# Tag Info

0

$\Gamma(z)=\dfrac1z+\displaystyle\sum_{k=0}^\infty a_k\cdot z^k,~$ where the terms $a_k$ form this “beautiful” sequence here. :-$)$

5

The following image shows half a torus mapping to a sphere via a scaled Weierstrass $\wp$ function. It was created in Mathematica. The four branch points lie on the two boundary circles of the half-torus, and map to the endpoints of the two "seams" on the resulting sphere. The other half of the torus maps to the sphere in the same way, with the deck ...

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the series $$\sum_{n} \frac{z^{2n}}{z^n} = \sum_n z^n$$ is convergent for every $z$ with $|z|<1$. It diverges for $z = 1$. Therefore the radius of convergence of this series is $1$ Similarly, (in case of a typo in your question) the series $$\sum_{n} \frac{z^{2n}}{2^n} = \sum_n \left(\frac{z}{\sqrt{2}}\right)^n$$ is convergent for every $z$ with ...

5

Here's another approach: Note that $z-\bar z=2iy.$ Thus $$\tag 1 |(z-\bar z)f(z)| \le 2$$ for all $z.$ Let's write $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Setting $z=re^{it},$ we get $$\tag 2 (z-\bar z)f(z) = (re^{it} - re^{-it})(\sum_{n=0}^{\infty}a_nr^ne^{int}).$$ Play with $(2)$ a little to see it equals $$\tag 3 \sum_{n=0}^{\infty}(a_nr^{n+1} - ... 0 The multiplicity of a zero of f at a is the number of vanishing coefficients in the Taylor expansion of f around z=a:$$ f(z) = f(a) + f'(a)(z-a) + \frac12 f''(a)(z-a)^2 + \cdots $$so the number of vanishing derivatives determines how large power of (z-a) we can factor out. Hence, the order of a holomorphic function at z=a is the multiplicity of ... 1 Your solution is fine. (Answered to keep the question off the unanswered list.) 1 Just use the Cauchy estimate.$$\left|\int_{C_R}Q(z)\,dz\right|\le 2\pi R\cdot \max_{z\in C_R}\left|{z^2\over (z-a_1)\ldots (z-a_N)}\right|<\!<{2\pi R^3\over R^4}=2\pi R^{-1}\stackrel{R\to\infty}{\longrightarrow} 0.$$since N\ge 4. 2 By the Fundamental Theorem of Algebra, we have p(z)=a(z-z_1)\dots(z-z_n) where a\neq 0 and n is the degree of p. (Clearly p(z) cannot be identically zero.) By the chain rule, we have$$ p'(z)=a\left(\sum_{i\neq 1}(z-z_i)+\dots+\sum_{i\neq n}(z-z_i)\right) $$and thus$$ \frac{p'(z)}{p(z)} = \sum_{k=1}^n\frac{\sum_{i\neq ...

2

With the branch cut $[-1,1]$, the domain $U = \mathbb{C}\setminus \{ z : \lvert z\rvert \leqslant 1\}$, the exterior of the unit disk, is contained in the domain of the branch of $h(z) = \sqrt{z^2-1}$ we choose. On the open unit disk, we have two branches of $\sqrt{1-w}$, let $g(w)$ denote the branch with $g(0) = 1$. Then we can write $$h(z) = z\cdot ... 5 By the assumption that f(0) = f'(0) = \dotsc = f^{(n-1)}(0) = 0, the Taylor series of f centred at 0 is$$f(z) = \sum_{k = n}^\infty a_k z^k.$$Hence the function$$g(z) = \frac{f(z)}{z^n} = \sum_{k = n}^\infty a_k z^{k-n} = \sum_{k = 0}^\infty a_{n+m}\cdot z^m$$is holomorphic on B(0,1). By the maximum modulus principle, for 0 < r < 1, ... 2 Differentiation is a real operator, that is, it commutes with conjugation,$$d\overline{\omega} = \overline{d\omega}$$for every k-form. Since you know that \overline{f(z)}\,dz is closed, i.e. d\bigl(\overline{f(z)}\,dz\bigr) = 0, the result follows. Alternatively, use the Wirtinger derivatives,$$dg = \frac{\partial g}{\partial z}\,dz + ...

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First approach: The first thing that comes to mind is Jensen's Formula. For any point $\omega \in \mathbb{R}$ and $r > 0$, we have: $$\log |f(\omega)| \le \frac{1}{2\pi}\int_0^{2\pi} \log |f(\omega + re^{i\theta})|\,d\theta$$ Since, $\displaystyle |f(\omega + re^{i\theta})| \le \frac{1}{|r\sin \theta|}$, we have: $$\log |f(\omega)| \le ... 1 For a basic treatment of elliptic functions I strongly recommend you the book Theory of Functions of a Complex Variable by A.I. Markushevich. Especially take a look at part III chapters 5 and 6. Chapter 5 is a complete treatment on Weierstrass theory, meanwhile chapter 6 introduces Jacobi's theory and the relation to the Weierstrass one. Here is a screen ... 0 Consider a cover of the image F of f (which is compact) this way: for every x\in F take an open ball B_x=B(x,\epsilon_x). You can take a finite subcover and let \epsilon the minimum of the selected \epsilon_x. Now you can pick (applying the uniform continuity) \delta>0 such that the image of every ball with a radius lesser than \delta is ... 2 Counterexample: Let p(z) = z^2 + 3z -1. Then |a_0| = 1, |a_1| = 3, |a_2| = 1. Clearly 3 = |a_1| > |a_0| + |a_2| = 2, so g(z) = z^2 - 1. But the roots of g(z) are \pm 1\in S^1. 0 Can you include the details of the limits for the cases \theta=\displaystyle\frac{\pi}{2} and \theta=\displaystyle\frac{\pi}{n}, please. This is a key step in the calculation. Thanks. 2 Another approach: A corollary of the Weierstrass approximation theorem says that if f:[0,1]\to \mathbb {R} is continuous, then there is a sequence of polynomials p_n converging uniformly to f on [0,1] such that p_n(0) = f(0),p_n(1) = f(1) for all n. Now let \alpha :[0,1]\to G be a continuous map (like your polygonal map say) such that \alpha ... 1 This is not at all correct. Try e.g. f(x) = x^2. Then (f(x)-f(y))/(x-y) = x + y, and the requirement on g is that \int_{\mathbb R} g(y)\; dy = 0 and \int_{\mathbb R} y g(y)\; dy = 0. Somewhat more generally, if f is a polynomial of degree d you need the moments \int_{\mathbb R} y^k g(y)\; dy = 0 for 0 \le k \le d-1. 0 This picture illustrates the situation: The interior of \Omega is all with gray colors, \Gamma (in blue) is the contour in the interior of \Omega and w is in the bounded part of \Gamma (a.k.a the "interior" of \Gamma as the others would like to call). 4 For simplicity, let's just consider modular forms (of weight k) of the modular group SL(2,\mathbb{Z}), i.e. holomorphic functions f \colon \mathbb{H} \to \mathbb{C} satisfying two other conditions, the second being the "holomorphicity at \infty" in which we're interested. The first condition is that f has a nice symmetry under the action of ... 1 \Omega \in {\rm Int}(\Gamma) means that \Omega is in the interior of \Gamma. Hope I helped! 0 This must be a typographical error: there is either a 1 instead of \Bbb i in the denominator, or the left bar of the modulus must also enclose the outer \Bbb i. Check for yourself: take z=2. Then clearly f(2) \notin \Bbb R, yet the computation they make gives a real result for every z \in \Bbb C. You are also right about the numerator. Your ... 0 You have tackled it correctly, all that is missing is that you see$$\lim_{\epsilon \to 0} \int_{\gamma(0,\epsilon)} f(z)(\ln z - i\pi)\,dz = 0.$$The standard estimate$$\biggl\lvert \int_\Gamma g(z)\,dz\biggr\rvert \leqslant L(\Gamma)\cdot \sup \{ \lvert g(\zeta)\rvert : \zeta \in \operatorname{Tr}(\Gamma)\}$$yields that, since f is bounded, and ... 0 Apply The Residue Theorem to the function$$f(w)=\frac{p(w)}{q(w)(w-z)}$$around a contour C, traversed counter-clockwise, that encircles the entire plane and thus includes both w=z and w=b_k for all k=1,2,\cdots n. We have$$\begin{align} \oint_C\frac{p(w)}{q(w)(w-z)}dw&=\lim_{R\to \infty}\int_{|w|=R}\frac{p(w)}{q(w)(w-z)}dw\\\\ &=0 ...

0

I'm not sure about the suggestion. Why the two variables $z,w$ and this function $f,$ which seems to be a function of both $z,w?$ Let's forget the $w$ business. Letting $a_k/(z-b_k)$ denote the principal part of $p/q$ at $b_k,$ we see $$\frac{p(z)}{q(z)}-\sum_{k=1}^{n}\frac{a_k}{z-b_k}$$ has removable singularities at each $b_k,$ hence extends to be ...

2

Assuming $\text{Re}(s)>0$ we have: $$\begin{eqnarray*} f(s)=\sum_{n\geq 1}\frac{(-1)^n}{(2n+1)^s}&=&\sum_{n\geq 1}\frac{(-1)^n}{\Gamma(s)}\int_{0}^{1}(-\log x)^s x^{2n}\,dx\\&=&-\frac{1}{\Gamma(s)}\int_{0}^{1}(-\log x)^s \frac{x^2}{1+x^2}\,dx\\&=&-\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{t^s}{e^t(1+e^{2t})}\,dt\end{eqnarray*}$$ ...

8

Write $f(z)=z^kg(z)$ with $k>0$, $g$ holomorhic, $g(0)\ne0$. Then $$2^kg(2z)=\frac{g(z)}{1+z^{2k}g(z)^2}$$ which gives a contradiction if we let $z=0$.

4

Since $f(0)=0$, write $f(z)=a_nz^n+O(z^{n+1})$ for some $n>0$. Then $$\frac{f(z)}{1+f(z)^2}=(a_nz^n+O(z^{n+1}))(1-a_n^2z^{2n}+O(z^{4n}))=a_nz^n+O(z^{n+1})$$ and $$f(2z)=2^na_nz^n+O(z^{n+1})$$ which after equating coefficients implies that $a_n=0$. Since $n$ was arbitrary, $f$ must be identically zero.

1

Yes. Since $1\in \mathfrak{I}$, if $f \in \mathcal{O}(\mathbb{C})$, $f = f1 \in \mathfrak{I}$.

2

Let $z_{r}=re^{2\pi ip/q}$, where $p$ and $q>1$ are coprime integers and $0<r<1$. We will show that \begin{align*} \lim_{r\rightarrow 1^{-}}\left(1-r\right)\left|f(z_{r})\right|=+\infty, \tag{1} \end{align*} whence $f$ admits no analytic continuation $g$ in a neighborhood of $z=e^{2\pi i p/q}$, since that would imply \begin{align*} ...

2

The big gap in your proof is that you've merely drawn one open set and one pair of points and shown by picture how to join the points by a path within the open set. In order to get a formal proof you need a strategy that depends on an open set $G$ and a pair of points $a$ and $b$ in $G$. Here's the usual approach: Let $U$ denote the set of points $x$ in ...

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The set of all functions $f:\mathbb C\to\mathbb C$ such that $f(z+1)−f(z)$ is entire is exactly the set of funcions $f(z)=e(z)+p(z)$ where $e$ is entire and $p:\mathbb C\to\mathbb C$ is $1$-periodic. Proof: Be $g(z)=f(z+1)-f(z)$. Since $g$ is entire, according to this Mathoverflow post (linked to by mathcounterexamples.net in the comments) there exists an ...

1

$$\int_{\gamma}\frac{-\cos(1/z)\,dz}{z^2\sin(1/z)} = \int_{|w|=\frac{1}{2}}\cot w\,dw = 2\pi i\cdot\text{Res}\left(\cot w,w=0\right)=\color{red}{2\pi i}.$$

1

(z/((z-i)(z-2)))=(z/((2-i)(z-i)))+(z/((2-i)(z-2))),1<∣z∣<2. The expansion in powers of z is (z/((2-i)z(1-i/z)))-(z/((2-i)2(1-z/2))) =(1/(2-i))∑₀((i/z))ⁿ-(1/(4-2i))∑₀((z/2))ⁿ.

1

Although the question is tagged (complex-analysis), here's another way to look at the problem: Setting $f(z)=z^4+iz^3+1$, we might as well look for the roots of the real polynomial $$g(z):=f(iz)=z^4+z^3+1,$$ in the quadrant $Q:=\{z\in\Bbb{C}:\operatorname{Re}(z)>0>\operatorname{Im}(z)\}$. As $g$ is everywhere positive it has no real roots, hence it ...

0

Put $f(z)=z^4+1$ and $g(z)=iz^3$. Step 1: On $|z|=2$, $|g(z)|=8<15=|z^4|-1\le|f(z)|$. What does this imply? Step 2: Consider $\gamma_1=\{z=x+iy:0\le x\le2,y=0\}$. Is it true that $|g(z)|<|f(z)|$ on $\gamma_1$? (Hint: $|g(z)|=x^3$ and $|f(z)|=x^4+1$, what is the minimum of $x^4+1-x^3$ on $[0,2]$?) Step 3: Do the same for $\gamma_2=\{z:|z|=2,0\le ... 0 For an isomorphism from upper half plane to the unit disk, suppose$i\mapsto0$. Then$-i\mapsto\infty$, since$i,-i$are symmetric (w.r.t. the$x$-axis) and$0,\infty$are symmetric (w.r.t. the unit circle). Thus$z\mapsto\frac{z-i}{z+i}$is what you want. 0 Hint: surely zeroes are inside some big quartercircle with radius R, than use argument principle. 1 Expanding the functions$\exp$and$\sinin Laurent series we get [ \begin{align} \exp \left( {\frac{1}{{z^2 - 1}}} \right) = \sum {\frac{1}{{\left( {z^2 - 1} \right)^n n!}}} = \sum {\frac{1}{{\left( {z + 1} \right)^n \left( {z - 1} \right)^n n!}}} = \sum {\frac{{\frac{1}{{\left( {z + 1} \right)^n }}}}{{\left( {z - 1} \right)^n n!}}} \end{align} ... 1 Letf: X \to \mathbb C$an holomorphic map, non-constant Then$f$is open (standard argument using Maximum principle for example). But$f(X)$is compact and open, and non-empty. Contradiction. Conclusion :$f$is a constant map. 0 There is a small list on Wikipedia: https://en.wikipedia.org/wiki/Riemann_hypothesis#Excluded_middle 3 I am aware that every solution is essentially equal to$\exp(g(z))$for some entire$g$, but that does not mean every answer needs to be in that form. Yes, every answer needs to be in that form. The only theorem you need is that an entire function has a Taylor series with infinite radius of convergence, so integrating it term by term gives again a ... 1 By the Weierstrass factorization theorem, an entire function that has no zeros must be of the form$\exp(g(z))$. 0 If you want to work in a classical pointwise way, the inversion is a Cauchy Principle Value integral $$f(x)=\lim_{R\rightarrow\infty}\int_{-R}^{R}\hat{f}(\xi)e^{2\pi i\xi x}d\xi .$$ If$f$is absolutely integrable and satisfies some condition such as having left- and right-hand derivatives at$x$, or is of bounded variation on an interval$[a,b]$with ... 1 yes they are the same. In fact in the first case you are writing$(u+iv)^2$in terms of real and imaginary parts but in the second case you are writing$(u+iv)^2$in terms of magnitude and phase. because$|u^2-v^2+2iuv|=\sqrt {(u^2-v^2)^2+(2uv)^2}=u^2+v^2$each complex number can be written in two forms: real and imaginary elements:$a+ib$polar form ... 0 Hint: In order to show that two complex numbers$z_1,z_2$are equal it is enough to show that$|z_1|=|z_2|$and that they create the same angel with the real axes. 0 Hint: (1) Show that$|1-\bar{z_1}z_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$. (2) Show that it is bijective on$\bar{\mathbb{C}}$by finding its inverse. (3) Use the fact that$(f/g)'=(f'g-fg')/g^2$. 1 First, you should know the following proposition: If$f$is analytic injection on a region$\Omega$, then its inverse function$f^{-1}:f(\Omega)\to\Omega$is also injective and analytic. (For a complete proof, see any book on complex analysis. Here is one possible outline: [1] Show that$f'(z)\ne0$for all$z\in\Omega$. [2] Show that$f^{-1}$is ... 1 Note that, if $$A(x,y)=\frac{f(x,y)}{xf(x,y)+yg(x,y)}$$ and $$B(x,y)=\frac{f(x,y)}{xf(x,y)+yg(x,y)},$$ you can derive and find that $$A_y=\frac{yf_yg-fg-yfg_y}{(xf+yg)^2}$$ and $$B_x=\frac{xfg_x-fg-xf_xg}{(xf+yg)^2}.$$ Theorem: If$\omega=Adx+Bdy$a 1-form is$C^1$such that$A_y=B_x$, then$\omega$is closed. Hint: Use that$f$and$g$are ... 3 This is a nontrivial task, and the best approach may depend on how the polynomial is specified (i.e., what kind of structure it has). In principle, the Routh–Hurwitz theorem can answer the question, as it gives the number of roots in the positive half plane$\{z:\operatorname{Re}z>0\}\$. Your problem amounts to determining whether this number is the same ...

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