The Principle of Deformation of Contours

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Recommends Mark Viola · 2017-07 · Mathematics Stack Exchange · on “Using a keyhole contour for this specific problem”

The arguments for $z$ and $1-z$ were chosen. But they are not unique. For example, we could have chosen $0<\arg(1-z)\le 2\pi$, as Felix Marin chose. As for contour deformation, Cauchy's Integral Theorem states that if $f$ is analytic on $D$, then $\oint_C f(z)\,dz=0$ for any rectifiable curve in $D$. This video tutorial [this video] explains the idea.

Recommends Torsten Schoeneberg · 2025-05 · Mathematics Stack Exchange · on “Evaluate $\int_{C}\frac{e^{z}}{(z-1)(z-2)}dz$ if $C$ is a circle of radius $3$ centered at the origin”

As for how the second approach works, that is an extremely useful consequence of the Cauchy-Goursat theorem (a.k.a. Cauchy Integral Theorem) which is often called "Deformation of Contours". Check e.g. section 49 (8th ed.) / 54 (9th edition) of the Brown/Chruchill textbook, or Video 1 [this video] or Video 2 [LINK] . To apply it here, just notice that your original integrand is analytic everywhere except at $1$ and $2$ , and convince yourself with a drawing that one can deform the original contour $C$ to the union of $C_1$ and $C_2$ without crossing either of the singularities.

Cites Accelerator · 2024-07 · Mathematics Stack Exchange · on “Complex integral with fractional singularities”

I believe your confusion stems from not understanding how exponentials, logarithms, and arguments come into play here. In this answer, we'll refresh ourselves with some basics and dive into the main problem after. PRIMER The definition of complex exponentiation $z^\alpha$ , where $\alpha \in \mathbb{C}$ , is $$ z^\alpha = \exp(\alpha \log z)\,. $$ In turn, the definition of $\log z$ is a multivalued function defined as $$ \log z = \ln |z| + i \arg z\,, $$ where $\arg z$ is the argument of a complex number $z$ , defined as the angle between the positive real axis and the line joining the origin. To avoid multivalued functions for $z^\alpha$ , we choose an interval for $\arg z$ . We may specif

Cites Accelerator · 2025-01 · Mathematics Stack Exchange · on “How to evaluate this integral use dumbbell contour? $I=\int_0^1 \frac{\sqrt[4]{x(1-x)^3}}{(1+x)^3}dx$”

This first part of my answer is me expanding on my comment. TLDR: the mistake is within your residue calculation. Using the formula for Residues at Mulitple-Order Poles [LINK] , we get $$ \begin{align} -\frac{2\pi i}{1+i}\mathop{\mathrm{Res}}_{z=-1}f(z) &= -\frac{2\pi i}{1+i}\left(\frac{1}{(3-1)!}\right)\lim_{z\to -1}\frac{d^{3-1}}{dz^{3-1}}(z-(-1))^3f(z) \\ &= -\frac{\pi i}{1+i} \lim_{z\to -1}\frac{d^{2}}{dz^{2}}z^{\frac{1}{4}}\left(1-z\right)^{\frac{3}{4}} \\ &= \frac{3\pi i}{16\left(1+i\right)}\lim_{z\to -1}z^{-\frac{7}{4}}\left(1-z\right)^{-\frac{5}{4}} \\ &= \frac{3\pi i}{16\left(1+i\right)}\lim_{z\to -1}\exp\left(-\frac{7}{4}\ln\left|z\right|-\frac{7i}{4}\arg z\right)\exp\left(-\frac{5