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- 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*}
- 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)/09%3A_%D0%A2%D0%B5%D0%BE%D1%80%D0%B5%D0%BC%D0%B0_%D0%BF%D1%80%D0%BE_%D0%B7%D0%B0%D0%BB%D0%B8%D1%88%D0%BE%D0%BA/9.04%3A_%D0%97%D0%B0%D0%BB%D0%B8%D1%88%D0%BA%D0%B8У цьому розділі ми розглянемо розрахунок залишків. Ми вже бачили достатньо, щоб знати, що це буде корисно. Ми побачимо це ще чіткіше, коли подивимося на теорему залишку в наступному розділі.
- 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)/08%3A_%D0%A1%D0%B5%D1%80%D1%96%D1%8F_%D0%A2%D0%B5%D0%B9%D0%BB%D0%BE%D1%80_%D1%96_%D0%9B%D0%BE%D1%80%D0%B0%D0%BD/8.09%3A_%D0%9F%D0%BE%D0%BB%D1%8F%D0%BA%D0%B8Поляки відносяться до ізольованих сингулярностей.