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1

Observe that $$\begin{align} \frac{1}{2\pi} \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty} \textstyle f\left(x'\right)\,\exp\left({-\text{i}kx'}\right)\, \text{d}x'\, \exp({\text{i}kx})\, \text{d}k &=\lim_{k_0\rightarrow\infty}\, \frac{1}{2\pi}\, \int\limits^{+\infty}_{-\infty} ...


1

Your reasoning is correct. I think it's better to focus on the polynomial $p$ itself rather than its real part. You are considering the preimage of the imaginary axis under $p$. In what follows, $p$ can be any holomorphic function, not necessarily a polynomial. Two facts about holomorphic functions: If $p'(a)\ne 0$, then $a$ has a neighborhood $U$ such ...


0

$I = \int_{-\infty}^\infty \text dz\ \frac{1}{z-i\varepsilon}e^{- a z^2+i b z} = \int_{-\infty}^\infty \text dz\ \frac{1}{z-i\varepsilon}e^{- a z^2+i b (z - i\varepsilon +i\varepsilon)} = e^{-b\varepsilon}\int_{-\infty}^\infty \text dz\ \frac{1}{z-i\varepsilon}e^{- a z^2+i b (z - i\varepsilon)}$, that is $$e^{b\varepsilon}I = \int_{-\infty}^\infty \text dz\ ...


0

Here is a solution avoiding the residu theorem or any change of variables over the complex plane. Let $g(\xi)$ be the left hand side :$$g(\xi)=\int_{\mathbb R} e^{-\pi x^2}e^{-2i\pi x\xi}dx.$$ One can easily check that $f:\mathbb R\times \mathbb C\rightarrow \mathbb C$ given by $f:(x,\xi)\mapsto e^{-\pi x^2}e^{-2i\pi x\xi}$ satisfies : $f$ is measruable, ...


4

The ring $R= \mathcal{H}(S^1)$ is both noetherian and factorial. a) Noetherianity follows from Theorème (I,9) page 123 of this Inventiones paper by J.Frisch. He proves that given a complex analytic space $X$ and a compact subset $K\subset X$, the ring $\mathcal{H}(K)$ is noetherian as soon as $K$ is real semi-analytic and has a basis of open Stein ...


2

By noticing that $$a x^{2} + b x = \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2} - \frac{b^{2}}{4 a}$$ then \begin{align} I &= \int_{-\infty}^{\infty} e^{-a x^{2} - b x} \, dx \\ &= e^{\frac{b^{2}}{4 a}} \, \int_{-\infty}^{\infty} e^{- \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2}} \, dx \end{align} Making the change $t = \sqrt{a} x + ...


0

Hint: $$-\pi x^2 - 2\pi i x \xi = -\pi(x^2+2ix\xi) = -\pi(x+i\xi)^2+(?)$$ Try to figure out what $(?)$ should be, and you should get something resembling a known integral. You might have to do some substitutions and/or contour deformations.


0

$$\int_{C}\frac{2z^2-z+1}{(2z-1)(z+1)^2}dz$$$$=\frac{1}{2}\int_{C}\frac{2z^2-z+1}{\left(z-\frac{1}{2}\right)(z+1)^2}dz$$ We see that the singularities $z=\frac{1}{2}$ & $z=-1$ are on the real axis inside the curve, $C: r=2\cos \theta$ hence there are the poles of first & second order respectively. Let's find out the residues at these poles as follows ...


1

Look at the original parametrization. Apparently your source allows for negative radius. This is the circle centered at $(1,0)$. It starts at $(2,0)$ for $\theta=0$ then travels back to $r=0$ at $\theta=\pi/2$ then back to $r=-2$ at $\theta=\pi$ which is identified with $(2,0)$ once more.


0

(Too long for a comment.) Hm, there may be a typo in the paper or an unqualified statement. When you have an equation in $x$ of form, $$P_1(x) = P_2(x)\sqrt{P_3(x)}$$ like your $(2)$ above, the straightforward way to get its degree is to square both sides, then equate it to zero, $$\big(P_1(x)\big)^2-\Big(P_2(x)\sqrt{P_3(x)}\Big)^2 = 0\tag3$$ As you ...


0

This is a very interesting question and one that points out how special holomorphic functions are! I don't know if the following is a complete answer, but let me adress some points: 1) The interpretation of "area under the function along a path" might be a little bit problematic in this case because the function takes complex values. However I think this is ...


0

While it is true that you can view the line integral as "the area under the function along a path", that alone is not enough to give an answer as to why closed line integrals over holomorphic functions are proportional to the residue(s) inside the path. You can write a holomorphic function $f(z)$ around $z_0$ as $f(z) = \sum_{n=0}^{\inf}a_n(z-z_0)^n + ...


1

Substituting $z\mapsto z+a$ and using the Binomial Theorem to get the coefficient of $z^{-1}$, we get $$ \begin{align} \frac1{2\pi i}\oint_{|z|=1}\frac{(z-b)^m}{(z-a)^n}\,\mathrm{d}z &=\frac1{2\pi i}\oint_{|z+a|=1}\frac{(z+a-b)^m}{z^n}\,\mathrm{d}z\\[6pt] &=\binom{m}{n-1}(a-b)^{m-n+1} \end{align} $$ Note that if $n\lt1$, $\binom{m}{n-1}=0$ and if ...


1

Notice, $$\frac{1}{2\pi i}\int_{|z| = 1}\frac{(z-b)^m}{(z-a)^n}dz=\frac{1}{2\pi i}\oint \frac{(z-b)^m}{(z-a)^n}dz$$ It is clear that $z=a$ & $z=b$ are two points out of which $z=a$ is lying inside the unit circle $|z|=1$ (given $|a|<1<|b|$) hence $z=a$ is a pole of $n$th order & $f(z)=(z-b)^m$ Now, using cauchy integral formula, as follows ...


3

Use Cauchy's integral formula properly. $$\frac{1}{2\pi i}\int_{|z| = 1}\frac{(z-b)^m}{(z-a)^n}dz$$ $$= \frac{2\pi i}{2\pi i}\frac{1}{(n-1)!}\lim_{z\to a}\frac{d^{n-1}}{dz^{n-1}}(z-b)^m =?$$


1

You were close $$f^{(2)}(1/2)=\frac{2!}{2\pi i}\oint_{|z|=1}\frac{(z-2)^3}{(2z-1)^3}dz=\frac{2!}{2\pi i}\oint_{|z|=1}\frac{\frac18(z-2)^3}{(z-1/2)^3}dz$$ where $f(z)=\frac18(z-2)^3$ and thus $f^{(2)}(1/2)=\frac18 (6)(-3/2)=-\frac98$ Finally, we have $$\bbox[5px,border:2px solid #C0A000]{\oint_{|z|=1}\frac{(z-2)^3}{(2z-1)^3}dz=-\frac98 \pi i}$$


0

Notice, $$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz=\frac{1}{8}\oint\frac{(z-2)^3}{\left(z-\frac{1}{2}\right)^3}dz$$ We see that $z=2$ & $z=\frac{1}{2}$ are two points out of which point $z=\frac{1}{2}$ is a pole of third order as it is lying inside the unit circle $|z|=1$. Hence, $f(z)=(z-2)^3$ hence, using cauchy integral formula, we have ...


4

$y$ and $0$ are harmonic functions on $\mathbb C$ (considered as $\mathbb R^2$ with coordinates $x, y$) that agree on the real axis. More generally, take the imaginary part of any analytic function that is real on the real axis.


3

Since $f$ is analytic at $z_0$, we write $$ f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n=\sum_{n=0}^\infty\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n $$ in some open disc $U\ni z_0$ of radius $r>0$. If $|f^{(n)}(z_0)|>n!n^n$ for all $n\in\mathbb{N}$, then $ |a_n|>n^n $ and so $$ \limsup_{n\to\infty}|a_n|^{\frac{1}{n}}>\limsup_{n\to\infty}n=\infty $$ By Hadamard's ...


7

Suppose the inequality is true. Let $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Let $g(z) = \exp (\bar z ).$ Note that $$g(re^{it}) = \sum_{n=0}^{\infty}r^ne^{-int}/n!.$$ Using the orthogonality of the exponentials, we then get $$(1/2\pi)\int_0^{2\pi}|f(re^{it}) - g(re^{it})|^2\, dt = |a_0-1|^2+\sum_{n=1}^{\infty}|a_n|^2r^{2n} + \sum_{n=1}^{\infty}r^{2n}/(n!)^2 ...


6

I don't think this has anything to do with the complex logarithm function. If the series $\sum a_i$ converges, then the left-hand side will exist (because $\exp$ is continuous): $\exp (\sum \limits _{i=1} ^\infty a_i) = \exp (\lim \limits _{N \to \infty} \sum \limits _{i=1} ^N a_i) = \lim \limits _{N \to \infty} \exp (\sum \limits _{i=1} ^N a_i) = \lim ...


2

First of all, in the case when $X : f(x,y) = 0$ is hyperelliptic there are $2g + 2$ Weierstrass points on $X$. I believe that in general the number of points is bounded above by $g^3 - g$. Keep this in mind in case you venture into non-hyperelliptic territory. Second, you may need to check more than just the places $P \in X$ with $x$-projections equal to ...


1

Considering $z=a+ib$, $$e^{2iz}=e^{-2b}e^{2ia}$$ So your function would be: $$f(a+ib)=e^{-2b}(cos(2a)+i sin(2a))$$ $$u(a,b)=e^{-2b}cos(2a)$$ $$v(a,b)= e^{-2b}sin(2a)$$


0

Hint: $e^z=e^{2i(x+iy)} = e^{-2y} \cdot e^{i \cdot 2x}$


0

Use the following property of definite integral $\color{blue}{\int_{-a}^{a}f(x)=2\int_{0}^{a}f(x)\iff f(-x)=f(x)}$ Now, we have $$\int_{-\infty}^{\infty}\frac{x^2 dx}{1+x^6}=2\int_{0}^{\infty}\frac{x^2 dx}{1+x^6}$$ Let, $x^3=t\implies 3x^2=dt$ $$=2\int_{0}^{\infty}\frac{\frac{dt}{3}}{1+(t)^2}$$ $$=\frac{2}{3}\int_{0}^{\infty}\frac{dt}{1+t^2}$$ ...


4

The residue theorem is kind of tedious to apply here, since the integrand resembles the derivative of the inverse tangent function, a quick substitution does the trick. Set $u=x^3$ then $du=3x^2dx$, your integral thus becomes: $$\frac{1}{3} \int_{-\infty}^{+\infty} \frac{1}{1+u^2}du=\frac{1}{3} [\tan^{-1}(u)]_{-\infty}^{+\infty}=\frac{1}{3} ...


14

There is an awesome physical interpretation! You can think of poles as "sources of outward pointing vector lines". The rest of this answer explains that statement in detail. Consider an integral of a complex function $f$ $$ I = \oint_C f(z) dz \, .$$ Let $f(z) = u(z) + i v(z)$ and $dz = dx + idy$. Then we have $$ I = \oint_C f(z) dz = \oint_C \left[ u dx - ...


0

Rather, the Dirichlet Eta function is generally used for analytic continuation in the critical strip: \begin{align} &\zeta(s)=\left(1-2^{1-s}\right)^{-1}\sum_{n=1}^{\infty}\dfrac{-1^{n+1}}{n^s},\quad 0<\Re(s)<1\\ \end{align} and to expand on Alex R.'s answer, assuming RH, one gets \begin{align} &\zeta(s)=\frac{\prod _{n=1}^{\infty} \left(4 ...


1

Yes and no, if you want something nontrivial. The analytically extended zeta function admits a Weierstrass factorization (called the Hadamard factorization in this case): $$\zeta(s)=\pi^{s/2}\frac{\prod_{\rho}(1-s/\rho)}{2(s-1)\Gamma(1+s/2)},$$ where the product is over the nontrivial zeros. You can also see the trivial zeros at $s=-2k$ from the Gamma ...


9

It is true. If a function is unimodular on the unit circle ($|f(z)|=1$), you can use Schwarz reflection principle that tells you that the zeros and poles are symmetric wrt the unit circle ($z\ \leftrightarrow\ \bar{z}^{-1}$). There are no poles inside $|z|=1$, so there are no zeros outside $|z|=1$ for the analytical continuation. All poles are outside ...


2

Let $z=e^{i\theta}$, so that $dz=ie^{i\theta}d\theta$ and the integral $I$ corresponds to the contour integral $$ \oint_C \frac{1}{z^{n}e^{z\rho\log n}}\frac{dz}{iz}=\frac{1}{i}\oint_C\frac{dz}{z^{n+1}e^{z\rho\log n}} $$ where $C$ is the unit circle. The function $f(z)=\frac{1}{z^{n+1}e^{z\rho\log n}}$ is meromorphic with a pole of order $n+1$ at the origin, ...


1

If you want to apply the Cauchy integral formula directly, you'll need to use partial fraction decomposition: $$ \frac{1}{z^2(z^2-z+1)}=\frac{1}{z^2}\left(z+1-\frac{z^3}{z^2-z+1}\right) =\frac{1}{z}+\frac{1}{z^2}-\frac{z}{z^2-z+1} $$ Let $a=e^{\frac{\pi i}{3}}=\frac{1+i\sqrt{3}}{2}$. Then \begin{align} ...


1

I have no idea where they managed to derive this formula from The Cauchy integral formula (for derivatives). Recall that $$f'(\zeta) = \frac{1}{2\pi i}\int_{\lvert z\rvert = r} \frac{f(z)}{(z-\zeta)^2}\,dz$$ for $\lvert\zeta\rvert < r$. Thus, setting $\zeta = 0$ we obtain the first summand, $$f'(0) = \frac{1}{2\pi i}\int_{\lvert z\rvert = r} ...


0

You were on the right track. We have $$\begin{align} \int_{-\infty}^0 \frac{\log x}{x^2+\alpha^2}dx+\int_{0}^{\infty} \frac{\log x}{x^2+\alpha^2}dx&=2\pi i \left(\frac{\log (\alpha)+i\pi/2}{2i\alpha}\right)\\\\ &=\frac{\pi\log \alpha}{\alpha}+i\frac{\pi^2}{2\alpha} \end{align} \tag 1$$ We change variables on the first integral on the left-hand ...


0

Let $\mathcal{D}^p(X)$ denote the space of compactly-supported smooth $p$-forms on a smooth manifold $X$ (topologized in the same way as the space of distributions), then a $p$-current is defined to be a continuous linear functional $\mathcal{D}^p(X) \to \mathbb{R}$. An important class of currents arise from the oriented $p$-submanifolds $S \subset X$ ...


1

Integrating $$\frac{\log^2 z}{z^2+\alpha^2}$$ around a keyhole contour with the branch cut of the logarithm on the positive real axis and the argument between $0$ and $2\pi$ we get from the two residues $$2\pi i \times \left(\frac{(\log\alpha+i\pi/2)^2}{2\alpha i} - \frac{(\log\alpha+i3\pi/2)^2}{2\alpha i}\right) \\ = \pi \frac{-\log\alpha \times 2i\pi + ...


1

The proof is correct, expect for 2 small points. First, the Maximum Modulus Theorem as stated in Complex Analysis II of Stein and Shakarchi is about a function defined on a open set, such as $\mathbb{D}$, and in $\mathbb{D}$ we still have $z^n \to 0$ as $n \to \infty$ for all $z \in \mathbb{D}$, so there is no need to go to a smaller ball, unless your ...


0

Let $\gamma_R$ denote the positively oriented contour $[-R,R]$ followed by $Re^{it},0\le t\le \pi.$ If $z$ is in the upper half plane, then for $R>|z|$ we can apply Cauchy's formula to get $$f(z) = \frac{1}{2\pi i}\int_{\gamma _R} \frac{f(w)}{w-z}\,dw$$ Now let $R\to \infty$ and use the assumed decay rate of $f$ to see $$f(z) = \frac{1}{2\pi ...


1

This proof rely on a previous theorem: Let $f$ be a function analytic in a region containing a rectangle $R$. Then $$\int_{\partial R} f(z) \mathrm d z = 0.$$ And its proof heavily use the analiticity of $f$ to bound something. An open disc is open and connected so any two points can be joined by a polygonal path with segments parallel to the axis. ...


1

Alternatively there's a topological approach, all the results rely in a topological index which is used to obtain the winding number of a curve around a point. The reference is Topological Analysis by Gordon Thomas Whyburn.


3

Let $R$ be greater than the absolute values of all the roots of $p$. Then $$ \int_{|z|=R}\frac{dz}{p(z)} = -\int_{|w|=1/R}\frac{dw}{w^2p(1/w)}, $$ using the change of variables $w=1/z$. But the singularity of $\frac{1}{w^2p(1/w)}$ at $w=0$ is removable: $$ \lim_{w\to 0} w^2p(1/w) = \lim_{z\to \infty} \frac{p(z)}{z^2},$$ which is finite since $p$ has ...


4

Let $r>0$ be such big that all roots of $p(z)$ lie withing the disc $|z| < r$. Then for every $R \geqslant r$ we have $$\int \limits_{|z| = R} \frac{1}{p(z)} \, \mathrm{d} z = \int \limits_{|z| = r} \frac{1}{p(z)} \, \mathrm{d} z.$$ But the left side tends to $0$ as $R \to \infty$, so $$\int \limits_{|z| = r} \frac{1}{p(z)} \, \mathrm{d} z = 0$$ ...


1

I got it wrong at first because i didn't notice that the degree of $p$ is greater than $1$. What you can easily observe is that $1/p$ has a holomorphic antiderivative in the annulus $D(0,R,\infty)$ where $D(0,R)$ contains all the roos of $p$. Thus the integral equals zero. To notice the difference, if $p$ had degree $1$ and assume $p(z)=z$ then $\int ...


0

To begin with, suppose that $f(z)$ has been defined in the square $(-\epsilon,1+\epsilon)\times(\epsilon,\epsilon)$. (By choosing a finite open cover.) By $f_1(z+1)=azf(z)+p(z)$, we define $f_1(z)$ in $(1-\epsilon,2+\epsilon)\times(\epsilon,\epsilon)$. Let's prove $f_1$ is a direct continuation of $f$. This follows from the fact that they are both ...


3

Here you go. This was ever so difficult to find :)


0

Alot of the time Vieta's formulas are used. For example, if we want to find where the roots of $f(z):=az^3-z+b$ ($a>0$ and $b>2$) are, we first note that $f(0)=b>2$, and $\lim_{x\to -\infty}f(z)=-\infty$ (for real $x$), so we must have at least one root of the negative real line. Now If $\alpha_1,\alpha_2,\alpha_3$ are the three roots (with ...


1

This is almost correct, but there is one small detail that I think needs to be addressed. You showed that $|g(z)| \geq |f(z)|$ on $\partial B(0, 1)$ and then concluded that $g$ and $f+g$ have the same number of zeroes in $B(0, 1)$. But this does not follow from Rouche's Theorem; the theorem requires $|g(z)| > |f(z)|$ on $\partial B(0, 1)$, not $|g(z)| ...


2

The Green's Function $G(\vec \rho,\vec \rho')$ that satisfies the $2$-D Helmholtz equation $$\nabla^2G(\vec \rho,\vec \rho')+k^2G(\vec \rho,\vec \rho')=\frac{\delta(\rho-\rho')\delta(\phi-\phi')}{\rho'} $$ inside a circle of radius $1$ and boundary condition $$\left.\frac{\partial G}{\partial \rho}\right|_{\rho =1}=0 \tag 1$$ can be decomposed into ...


2

We have $$\begin{align} f(\theta)&=\sin\theta + \sum_{k=2}^{\infty}\frac{\sin k\theta}{2^k} \tag 1\\\\ &=\sin\theta+\text{Im}\left( \sum_{k=2}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^k\right) \tag 2\\\\ &=\sin\theta+\text{Im}\left(\frac{e^{i2\theta}}{2-e^{i\theta}}\right) \tag 3\\\\ &=\sin\theta+\frac12 \frac{2\sin 2 \theta-\sin ...


2

$$\sum_{n=1}^{+\infty}\frac{\sin(n\theta)}{2^n}=\text{Im}\sum_{n\geq 1}\left(\frac{e^{i\theta}}{2}\right)^n =\text{Im}\left(\frac{e^{i\theta}}{2-e^{i\theta}}\right)=\text{Im}\left(\frac{e^{i\theta}(2-e^{-i\theta})}{5-4\cos\theta}\right)$$ hence: $$\sum_{n=1}^{+\infty}\frac{\sin(n\theta)}{2^n}=\frac{2\sin\theta}{5-4\cos\theta}.$$



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