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2

Sometimes referred to as Clairaut’s Theorem, it is a standard fact from calculus that if $f\in C^2$, then $f_{xy} = f_{yx}$.


2

Why does $C^2$ "obviously" imply harmonic? There are plenty of twice differentiable functions that are not harmonic. Eg: $f(x,y)=x^3$. The argument they are using here is called Clairaut's theorem which says that if a function is twice differentiable, then the second order mixed partial derivatives are equal. Which is exactly the statement that $A$ in your ...


0

The answer is no, Suppose on the contrary there exists such $C$ would satisfies the inequality. Since the Laplace transform of $f$ $$F(s)= L\left( {f\left( t \right)} \right) = \int_0^\infty {f\left( t \right)e^{ - pt} dt},$$ where, $ p = s + id $, $\Re (p)=s>0$, exists. Then $f$ is of exponential order i.e., there exists $K,a>0$ such that ...


2

Alternatively: $f$ is holomorphic if and only if $f$ is complex-differentiable everywhere in its domain, if and only if the limit $$\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$$ exists for all $z_0$ in the domain. Now, applying a simple property of limits, note that the limit $$\lim_{z \to z_0} \frac{k \cdot f(z) - k \cdot f(z_0)}{z - z_0} = k \cdot ...


0

Yes. Recall the Cauchy-Riemann equation: A function $$f:\Bbb C\to\Bbb C,\quad f(x+iy)=u(x,y)+i\,v(x,y)$$ is holomorphic if and only if $u$ and $v$ are continuously differentiable and satisfy $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} .\tag{1}$$ Now multiplying ...


0

As noted in the comments there are some mistakes in the question, and I don't understand what the Fourier series have to do with the problem. Anyway, we have: $$ -\pi<x<0 \Rightarrow \sin x<0 \Rightarrow -i\sin x =i |\sin x| \Rightarrow \mbox{arg}(i |\sin x|)=\frac{\pi}{2} $$ $$ x=0 \Rightarrow \sin x=0 \Rightarrow -i\sin x =0 \Rightarrow ...


0

This is true $\text{Arg}(-i\sin(x))=\frac{\pi}{2}\cdot \text{sgn}(x)$, $x\in(-\pi,\pi)$ Recall that the principal argument $\text{Arg}(z)$ lies in the domain $-\pi<\text{Arg}(z)\le \pi$. Setting $y=\sin x$, then $y \in [-1,1]$. Note that $z=-iy$ is a pure imaginary complex number, and the sign of the imaginary part is $\pm$ (liying on the $y$-axis) ...


2

Hints: 1. Review the proof of the Schwarz Lemma; there you have $g(0) = 0,$ and you consider $g(z)/z.$ Here you look at $g(z)/z^2.$ Because of 1., the maximum modulus theorem finishes this off in one line.


1

That function certainly can't solve that equation, since the parameters $\hbar$ and $m$ don't occur in it. It's quite usual to set various constants or combinations thereof to $1$ in physics to simplify expressions. You can either do this simply by choosing units such that the constant expression takes the value $1$, or you can explicitly rescale time and ...


0

As the solution says, let $a=|a|e^{i\theta_0}$, for some $\theta_0\in[0,2\pi)$. The first quadrant $A_1:=\{z=x+iy:x>0\,\wedge\,y>0\}$ can be described equivalently as $A_1=\{re^{i\phi}\in\mathbb{C}:r\in\mathbb{R}_{>0},\wedge\,\phi\in(0,\frac{\pi}{2})\}$. Then \begin{equation} \begin{split} \{az+b:z\in ...


0

Intuitively, you can draw a diagram and use some elementary geometry to determine the angle. If you want to prove your conjecture rigorously, notice that $\text{Arg}$ is multi-valued. To make it into a single valued function so that we can talk about periodicity, we need to consider it as a single-valued function from its Riemann surface $\mathscr S$ to ...


1

Hints/Ideas: $xe^{ix} = x\cos x + ix\sin x$, and you integrate on an interval symmetric around $0$. The function $x\mapsto \frac{x\cos x}{1+\cos^2 x}$ is odd, and the function $x\mapsto \frac{x\sin x}{1+\cos^2 x}$ is even. $$ \int_{-\pi}^\pi f(x) dx = i\int_{-\pi}^\pi dx\frac{x\sin x}{1+\cos^2 x} = 2i\int_{0}^\pi dx\frac{x\sin x}{1+\cos^2 x} $$ Now, ...


5

This can be written as $$\int_{-\pi}^{\pi}\frac{x(\cos x+i\sin x)dx}{1+\cos^2 x}$$ $$=\int_{-\pi}^{\pi}\frac{x\cos xdx}{1+\cos^2 x}+i\int_{-\pi}^{\pi}\frac{x\sin xdx}{1+\cos^2 x}$$ The first integral evaluates to $0$ (Odd function) Whereas the second can be written as $$2i\int_{0}^{\pi}\frac{x\sin xdx}{1+\cos^2 x} \space\space\text{(even function)}$$ ...


2

It is indeed simply connected. In fact it is even a star domain : if you define $z_0=1-i$, then for any $z\in \mathbb C \setminus \{z |\Re z \le 0 \text{ and }\Im z=-1\}$ the segment $\{z_0+t(z-z_0)|t\in [0,1]\}$ lies in $\mathbb C \setminus \{z |\Re z \le 0 \text{ and }\Im z=-1\}$. This implies that it is also contractible, and thus simply connected.


0

You can proceed similarly as in the proof of the "ordinary" argument principle. Write $$ f(z) = \frac{\prod_{j=1}^n (z-z_j)}{\prod_{k=1}^m (z-p_k)} h(z) $$ where $h$ is holomorphic and $\ne 0$ in $G$. Now take the logarithmic derivative $$ \frac{f'(z)}{f(z)} = \sum_{j=1}^n \frac{1}{z-z_j} - \sum_{k=1}^m \frac{1}{z-p_k} + \frac{h'(z)}{h(z)} $$ and multiply ...


1

Let $z=-1$ then Arg$(\frac{1}{z})$=Arg$(\frac{1}{-1})$=Arg$(-1)$=$\pi$. On the other hand Arg$(-1)$=$\pi$ so that $$Arg(\frac{1}{-1})=\pi \ne -(Arg(-1))=-(\pi)$$


3

Suppose that $f(z)=u(z)+iv(z)$. Consider the functions $$g(z) = e^{f(z)},~ ~~~ h(z) = e^{-f(z)}.$$ Note that, since $u|_{|z|=1} \equiv 0$, it follows that $$|g(z)|=|h(z)| = 1$$ on $|z|=1.$ Thus, by Maximum Modulus Principle, we have, $|g(z)| \leq 1$ and $|h(z)| \leq 1$ on $|z| \leq 1$. However, $$|g(z)|=|e^{u(z)}|$$ and $$|h(z)| = |e^{-u(z)}|$$ which ...


0

In the Wolfram Alpha definition, it is required that the domain be path-connected to begin with. I guess it depends on which definition you're using; if only your condition (on loop shrinking) were imposed, then $3$ would be simply connected.


1

Let $g\colon \mathbf C\setminus \{-i\} \rightarrow \mathbf C\setminus\{1\}$ be a Moebius transformation mapping the real line $\mathbf R$ onto the unit circle minus the point $1$. Let $h=i\cdot f\circ g$. Then $h$ takes real values on $\mathbf R$. It follows that $h(\bar z)=\overline{h(z)}$ for all $z\in\mathbf C\setminus\{\pm i\}$. Since $h$ can be extended ...


2

The harmonic function $u(z)=\textrm{Re}f(z)$ satisfies $u=0$ on $|z|=1$, so $u\equiv 0$ by the uniqueness of solutions to the Dirichlet problem $\Delta u=0$ on $D$, $u=0$ on $\partial D$.


2

The problem you address is already manifest for the simple "function" $z\mapsto \log z$. We want $e^{\log z}=z$ in a neighborhood of $z=1$, say. We know that the equation $e^w=1$ has the infinitely many solutions $2k\pi i$, and we then for sake of simplicity choose $w=0$ as our "target" solution. After some computation with real and imaginary parts, etc., we ...


3

Of course, you can't define $\log 0$ since there's no complex number $z$ with $e^z=0$. But as long as $f(z)\neq 0$, there's no problem defining $\log f(z)$. But it may not be the case that $\log f$ is continuous.


2

We notice that $T_{a,b,c,d}$ maps $\Bbb R\cup\{\infty\}$ to itself. Hence (a) and (b) are possible options. As $T_{a,b,c,d}(i)=\frac{ai+b}{ci+d}=\frac{(ai+b)(-ci+d)}{(ci+d)(-ci+d)}=\frac{(bd+ac)+(\color{red}{ad-bc})i}{c^2+d^2}$ has positive imaginary part, we see that (a) is certainly valid and (b) false. On the other hand, (c) and (d) may or may not be ...


1

$$f(z)=\frac {(z-z_1)\dots(z-z_n)}{(z-p_1)\dots (z-p_m)}g(z)$$ And $g$ has no zero and no pole since $z_i$ and $p_i$ are all the zeros and poles of $f$. Let's start, we let $a_1(z)=\frac {(z-z_2)\dots(z-z_n)}{(z-p_1)\dots (z-p_m)}g(z)$ : $$f(z)=(z-z_1)a_1(z) \Rightarrow \frac{f'(z)}{f(z)}=\frac {1}{z-z_1}+\frac{a_1'(z)}{a_1(z)}$$ Let $a_2(z)=\frac ...


1

Write $z-2 = z+1-3$ so we can break $$\frac{z-2}{z+1} = -\frac{3}{z+1}+1 = -3(z+1)^{-1} + 1$$and you're done.


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


0

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


2

From the notation and complex analysis tag I would guess we should assume $z\in \mathbb C,|z|\le r.$ For $k>0,$ we can then say $|1+z^k|\ge 1-|z^k| = 1-|z|^k \ge 1- |z| \ge 1-r.$ Thus $$\left |\frac{z^k}{1+z^k}\right | = \frac{|z^k|}{|1+z^k|} \le \frac{r^k}{1-r}.$$ Since $\sum_{k=1}^{\infty} r^k/(1-r) < 1/(1-r)^2,$ we have uniform convergence in ...


3

For $z\in [0,r]$, with $0\le r<1$, we have $$\left|\frac{z^k}{1+z^k}\right|\le r^k$$ and $\sum_{k=0}^\infty r^k=\frac{1}{1-r}<\infty$


2

When $\beta_1+\beta_2<1$ then the upper half plane is mapped onto an infinite truncated sector (a sector with the tip cut off). The apex angle of the sector is $(1-\beta_1-\beta_2)\pi$. When $\beta_1+\beta_2=1$ then the upper half plane is mapped onto half a strip with a slanted edge (an infinite strip cut in two at an angle).


1

We have that $\;z=0\;$ is clearly a triple pole of $$f(z):=\frac{e^{i2z}}{z^3} ,\;\;\text{and in this case it is probable easier to use power series for the residue:}$$ $$\frac{e^{2iz}}{z^3}=\frac1{z^3}\left(1+2iz-\frac{4z^2}{2!}-\ldots\right)=\frac1{z^3}+\frac{2i}{z^2}-\frac2z-\ldots\implies\text{Res}_{z=0}(f)=-2$$ so taking the usual contour with a ...


0

It is not difficult to show (through the Laplace transform, for instance) that $$ \int_{-\infty}^{+\infty}\frac{\sin(2z)-2z}{z^3}\,dz = -2\pi \tag{1}$$ so the principal value of your integral is given by $-2\pi$ plus twice the principal value of $$ \int_{-\infty}^{+\infty}\frac{dz}{z^2} \tag{2} $$ that is just $+\infty$, since the integrand function is even ...


0

$$ \frac{\sin (2x)}{x^3} = \left( \vphantom{\frac2{x^2}} \right.\underbrace{\frac{\sin(2x)}{2x}}_{\begin{smallmatrix} \text{This} \\ \text{approaches} \\[2pt] \text{$1$ as $x\to0$} \\[2pt] {} \end{smallmatrix}} \cdot \left. \frac 2 {x^2} \right) $$ For $x$ close enough to $0$ you have $\dfrac{\sin(2x)}{2x}> 0.9$ and then $$ \int_{-a}^a ...


1

Since $$ \sin z=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{2k+1} $$ we have, for $z\ne0$, $$ \frac{\sin z}{z}=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{2k} $$ Since this power series has radius of convergence infinite, it defines an entire function, which is an analytic continuation of $(\sin z)/z$ at $0$, where it has the value $1$. So, as written, $(\sin z)/z$ ...


1

Once the integral is in the form $$ I_p = \frac{1}{p}\int_{0}^{+\infty}\frac{\sin x}{x^{2-1/p}}\,dx \tag{1}$$ we may use the Laplace transform, giving $$ \mathcal{L}(\sin x)=\frac{1}{1+s^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{x^{2-1/p}}\right)=\frac{1}{\Gamma\left(2-\frac{1}{p}\right)}\,s^{1-1/p}\tag{2}$$ and $$ I_p = ...


1

This interesting question reminds us that Residue theorem is correct only when $$ \lim_{z\to a+i\infty}f(z) = 0. $$


6

As defined, no, as it isn't well-defined at $z=0$. It does, however, have an analytic continuation to the entire plane. What you can observe is that $\sin(z)$ has a zero of order $1$ at $z=0$, and therefore the singularity of $\sin(z)/z$ has order 0 at $z=0$, meaning it is removable and an analytic continuation at $z=0$ exists (and is given by the limit). ...


4

First, the obvious substitution is: $$t=x^p$$ Not thinking about the conditions for convergence for now, we have the integral: $$I=\int_0^\infty \frac{\sin(x^p)}{x^p}\mathrm{d}x=\frac{1}{p}\int_0^\infty \frac{\sin(t)}{t^{2-1/p} }\mathrm{d}t$$ Now let's put $2-1/p=q$. The trick is to turn this into a double integral. Notice that: $$\int_0^{\infty} ...


0

We can look at the power series for the difference: $$ \begin{align} u-v &=\frac{\log\left(z+\frac12\right)}z-\frac{\log(z)}z\\ &=\frac{\log\left(1+\frac1{2z}\right)}z\\ &=\frac1{2z^2}-\frac1{8z^3}+\frac1{24z^4}-\frac1{64z^5}+\cdots+\frac{(-1)^{n+1}}{n2^nz^{n+1}}+\cdots \end{align} $$ This series converges for $\left|z\right|\gt\frac12$ and has ...


0

First, $u$ and $v$ are not defined at $-\frac 12$ and $0$ respectively so you can't pretend they are functions $\Bbb C \to \Bbb C$. From the properties of the principal branch of $\log$, $u$ and $v$ are holomorphic on $U = \Bbb C \setminus \Bbb R_{\le 0}$. Their difference $u-v$ is also holomorphic on $U$. Next, $\log(1+\frac 1{2z})$ is holomorphic on $V ...


1

Consider $f(r e^{i \pi/4})$ for $r \to 0$ and conclude that $f$ is not even continuous at $z=0$.


1

The difference between $s_n$ and $s_{n+1}$ has an absolute value of $n+1$ when $z$ has an absolute value of $1$. So no convergence. Actually, the partial sums must be unbounded because one of $s_n$ and $s_{n+1}$ must have an absolute value greater than $\frac{n}{2}$, which is stronger than just not converging.


5

For $z \in S:=\{z \in \mathbb{C}:|z|=1\}$ it also holds $z^n \in S$ for all $n \in \mathbb{N}$ (since in this case $|z^n|=|z|^n=1^n=1$). Thus, the sequence $(a_n)_{n \in \mathbb{N}}$ with $a_n=nz^n$ does not converge to zero which is necessary for the corresponding sum $\sum_{n \in \mathbb{N}}{a_n}$ to be convergent. Hence this sum does not converge.


3

Assuming you start the sum at $n=1$ (otherwise your first term involves division by zero), this is one of the polylogarithms. $$ \mathrm{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k}{k^s} $$ So in particular, you have asked about $\mathrm{Li}_2(z)$, also known as the dilogarithm. See also the Dilogarithm page at MathWorld and Dilogarithms at the NIST Digital ...


1

For continuity away from the origin, consider points $z = re^{i \theta}$ and $z_0 = r_0e^{i \theta_0}.$ Suppose $|z - z_0| < \delta.$ In this case, $z$ is in a disk of radius $\delta$ with center $z_0$. Using some geometry we find $$|\theta - \theta_0| \leqslant \arcsin \frac{\delta}{r_0},$$ and $$| \sin \theta - \sin \theta_0| = 2 \left|\sin ...


0

Using Euler's identity you can write a direct formula for $f$, which makes continuity obvious where polar coordinates are unique. Geometrically speaking, you function works like this: draw a line $L_z$ through your given point $z$ and the origin, and then project the intersection of $L_z$ and the unit circle onto the y-axis. In particular, $f$ is ...


0

For discontinuity at $0$, approach $0$ along the lines $\theta = \pi/3$ and $\theta = \pi/4$. For continuity, let $\varepsilon > 0$ and $x = r_0(\cos \theta_0 + \mathrm{i} \sin \theta_0)$ be given. Can you find a $\delta$ (depending on $x$ and $\varepsilon$) such that for any $y = r(\cos \theta + \mathrm{i} \sin \theta)\in \Bbb{C}$ with $|x-y| < ...


0

Find automorphisms of $D$ that map $\frac 34\mapsto 0$ and $0\mapsto \frac 34$, respectively. From these and $f$ construct a holomorphic $g\colon D\to D$ with $g(0)=0$ and $g'(0)=???$. Similarly for 4.


0

As a supplement to the first answer given: When $|x|>1$ , let $|x|=1+y$ with $y>0.$ Then for $n\geq 2,$ use the binomial theorem: $|x|^n=(1+y)^n=1+ n y +n(n-1)y^2/2+...>n(n-1)y^2/2.$ So $|x^n/n|>(n-1)y/2,$ which goes to $\infty$ as $n\to \infty.$


0

For the principal branch of $\log z $ : Let $z+1+y.$ Then for $0<|y|<1$ we have $$(z-1)^{-3}\log z=y^{-3}\log (1+y)=y^{-3}\sum_{n=1}^{\infty}y^n/n= y^{-2}+y^{-1}/2+\sum_{n=0}^{\infty}y^n/(n+3).$$ So,as you said, the order of the pole at $z=1$ is $2.$



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