# Tag Info

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{split} \{az+b:z\in ...

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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.

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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


0


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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