# Tag Info

11

...This happens exactly when $\mathrm e^{2z}+1=0$ (why?), that is, when $\mathrm e^{2z}=-1=\mathrm e^{\mathrm i\pi}$, that is, when $2z=\mathrm i\pi+2n\mathrm i\pi$ for some integer $n$ (why?), which is equivalent to the fact that $z=\mathrm i\pi(n+\frac12)$ for some integer $n$.

8

With a closed Riemann surface $\Sigma$ of genus $g>0$ there is naturally associated a pair $(\Lambda, H)$, where $H$ is a $g$-dimensional complex vector space and $\Lambda\subset H$ is a full rank lattice (an abelian subgroup isomprphic to $\mathbb Z^{2g}$). For $H$ you can take $H^1_{DR}(\Sigma; \mathbb R)$ (the degree 1 DeRham cohomology of $\Sigma$ ...

6

When $0<x<1$, we have the identity $K(x^{-1})=x(K(x)-iK(\sqrt{1-x^2}))$. Therefore $$\int^{\infty}_{1}K^2(x)\frac{dx}{x}=\int^{1}_{0}K^2(x^{-1})\frac{dx}{x}\\ =\int^{1}_{0}\left(x(K(x)-iK(\sqrt{1-x^2}))\right)^2\frac{dx}{x}\\ =\int^{1}_{0}x\left(K(x)-iK(\sqrt{1-x^2})\right)^2dx\\ =\int^1_0(xK^2(x)-xK^2(\sqrt{1-x^2})-2ixK(x)K(\sqrt{1-x^2}))dx.$$ We ...

5

The level sets of holomorphic functions $\mathbb{C}^n \to \mathbb{C}^k$ are exactly the Stein spaces of finite embedding dimension. Indeed, any Stein space of finite embedding dimension (for example a connected Stein manifold) can be embedded into some $\mathbb C^n$. And then any analytic subspace of $\mathbb C^n$ can be defined by $n$ global holomorphic ...

5

It depends on what is meant by "polynomial". If only $\sum c_n z^n$, then every function that is uniformly approximable by polynomials must be holomorphic on the interior of $J$. Although that condition is trivially satisfied if $J$ has empty interior, that doesn't mean that for such $J$ every continuous function is the uniform limit of polynomials. For ...

5

This is a consequence of Rouché's Theorem: Let $g(z)=-z$ and $h(z)=f(z)-z$. Since $$\lvert h(z)-g(z)\rvert=\lvert f(z)\rvert <1=\lvert g(z)\rvert$$ for every $z$ on $\gamma$, then the functions $g$ and $h$ have the same number of zeros inside $\gamma$. Clearly $g$ has exactly one zero.

4

I've given the skeleton of my work below. Fill in any missing pieces and check your answer against mine. Using $\gamma=[-R,R]\cup Re^{i[0,\pi]}$ and the simple poles at $\frac{1+i}{\sqrt2}$ and $\frac{-1+i}{\sqrt2}$ inside $\gamma$ \begin{align} \int_{-\infty}^\infty\frac{\cos(5x)}{x^4+1}\mathrm{d}x ... 3 This exercise is solved using Cauchy's Integral Formula: f^{(m)}(0)=\frac{m!}{2\pi i}\int_{|z-z_0|}\frac{f(z)\,dz}{z^{m+1}} and hence, for m\ge n: \begin{align} \lvert f^{(m)}(0)\rvert&=\frac{m!}{2\pi}\left|\int_{|z-z_0|=R}\frac{f(z)\,dz}{z^{m+1}}\right| \le \frac{m!}{2\pi}\cdot \frac{2\pi R}{R^{m+1}}\max_{|z|=R}\lvert f(z)\rvert ... 2 First recognize that since your integrand is even, you have\frac{1}{2}\int_{-\infty}^{\infty} \frac{1}{(1+x^2)^2}dx = \int_0^{\infty}\frac{1}{(1+x^2)^2}dx.Then use the residue theorem with a semicircular contour in the upper (or lower) half plane. Of course you will need to argue that the integral along the semicircular arc goes to zero. If you look ... 2 In response to @Cameron Williams' hint and comments, I am going to attempt the solution. We have f(z) = \frac{1}{(z^2+1)^2}. Let C be the half circle as described by @Cameron Williams. Now, we have z = i to be the singularity point inside C. In finding the residue, \begin{align} \text{Res}_{z = i} f(z) &= \text{Res}_{z = i} \frac{1}{(z^2+1)^2} ... 2 Look at the terms of the series. Ignoring the (-1)^n for the moment, we have\frac{n^{2x-1}-1}{n^x}n^{iy} = n^{x-1}n^{iy} - n^{-x}n^{iy} = n^{z-1} - n^{-\overline{z}}.$$The first term is holomorphic, and hence harmonic. The second term is antiholomorphic, and hence harmonic. Thus the difference of the two terms is harmonic. So every term in the ... 2$$A = { (e^{i\pi k} -1) (e^{-i\pi kh} -1) \over (e^{i\pi kh} -1) (e^{-i\pi kh} -1) } = { e^{-i\pi k(h-1)} - e^{-i\pi kh} - e^{i\pi k} + 1 \over 2(1-\cos (kh \pi)) }$$From which we obtain:$$ \operatorname{re}A = { \cos({\pi k(h-1)}) - \cos({\pi kh}) - \cos({\pi k}) + 1 \over 2(1-\cos (kh \pi)) } $$As Robjohn mentioned, the basic idea here is that if z ... 2 We have$$\partial_x [h(x^2+y^2)]=2x\cdot h'(x^2+y^2),$$hence$$\partial_{xx}[h(x^2+y^2)]=2 h'(x^2+y^2)+4x^2h''(x^2+y^2).$$Similarly, we obtain$$\partial_{yy}[h(x^2+y^2)]=2 h'(x^2+y^2)+4y^2h''(x^2+y^2).$$The Laplacian of u at (x,y) can be expressed as a function of x^2+y^2, hence we get a differential equation that h needs to satisfy on ... 2 I think you need to assume that C is a simple closed curve. Otherwise, take a long skinny ellipse (a "sausage") and bend it until the two ends just touch, like your thumb touching your forefinger when you use them to make a circle. There's clearly a curve that bounds the sausage (even after the ends touch at a single point), and this curve is clearly ... 2 Absolute value of polynomial tends to infinity for \left|z\right|\to\infty. That is, for each M>0, there exists R>0 such that for \left|z\right|>R we have \left|p(z)\right|>M. Take sufficiently large closed disk, so that \left|p(z)\right|>1 for z outside the disk. The disk is compact, so it's image by \left|p(z)\right| is ... 2 It follows from the duplication formula for the gamma function.$$B(z,z) = \frac{\Gamma(z) \Gamma(z)}{\Gamma(2z)} = \Gamma(z) \Gamma(z) \frac{\sqrt{\pi}}{2^{2z-1} \Gamma(z) \Gamma (z+1/2)}= 2^{1-2z} \frac{\Gamma (z) \Gamma(1/2)}{\Gamma(z+1/2)} = 2^{1-2z}B \left(z, \frac{1}{2} \right)$$http://mathworld.wolfram.com/LegendreDuplicationFormula.html 2 Note that \operatorname{re} f'(z) >0 for all z \in B(0,1). Suppose f(z_1) = f(z_2) with z_1 \neq z_2. Let \gamma:[0,1] \to B(0,1) be given by \gamma(t) = z_1+t(z_2-z_1), then f(z_2)-f(z_1) = 0 = \int_0^1 (f \circ \gamma)'(t) dt = \int_0^1 f'(\gamma(t)) (z_2-z_1) dt = \left( \int_0^1 f'(\gamma(t))dt \right) (z_2-z_1), from which we get ... 2 Try to read it like$$ \sum_{i_1}z^{i_1}...\sum_{i_n}z^{i_n}=\sum_k(\sum_{i_1+...+i_n=k}1)z^k $$If you going to write out the LHS then you want the know the coefficient in front of z^k you want to know how manny different ways z^{i_1}\cdot...\cdot z^{i_n}=z^{i_1+...+i_n}=z^k so how many ways i_1+...+i_n=k. Which is what we have on the RHS. Addition: ... 2 To start, recall that \sin{z} = \frac{1}{2i}(e^{iz}-e^{-iz}). So we are looking for z such that $$e^{iz}-e^{-iz} = -\frac{5i}{2}.$$ Multiplying both sides by e^{iz} and bringing all terms to the left yields $$(e^{iz})^2+\frac{5i}{2}e^{iz}-1=0.$$ Setting w = e^{iz}, we have the quadratic ... 2 To determine a point's winding number draw a ray starting in that point in an arbitrary direction. For 2i, draw a ray meeting the x-axis in 2, for example. Now this ray meets the curve in three points. The cutting number of such a point is 1 if the curve passes the ray from left and -1 otherwise. In our case the cutting numbers are -1, 1 and ... 2 we can prove this limit$$\lim_{n\to\infty}\dfrac{\log^2{n}}{n}\sum_{k=2}^{n-2}\dfrac{1}{\log{k}\log{(n-k)}}=1$$This is a International Competition in Mathematics for Universtiy Students in Plovdiv, Bulgaria 1994 last problem: can see this solution:http://www.imc-math.org.uk/imc1994/prob_sol.pdf 1 We are given the differential equation:$$\tag 1 y'' + 2y' + 5y = f(t), ~ y(0) = 1, y'(0) = 0 $$Taking the Laplace Transform of (1) and simplifying, we end up with the system:$$Y(s) = \dfrac{1}{(s+1)^2 + 2^2} \left(F(s) + s + 2 \right)$$We need to find the Inverse Laplace Transform for those three terms on the RHS. For the general F(s) term, we ... 1 It can be done by parts, without the trig. Let I_n=\int \frac{x^n \ dx}{\sqrt{1-x^2}}, and consider the integral$$J_n=\int \frac{x^n(1-x^2)}{\sqrt{1-x^2}}.$$On the one hand it is I_n-I_{n+2} by expanding the numerator. On the other hand, simplifying the integrand of J_n to x^n \sqrt{1-x^2}, we can integrate J_n by parts with u=\sqrt{1-x^2} and ... 1 An alternative, and imo much simpler, way. Put$$a_n:=\frac{z^{2n}(-1)^n}{(2n)!}\implies\;\frac{a_{n+1}}{a_n}=\left|\frac{z^{2n+2}(-1)^{n+1}}{(2n+2)!}\cdot\frac{(2n)!}{z^{2n}(-1)^n}\right|=|z|^2\frac1{(2n+1)(2n+2)}\xrightarrow[n\to\infty]{}0and thus the series' radius of convergence is infinite, which means the series defines an analytic function ... 1 \begin{align*} e^z &= e^x (\cos y + i \sin y) \\ \left| e^z \right| &= \sqrt{e^{2x} \cos^2 y + e^{2x}\sin^2 y} \\ &= e^x \sqrt{\cos^2 y + \sin^2 y} \\ &= e^x. \end{align*} Also, you should be careful--\arg e^z is only equal to y when y \in (-\pi, \pi] (or whatever branch you're using for the \arg function.) 1 In general,\int_\gamma f(z)\ dz = \int_0^1 f(\gamma(t)) \gamma'\ dt$$where \gamma is a parameterized curve between A and B such that \gamma(0) = A and \gamma(1) = B. Choose any such \gamma that makes your life easy. Perhaps a straight line between A and B will work:$$\gamma(t) = (1-t)A+tB.$$The next step:$$\gamma'(t) = B-A.\\ ...

1

As far as I can tell, you have a set of the form $$\eta + \overline{\eta} e^{it}$$ where $\eta$ is a fixed complex number and $t$ varies in $\mathbb{R}$. Did I understand correctly? This is always a circle. Indeed, we can rewrite it as $$\eta \left(1 + \frac{\overline{\eta}}{\eta} e^{it}\right)$$ which is just $$\eta \left(1 + e^{i(t+\alpha)}\right)$$ ...

1

Set $$f(z)=\exp(\pi z/2),$$ Then $f'(z)\ne 0$, $f$ is one-to-one on $$S=\{z\in\mathbb C : 0<\mathrm{Im}\, z<1\},$$ and $$f[S]=Q=\{z\in\mathbb C: \mathrm{Re}\,z>0,\,\mathrm{Im}\,z>0\}$$ The harmonic function on $S$ is $v=\frac{\pi}{2}\mathrm{Im}\, z$ and $u=v\circ f^{-1}$ which is constant on the boundaries of $Q$. In particular, ...

1

Imagine you have a annular region $1<\vert z\vert<3$ and take a function $f\left(z\right)=\frac{1}{z}$ and a contour $\vert z\vert=2$. Then the integral of this function over this contour is equal $2\pi i$ times residuum at 0 and this is certainly not zero. We just used Cauchy residuum theorem.

1

Part of the problem is that $z\mapsto a^c$ is "multiple-valued". The simplest example is when $c=1/2$. You learned at your mother's knee that each positive real number $z$ has two square roots, called $\pm\sqrt{z}$, and each negative real number $z$ has two square roots, called $\pm i\sqrt{-z}$ (where of course $-z$ is a positive number, and so is ...

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