Difference between revisions of "Lentarot"

Revision as of 16:01, 30 January 2025

This article has been proposed for deletion. The reason given is: not in user namespace, move here. Sysops: Before deleting this article, please check the article discussion pages and history.

こんにちは！

Hello

$f(z)=\sum_{j=-\infty}^{\infty} C_n (z-\alpha)^n$

$C_n=\frac{1}{2\pi i}\int\frac{f(\xi)}{(\xi-\alpha)^{n+1}}$

$\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\oint_{\gamma_k}f(z)dz$

$\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\oint_{\gamma_k}\sum_{j=-\infty}^{\infty}C_n (z-\alpha_k)^n$

$\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n\oint_{\gamma_k} (z-\alpha_k)^n$

$z(\theta)=\alpha_k+ae^{i\theta}$ $(0\leq\theta\leq 2\pi)$

$dz=iae^{i\theta}d\theta$

$\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n\int_{0}^{2\pi} (ae^{i\theta})^j iae^{i\theta}d\theta$

$\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n ia^{j+1} \int_{0}^{2\pi}e^{i(j+1)\theta}d\theta$

$\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n ia^{j+1} [\frac{1}{i(n+1)}e^{i(j+1)\theta}]_{0}^{2\pi}$

$\int_{0}^{2\pi}e^{i(n+1)\theta}d\theta =\begin{cases}0 & n\neq -1\\2\pi i & n=-1\end{cases}$

$\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}2\pi i C_{-1}$

$\boxed{\oint_{\gamma}f(z)dz = 2\pi i\sum_{k=1}^{n}res(f(z),\alpha_{k})}$

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+{{delete|not in user namespace, move [[User:Lentarot|here]]}}
 ==HELLO GUYS==
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 <math>dz=iae^{i\theta}d\theta</math>
-<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n\int_{0}^{2\pi} (ae^{i\theta})^n iae^{i\theta}d\theta</math>
+<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n\int_{0}^{2\pi} (ae^{i\theta})^j iae^{i\theta}d\theta</math>
-<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n ia^{n+1} \int_{0}^{2\pi}e^{i(n+1)\theta}d\theta</math>
+<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n ia^{j+1} \int_{0}^{2\pi}e^{i(j+1)\theta}d\theta</math>
-<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n ia^{n+1} [\frac{1}{i(n+1)}e^{i(n+1)\theta}]_{0}^{2\pi}</math>
+<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}\sum_{j=-\infty}^{\infty}C_n ia^{j+1} [\frac{1}{i(n+1)}e^{i(j+1)\theta}]_{0}^{2\pi}</math>
 <math> \int_{0}^{2\pi}e^{i(n+1)\theta}d\theta =\begin{cases}0 & n\neq -1\\2\pi i & n=-1\end{cases} </math>
-<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}2\pi i C_n</math>
+<math>\oint_{\gamma}f(z)dz = \sum_{k=1}^{n}2\pi i C_{-1}</math>
 <cmath>\boxed{\oint_{\gamma}f(z)dz = 2\pi i\sum_{k=1}^{n}res(f(z),\alpha_{k})}</cmath>
 ==contributions==
 [[2016 AIME I Problems/Problem 10]] Solution 4