# Questions tagged [branch-cuts]

A branch cut is curve in the complex extending from a branch point of the function.

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### $\log(z_1z_2) = \log(z_1) +\log(z_2) + i2n\pi$ where log is the analytic branch of complex log.

In a problem provided in my course (the assignment was due a while ago, this isn't a homework help request), we needed to show : $$\log(z_1z_2) = \log(z_1) + \log(z_2) + i2n\pi$$ I seem to understand ...
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### Limit of Complex logarithm towards branch cut

The textbook asks me to verify whether or not this statement is true: The limit of $Log(-1 + \frac{i}{n}) + Log(-1 - \frac{i}{n})$ as $n \rightarrow \infty$ is $0$, given that $Log(z)$ is the main ...
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### Can a function have a branch cut along the real and also imaginary axis?

Is it possible for a complex function to have a branch cut along the real axis and also the imaginary axis, that cross over like a + sign?
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### Traveling Around a Complex Number

In functions with complex arguments it's been said that to determine whether a point is a branch point or not we have to travel around that point and see if we get the same answer. How do I travel ...
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### How to evaluate this contour integral with branches?

I want to evaluate this integral below using contour integral. $$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}(1+x^2)}$$ I know this can be done transforming x to sin or cos, but I want to solve this by ...
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### Integral with two branch cuts

I need help calculating the following integral $$\int_0^\infty\!\mathrm{d}u\frac{u^{-s}(u+i\omega_n)^s}{u-w+i\omega_n+i\omega_m}$$ where $0<s<1$. What I have thought so far is to do contour ...
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### Branch cut of $\sqrt{z^2-1}$.

I was reading something that defined the function $f(z)=\sqrt{z^2-1}$ on $\mathbb{C}\setminus [-1,1]$ where the branch cut is such that the argument of $z$ and $\sqrt{z^2-1}$ are in the same quadrant. ...
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### Branch Cut/s for $f(z)=\operatorname{Log}(\frac{1}{z(z-1)})$

I've been trying to arrive at a branch cut that works for this function. Firstly, I find the branch points which are, if I'm not mistaken, $z_1=0$ and $z_2=1$. I then tried encircling them to see ...
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### Integrating a function with two branch point ${\int_{1}^\infty} \frac{dx}{{x(x^2-1)}^{1/2}}$

I was studying a book about residue intigrating then I get into trouble solving this : ${\int_{1}^\infty} \frac{dx}{{x(x^2-1)}^{1/2}}$ I know that I have to choose a contour like this: but I don't ...
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### Branch of incomplete gamma function

The incomplete gamma function is defined as follows: $$\Gamma(s,z)=\int_z^{\infty}t^s e^{-t}\frac{dt}{t}, \qquad Re(s)>0, \quad z\in \mathbb C$$ Since $t^s$ is a multi-valued function, this ...
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### Integral involving a Multi-Valued function

I don't really have any idea about how does one calculate integrals involving Multi-valued functions, all I know about Multi - valued functions, are the closely associated concepts of Branch Points ...
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### How should I chose the branch of a multivalued intagrand?

I need to calculate the integral of $$\oint_C \frac{log(z^2-1)}{z^2+1}$$ Where $C:2x^2+3y^2=1$ The singularity of the denominator lies outside the contour. Consequently, I thought Cauchy ...
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### How to calculate the contour integration $\int_{2}^{\infty} \dfrac{1}{x \sqrt{x^{2} - 4}}dx$?

How could I calculate the integration $\int_{2}^{\infty} \dfrac{1}{x \sqrt{x^{2} - 4}}dx$ using the following contour? I tried to use $f(z) = \int_{2}^{\infty} \dfrac{\log(z)}{z \sqrt{z^{2} - 4}}dz$ ...
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### Why $\log(1/z) = -\log(z)$ is true for $- \pi < \arg(z) < \pi$ and it's false for $0 < \arg(z) < 2 \pi$?

I don't underestand why this argument is false: We take $$z = re^{i\theta}.$$ Then $$\log(z^{-1}) = \log(r^{-1}e^{-i\theta}) = -\log(r) - i\theta = -\log(re^{i\theta}) = -\log(z).$$ And why it ...
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### Heterodox asymptotics for the square root at the origin

(TLDR: Need an asymptotic expansion for the square root which works at the origin, specifically. The classical Taylor series won't do: i need it to be exact at zero, outside of which it can have ...