[C15] 「変身」

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高清赞助4K/60FPS is on Patreon/面包多/Fanbox/Fansky您的支持是我创作的最大动力❤️ Your support will be the greatest motivation for me❤️啪啪还在k，先发纯舞版吧。最近期末压力大，更新会慢一些请见谅...I\'m still working on the sex part, so I post the dance first. I\'m quite stessed with the schoolwork lately at the end of the semester. Apologies for the delay...你怎么能直接瞪眼写出积分结果啊？！复变函数不是这样的！你应该先观察f(z)的格式，然后从分母中找到所有孤立奇点，比如叫z0,z1...然后你把你的孤立奇点列出来，并判断它们是否在积分路径C的内部，确保能使用留数定理。接着你要求极限，当z→z0时如果f(z)＝C，那么可去奇点留数Res＝0；当z→z0时如果f(z)＝∞，那么你要判断奇点的级数。然后你要用(z-z0)^m乘上f(z)求z→z0的极限，如果＝C那么就是m级极点。或者你要用零点判断，如果z0是分子的n级零点是分母的m级零点，那么就是f(z)的m-n级极点。之后你把这几个m阶奇点代入留数公式里，然后用Res(m)＝(1/(m-1)!)*lim_{z→z0}{[z-z0]^{m}f(z)}^{(m-1)}求出留数。最后你需要用留数定理∮_C f(z)dz＝2πiΣ_{k＝1}^{n}Res[f(z),z_k]才能求出积分结果！你怎么直接上来就想瞪眼看出答案？！复变函数根本不是这样！我拒绝作答！How could you just stare blankly and write down the integral result?! Complex functions don\'t work like that! You should first examine the form of f(z), then identify all isolated singularities in the denominator—let\'s call them z₀, z₁, and so on. Next, list these isolated singularities and determine whether they lie inside the integration path C to ensure the residue theorem applies. Next, evaluate the limit: if f(z) = C as z → z₀, then the removable singularity residue Res = 0. If f(z) = ∞ as z → z₀, determine the singularity\'s order. Then, multiply (z - z₀)^m by f(z) to find the limit as z → z₀. If this equals C, it\'s an mth-order pole. Alternatively, use the zero-order test: if z₀ is an nth-order zero in the numerator and an mth-order zero in the denominator, it\'s an m - nth-order pole of f(z). Substitute these m-order singularities into the residue formula. Calculate the residue using Res(m) = (1/(m-1)!) * lim_{z→z0} [z-z0]^m * f(z)^(m-1). Finally, you must apply the residue theorem: ∫_C f(z) dz = 2πi*Σ_{k=1}^{n} Res[f(z), z_k] to obtain the integral result! How dare you expect to stare the answer out of me?! Complex functions don\'t work that way! I refuse to answer!Q群:807984715 or Discord