Questions tagged [branch-cuts]

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

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Understanding where branch cuts live

I understand how we define $e^z = e^x e^{iy}$ and use this to define the multi-valued function $\log(z) = \ln(r) + i(\theta + 2\pi n)$. I think I also understand how we may take any branch by setting ...
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Branch points: Find value of $f(z)=\sqrt{\pi^2 + \log^2(z)}$ at $f(i)$

I have the function (and the problem) mentioned in the title and I am given the initial condition of $f(1)=\pi$. The cut is a rather strange one: It is the portion of the unit circle that connects $-1$...
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Understanding branch cut of the following function $(z^3+1)^{1/3}$

I'm trying to understand branch cuts in complex analysis. Let's say I have the following function $$f(z)=(z^3+1)^{1/3}.$$ I see that the branch points in this case are the solutions to the equation $z^...
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Find $\operatorname{Log}(1-i)$ where $\operatorname{Log}z$ is the principal value.

My attempt starts with $\log z=\operatorname{Log}|z| + i\operatorname{Arg}z$. I start with the $\operatorname{Arg}$. We know $$|1-i|=\sqrt{1^2+1^2}=\sqrt{2}\implies\sin\theta=-\frac{\sqrt{2}}{2}\...
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Integral Representation for the Legendre Function of the First Kind

In the text Special Functions and their Applications by N.N. Lebedev (pp.172-173), a derivation is presented for a certain integral representation of the Legendre function of the first kind $P_{\nu }(\...
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Evaluating $\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\,\mathrm{d}y$; how do you avoid using a complex substitution?

$\newcommand{\d}{\mathrm{d}}$The given exam question - I provide the beginning for context: Let: $$I=\int\frac{1}{(b^2-y^2)\sqrt{c^2-y^2}}\d y$$Where $b,c\gt0$, and employ the substitution $y=\frac{...
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Evaluate $\int_{-\infty}^{\infty}\frac{\sinh\left(y\sqrt{\alpha^2-\omega^2}\right)}{\sinh\left(H\sqrt{\alpha^2-\omega^2}\right)}e^{i\alpha x}d\alpha$

I need to evaluate the Fourier inverse integral $\displaystyle \int_{-\infty}^{\infty}\frac{\sinh\left(y\sqrt{\alpha^2-\omega^2}\right)}{\sinh\left(H\sqrt{\alpha^2-\omega^2}\right)}e^{i\alpha x}d\...
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Removable branch point discontinuity?

While reading this paper https://arxiv.org/pdf/1105.3426.pdf on Fourier Extensions, I came across something I can't get my head around. Here is the thing: Suppose you have an entire function $f : \...
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analytically continuing a function into a two-leaf function

My function is defined as $$f(z) = \int_0^\infty \frac{\rho(x) dx}{x-z } , $$ where $\rho(x)$ is a real function. It is easy to see that $f$ has a branch cut along the positive real axis. Now let us ...
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Prove or disprove: analytical function

Prove or disprove: There exist an analytical function $f(z)$ in a deleted neighbourhood of $z=0$ such that $f^2(z) = z$. I disproved it with the following argument: the function $f(z)$ is given by ...
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Integration along a contour containing a branch cut

Consider the following integral $I = \int_{-\sqrt{b^{2} + a^{2}}}^{\sqrt{b^{2}+a^{2}}} (a^{2} - z^{2})^{\lambda}e^{- i \omega z}\mathrm{d}z$ In this integral, $a$, $b$ and $\omega$ are real numbers. ...
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Determine a branch of $f(z)=\log(2iz−z^2)$

Determine a branch of $f(z)=\log(2iz-z^2)$ that is analytic at $z=1$. Then find $f(1)$ and $f'(1)$. First we note that $g(z)=2iz-z^2$ and recall $\mathcal{L}_\tau:=\log|z|+i\arg_\tau z$. So a branch ...
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$z/\sqrt{-z^2}=-i$ when $\operatorname{arg}(z)\leq 0$?

Looking at asymptotic expansions for the imaginary error function I find the following for $z\in\Bbb C\setminus\{0\}$: $$ \tag{1} \frac{z}{\sqrt{-z^2}}= \begin{cases} -i, &\operatorname{arg}(z)\...
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Which branch of the square root allows $(z^2)^{1/2} = z$

I am aware that $(z^2)^{1/2} = z$ does not hold for every branch of the function $f(z) = z^{1/2}$. For example if we do not take a branch and consider $z=-1$, we get $$f((-1)^2) = \exp(\frac{1}{2}Log((...
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How to find a branch which is analytic on the exterior of the unit circle for $\sqrt(z^2 +1)$, $|z| > 1$

I know we can rewrite $\sqrt{z^2 +1}$ as $z^2 (1+z^{-2})$ and use this by looking at the principal branch of the function $\exp{\left(\frac{1}{2} \log(1+z^{-2})\right)}$. However I am struggling to ...
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Difficulty evaluating $\sum_{n=1}^\infty\frac{1}{n^2}\sqrt{\alpha n-1}$ for a real $\alpha\gt1$

$\newcommand{\d}{\,\mathrm{d}}$I've done my best with this series, but I've never actually seen a sum of square roots before! The following question was given to me by a friend - apparently some ...
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How to find branch cuts and branch points of $\sqrt(z^a +1)$

I have tried to find out the branch cuts and branch points of the function $\sqrt(z^a +1)$, where $0<a<2$. The function can be written as $e^{\frac{1}{2}(\log(z^a+1))}$ and from here I have ...
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Finding branch cuts

Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$. I would like to find branch cuts so that the complex function $$f(z)=\sqrt{z(z+1)(z-\omega)}$$ can be defined continuously off the branch cuts. I ...
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Confusion about the change of variable $z \to \frac{1}{z}$ for a multivalued function

I'm currently struggling with something that came up in my studies. I'm trying to integrate a multivalued function like the square root on a given path, specifically a function with two branch points, ...
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How to improve the solution on this problem of branch cut

I was given the following problem. Consider the function $w=(z^2-4)^{1/2}$. Insert branch cuts in the z-plane such that $\pi/4>\arg{w}>-3\pi/4$. Find the equations of the branch cuts in the $z$-...
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Find branch cut of $w=(z^2-4)^{1/2}$ such that $\pi/4>\arg{w}>-3\pi/4$

I was given the following problem. Consider the function $(z^2-4)^{1/2}$. Insert branch cuts in the $z$-plane such that $\pi/4>\arg{w}>-3\pi/4$. Find the equations of the branch-cuts in the $z$-...
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Branch cut contour integral - contour understanding [closed]

Im trying to understand the contour when there are two poles on the real axis without poles on Im axis, because their residue cancle each other. contour $$I=\int_0^\infty\frac{dx}{(x+1)^2(x+2)}$$ For ...
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Deducing $\int_0^{\pi}\log \sin x dx =-\pi\log 2$ from $\int_0^{\pi}\log (-2ie^{ix}\sin x) dx = 0$

On Ahlfors Complex Analysis, chpter 5.3, the author explains how to evaluate $\int_0^{\pi}\log \sin x dx =-\pi\log 2$. He does so by using Cauchy's formula to deduce that $$\int_0^{\pi}\log (-2ie^{ix}\...
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Square root of continuous function on simply connected domain

It is well that every holomorphic function on a simply connected domain which is everywhere non-zero has a holomorphic square root. But is it true that every continuous function on a simply connected ...
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Square root branch selection of a polynomial

The complex function $$ f(z) = \sqrt{(z-(a+b)) (z-(b-a))(z-(a-b)) (z+(a+b))}, \quad a,b\in \mathbb{R}, \quad a,b >0, \quad a > b $$ has 2 branch cuts located between $z \in [-a - b,b - a ]$ and $...
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Branch points of a complicated function

I would like to study the branch points of the function $q(z)=\frac{z^{\frac{1}{3}}+\frac{1}{2}z^{\frac{-1}{3}}}{[1+z^{-\frac{2}{3}}]^{\frac{1}{2}}}$ . I plotted its real part with Mathematica and it ...
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Contour integral of (-z)^1/3 over unit circle depending on branch cut

I do not understand how to integrate $(-z)^{\frac{1}{3}}$ over unit circle if branch cut is (for example) along the positive x axis. How to choose boundaries for $\theta$ if we change variable $z=e^{i\...
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Show that $ \int_{-1}^{1} z^{i} dz= \frac{1 + e^{- \pi}}{2} (1 -i)$

The problem says: Show that $ \int_{-1}^{1} z^{i} dz = \frac{1 + e^{- \pi}}{2} (1 -i)$ with $z^{i}$ the main branch $z^{i} = exp(i log z)$ $(|z|>0, -\pi < Arg z < \pi)$ and where the path of ...
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Branch cuts: what makes the use of the Log-Gamma function favourable over the use of the logarithm in contour integration?

I cite this paper by Iaroslav Blagouchine. As mentioned in the comments under this post, Blagouchine erroneously claims that the Log-Gamma function has no branch cuts (fortunately none of his contours ...
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Legendre relations for elliptic integrals with null imaginary part

I am trying to compute the transformation of $K(k)$ as $k \to 1/k$, where $K(k)$ is just the Legendre integral $$K(k) := F\left(\frac{\pi}{2}, k\right).$$ The Digital Library of Mathematical Functions ...
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How to prove there is a branch point in complex function?

I'm working with the function $$f(z)=(-z)^{1/3}$$ And I've been asked to prove that the function has a branch point at x=0 by calculating the difference: $$\frac{f(x+i \theta)-f(x-i\theta)}{r^{1/3}}$$ ...
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calculate the integral $\int_{\gamma}^{} \operatorname{Log}^2(z) \,dz$ where $\gamma$ is the given path:

Let $\gamma : [0,1] \to \mathbb{C}$ be a $C^1$ curve with $\gamma(0) = i, \gamma(1) = -i$ such that $\gamma$ does not intersect $(-\infty,0] $. Calculate the integral $$ \int_{\gamma}^{} \...
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How do we know that this function is multivalued?

So I have an integral $$ \int_{0}^{2\pi} \frac{1}{2}\left(e^{e^{ix}} + e^{e^{-ix}}\right) \text{ d}x$$ I am told that I am able to substitute $z=e^{ix}$ into this and convert it into a contour ...
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Log branch cut integration

I encountered with this problem: (It's actually calculating the Uehling potential from Fourier transformation of propagator, but there is no technical physics in this problem and I just stuck at ...
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3 votes
2 answers
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Conjecture: $\lim\limits_{x\to\infty}\operatorname{Re}\text W_x(x)\mathop=\limits^?-\ln(2\pi)$

The inspiration for the question is Closed form of $$\frac{d}{dk}\text W_k(z)$$ Derivative of W-Lambert function with respect to its branch cuts experiment. I also like making functions central to ...
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How to determine if $z+\sqrt{z-1}$ and $\frac{\sin{\sqrt{z}}}{\sqrt{z}}$ is a multi-valued function?

What I know: I understand that for a complex function to be a multi-valued funciton they must have some branch points. I understand branch points as this definition (translated into English from my ...
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5 votes
1 answer
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Closed form of $\frac{d}{dk}\text W_k(z)$. Derivative of W-Lambert function with respect to its branch cuts experiment.

For a change, I will ask a derivative question. Please consider the Generalized W-Lambert/Product Logarithm function $\text W_k(z)$. Let’s see what happens when we try to differentiate with respect to ...
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REVISED Proving $-1,1$ are branch points of $\sqrt{z^2-1}$

On some lecture notes that I am working on there is an exercise to prove that $-1,1$ are branch points of the multi-function $\sqrt{z^2-1}$. I know that the branch $f=\sqrt{rs}e^{i\frac{1}{2}(\theta_1+...
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Understanding contour integral with branch cuts

I was trying to understand contour integral which have or those involving branch cuts. Like $$I=\int_0^\infty\frac{dx}{x^3+1}$$ because the integrand is not even we cannot extend the integration to ...
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3 votes
1 answer
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How can we mechanically rotate a branch-cut of a complex function?

I'm looking for an algorithmic and mechanical way to define the branch cuts of an arbitrary (or not-so arbitrary, depending on how doable this is) complex function. Given a function $f(z)$ that has a ...
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Branch of Complex Logarithm

I am reading a mathematical physics textbook and came across this line that I didn't Understand: "$$\phi= \frac{1}{2i}\ln\frac{\Gamma(\frac{1}{2}+ia)}{\Gamma(\frac{1}{2}-ia)}$$ with the branch of ...
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3 votes
2 answers
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Determining $A = \{n\in\Bbb N : n\ge2$ and there exists a branch of $\sqrt[n]{f}$ in $D \}$

Consider $D = \Bbb C\setminus(\{e^{i\theta} : -\frac\pi2\le\theta\le\frac\pi2\}\cup(-\infty,0])$ and $f(z) = z(z^2+1)$. We know that $f$ can't be zero for any $z\in D$. If we consider now $A = \{n\in\...
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Square root integral with complex analysis

I'm currently trying to solve an integral of this type $$ I=\int_{x_0}^\infty dx \frac{1}{x^2 \sqrt{1+\frac{a}{x}+\frac{b}{x^2}}}, $$ where $x$ and $x_0$ are positive but $a,b$ can be both positive ...
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Understanding the square roots of polynomials: $f(z)=[z(z^2-1)]^{1/2}=\sqrt{z(z+1)(z-1)}$

The book I'm studying asks: Writing $z=r\exp(i\Theta),\;z-1=r_1\exp(i\Theta_1),\;z+1=r_2\exp(i\Theta_2), \;-\pi<\Theta<\pi$. Define an extension $F$ of the multi-valued function $f(z)=[z(z^2-1)...
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Evaluating $\left(\frac{1}{u}\right)^{i\eta/2} u^{i\eta/2}$ for real $u<0$ and $\eta>0$, where $i$ is the imaginary unit [closed]

Let $u$ and $\eta$ be real numbers such that $u<0$ and $\eta>0$. What is the value of this expression? $$\left(\frac{1}{u}\right)^{i\eta/2} u^{i\eta/2}$$ This is probably trivial, but the ...
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Question on convergence of the infinite product.

I am going to present the proof of Weierstrass Factorization Theorem in a student seminar in the next week. For that purpose I am following Conway's book on complex analysis for the essential details ...
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1 vote
1 answer
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An integral solves by complex analysis

I try to compute $$ I:=\int_{-\infty}^{\infty}\frac{\arctan^2x}{x^2}\text{d}x $$ by complex analysis. This is my attempt: Consider $$ f(z)=\frac{\arctan^2z}{z^2},\frac{\pi }{2}\ge\text{arg}(z)>-\...
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6 votes
1 answer
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Pole lying in the middle of a logarithm's branch cut in $\int_{-\infty}^{\infty} dz \frac{\ln(a+z^2)}{1+z^2}$

Consider the integral, for $a\geq0$, $$I = \int_{-\infty}^{\infty} dz \frac{\ln(a+z^2)}{1+z^2} = 2 \pi \ln(1+\sqrt{a})$$ This is straightforward to do with real methods via differentiation with ...
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2 votes
2 answers
114 views

Evaluating $\int_{-\infty}^{\infty} \frac{z\ln(1+e^z) }{(1+z^2)^2}dz = \frac{\pi}{4}$ - how to handle branch cuts?

My aim is to evaluate the integral $$I = \int_{-\infty}^{\infty} \frac{z\ln(1+e^z) }{(1+z^2)^2}dz = \frac{\pi}{4}$$ directly with contour integration. How can I do this? In particular, how should I ...
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1 vote
1 answer
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Branch cuts imaginary part

This is a very simple gap in my intuition about branch cuts. I have heard informally a statement of the type: if a complex function suddenly acquires an imaginary part at the point $x_0$ in the real ...
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