i e^{-i \theta}\ d \theta = -e^{-i \theta} \vert_{0}^{2\pi} = 0.$