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By "a Möbius transformation from $\mathbb{D}$ to $\mathbb{D}$" you seem to mean a Möbius transformation whose restriction to $\mathbb{D}$ is an automorphism of $\mathbb{D}$. In this setting, the result is true. However, if $T$ is just a Möbius transformation such that $T(\mathbb{D}) \subseteq \mathbb{D}$, then the claim is false: take $T(z) = \frac{1}{2}z$. ...

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 2 Let$\alpha=e^{2\pi i/3}=\frac{-1+i\sqrt3}2$, that is$\alpha^3=1$, then if$k\not\equiv0\pmod3$,$\alpha^k+1+\alpha^{-k}=\frac{\alpha^{3k}-1}{\alpha^k\left(\alpha^k-1\right)}=0$. If$k\equiv0\pmod3$, then$\alpha^k+1+\alpha^{-k}=1+1+1=3$. That is, $$\frac{\alpha^{k-1}+1+\alpha^{1-k}}3=\left\{\begin{array}{} 1&\text{if }k\equiv1\pmod3\\ 0&\text{if }... 1 There is an error towards the end. When you apply the Weierstrass theorem, you are assuming that the derivative f' is continuous, but there is no way to determine that from your hypotheses. If you change your conditions to include that f is continuously differentiable, then I believe you are correct. EDIT I found a proof of your result, which uses ... 1 For rational functions, the idea is essentially always the same: transform the given annulus boundary conditions r < \lvert z - a \rvert < R into the two equivalent conditions \lvert R^{-1} (z - a) \rvert < 1 and \lvert r(z - a)^{-1} \rvert < 1, and then typically use partial fraction decomposition to separate your function into terms ... 1 This is an interesting question. I hope others can also give responses. The key idea behind all the explicit formulas is to use contour integrals of L-functions to relate sums over the primes to sums over the zeros. The Weil and Riemann explicit formulas have similar statements and proofs, but neither is a strict specialization of the other. Briefly, ... 2 What you are probably looking for is a holomorphic function mapping the unit disk \Bbb D into itself with f(z_1) = w_1 and f(z_2) = w_2. Yes, you can proceed similarly as in the proof of the Schwarz–Pick theorem by "normalizing" one source point and its image to the origin:$$ T(z) = \frac{z - z_2}{1 - z\overline{z_2} } \quad \text{and} \quad S(w) ... 0$f(x)=\frac{1}{x}$is not a$L^1$function but just a locally integrable function, hence its Fourier transform is usually defined as $$\widehat{f}(\xi)= \lim_{r\to 0^+}\lim_{R\to +\infty}\int_{r\leq |x|\leq R}\frac{e^{i\xi x}}{x}\,dx = 2i \lim_{r\to 0^+}\lim_{R\to +\infty}\int_{r}^{R}\frac{\sin(\xi x)}{x}\,dx=\color{red}{\pi i\, \text{sign}(\xi)}.$$ 1 First, your notation with the Big-O symbol is not the most conventional; to denote that the time complexity of$4n$grows slower than that of$n\log{n}$, we would use the Little-O symbol: $$4n \in o(n \log{n})$$ Proving this is quite trivial: since $$\lim_{n \to \infty} \frac{4n}{n \log{n}} = \lim_{n \to \infty} \frac{4}{\log{n}} = 0$$ by definition we ... 0 If$x\in\mathbb{Z} $obviously we have $$\sum_{n\geq1}\frac{e^{2\pi inx}}{n}=\sum_{n\geq1}\frac{1}{n}=\infty$$ if$x\notin\mathbb{Z} $we observe that, from the Taylor series of the complex log $$\textrm{Log}\left(1-z\right)=-\sum_{n\geq1}\frac{z^{n}}{n},\,\left|z\right|<1$$ that $$\sum_{n\geq1}\frac{e^{2\pi inx}}{n}=\color{red}{-\textrm{Log}\left(1-... 1 Multiply both sides by (n+1)/(2\pi i) to obtain$$ \frac{1}{2\pi i}\int_{|z|=1} \frac{f(z)}{z-\frac{1}{(n+1)}}\,dz = 0. $$Then notice that by the Cauchy integral formula, the integral on the left is precisely f(1/(n+1)). So the zero set of f(z) has an accumulation point inside D(0,2), so it must be identically zero there. 2 No. Let f_k be your rotation, and let g be literally any homeomorphism S^1 \to S^1. Then gf_kg^{-1} is also an order n homeomorphism; it's easy to see you can pick g so that this new homeomorphism doesn't, e.g., send 1 to e^{2\pi ik/n}. On the other hand, every free homeomorphism of order n is conjugate to one of yours. This follows from a ... 2 If the limit were to exist, it would have to exist and be the same no matter what direction you approached from. Let us consider z_0 = x_0+iy_0 and see what happens when we approach along the line y = y_0:$$f'(z_0) = \lim_{z\to z_0} \frac{ix}{x} = i.$$If instead we approach along the line x=x_0, we have$$f'(z_0) = \lim_{z\to z_0} \frac{y}{iy} = -... 0 Consider$O$to be the origin, then$|a|=|b|=|c|=r$We know that the equation of a line passing through points$z_1$and$z_2$is represented by $$\begin{vmatrix} z & \overline{z} & 1 \\ z_1 & \overline{z_1} & 1 \\ z_2 & \overline{z_2} & 1 \\ \end{vmatrix}= 0$$ Now as per the question$a, b$and$c$, lie on a circle, now lets ... 0 HINT: Denoting a vector as difference of position vectors,so angles$ bcz = cab,$so $$bc \cdot cz / ( |bc|\, |cz| ) = ca \cdot ab / ( |ca|\, |ab| )$$ 0 If$\lvert z \rvert < 1$, then$\lvert z^n \rvert = \lvert z \rvert^n \to 0$, which is equivalent to$z^n \to 0$. On the other hand, if$\vert z \rvert > 1$, then$\lvert z^n \rvert = \lvert z \rvert^n \to \infty$, so the sequence$(z^n)$cannot converge. Now for the behavior on the unit circle. Convergence at$z = 1$is clear. Suppose that$\lvert z \...

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If it was differentiable in some $z_0=r_0e^{it_0}$, use that $\displaystyle\lim_{r\to r_0}\frac{Arg(re^{it_0})-Arg(r_0e^{it_0})}{re^{it_0}-r_0e^{it_0}} = \lim_{t\to t_0}\frac{Arg(r_0e^{it})-Arg(r_0e^{it_0})}{r_0e^{it}-r_0e^{it_0}}$ Edit: As requested, here a further evaluation of RHS. $\displaystyle\lim_{t\to t_0}\frac{Arg(r_0e^{it})-Arg(r_0e^{it_0})}{r_0e^... 3 As you already noticed, Picard and tha fact that more than one value is left out tells us that the singularities at points in$\Bbb Z\cup \{\infty\}$are non-essential. On the other hand, if$n\in\Bbb Z$is a pole of order$k>0$, then the image under$f$of a sufficently small circle around$n$will wind$k$times around$\infty$and thus intersect the ... 1 There is no such function: First, we observe, that$f$maps interely into$\{z\in\mathbb C: Im(z)>0\}$or$\{z\in\mathbb C: Im(z)<0\}$, since its image is open (and therefore$f$has to ommit the value$0$) and path-connected. We assume that$f$maps into$\{z\in\mathbb C: Im(z)>0\}$. As you already mentioned,$f$can't have an essential ... 1 what am I missing in my solution After having$\cos\left(2\theta+\frac{\pi}{4}\right)\gt 0$, you have $$-\dfrac{\pi}{2}\lt\left(2\theta+\dfrac{\pi}{4}\right)\lt\dfrac{\pi}{2}$$ which is incorrect. To make it easy to understand why this is incorrect, let$\alpha:=2\theta+\frac{\pi}{4}. Then, we want to solve \cos\alpha\gt 0\quad\text{and}\quad -\... 1 I would find the general solutions of the inequation first: \begin{align*} \cos\Bigl(2\theta+\frac\pi4\Bigr)>0&\iff-\frac\pi2<2\theta+\frac\pi4<\frac\pi2\iff-\frac{3\pi}4<2\theta<\frac\pi4\color{red}{\mod2\pi}\\ &\iff-\frac{3\pi}8<\theta<\frac\pi8\color{red}{\mod\pi} \end{align*} Now that if conventionally, we choose \;-\pi<\... 1 A holomorphic function G(z) vanishes to order (at least) n at z_0 if G^{(m)}(z_0)=0 for all m<n. To get f(z)-f(z_0)=a(z-z_0)^k+G(z), just let a(z-z_0)^k be the first nonzero term (after the constant term) in the Taylor expansion of f(z) around z_0, and G(z) be the rest of the terms. We have k\geq 2 since the degree 1 Taylor ... 0 I think this really depends on the specific course, but in general a course on (elementary) real analysis is a prerequisite to really appreciate the material that is covered in a course on complex analysis. However, if your course is at the same level as Serge Lang's book on Complex Analysis, then a calculus course might be enough. This book starts with ... 0 The answer should be a\ge 6+2\sqrt 2. (a\gt 9 is not correct since, for a=9, z=-3i satisfies the conditions.) In your Algebraic Solution, it seems that you have some errors. Case 1:-8 \lt \sqrt{a^2-12a+28} + a \implies a\gt 9$$Case 2:-$$8 \gt \sqrt{a^2-12a+28} -a \implies a\gt -\frac{9}{7}$$Case 3:-$$8\lt a-\sqrt{a^2-12a+... 1 Well, I would consider using the fact that there exists such anF$to be using Cauchy's theorem, since the usual proof of this for arbitrary holomorphic functions on simply connected domains uses Cauchy's theorem. But in this case you can avoid that by just writing down such an$F$explicitly:$F(z)=\log(z-a)$, for some branch of$\log$defined in the ball ... 0 I'm not quite sure what's going on in your first statement, with a set$U$and closed balls inside$U$. All that matters is that$f$be holomorphic on an open disk centered at the point in question - this is the disk on which the power series representation of$f$will converge. Maybe you are worried about the kind of convergence? For both Taylor and Laurent ... 1 If the derivative is nonzero then the complex derivative at a point is simply given by$d_pf(h) = ch$. In other words, the complex derivative is given by multiplication by a complex number. Any complex number which is nonzero can be written in polar form$c=re^{i \theta}$hence the action of multiplying by$c$is just a rescaling (dilation) and rotation of$...

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Hint: Try integrating $1/z-f(z)$ around the unit circle.

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Sure; you can even additionally require that $\rho$ map the entire unit disk to itself. Just note that for any point $p$ in the open unit disk, there is a Mobius tranformation $\sigma$ such that $\sigma(p)=0$ and $\sigma$ maps the unit disk to itself. Now pick any $p$ such that $0<|p|<1$ and $1/p$ is not a root of $f$ and let $$\rho(z)=\frac{1}{\... 1 Your argument has the right idea, but it is unclear how you are deducing the last step. Why must (2+z)Arg(z) have no removable discontinuities just because Arg(z) has no removable discontinuities? For instance, notice that f(z)=1/z has no removable discontinuities, but zf(z) does have a removable discontinuity at 0. 3 Different approach, hint: Consider g(z) =f(z)f(iz)f(-z)f(-iz). 3 The credit belongs to @user64494 and @Evan (and Phragmén–Lindelöf). Write z = x + iy. It is enough to consider x such that (say) x^2 > y^2, since otherwise$$ c_1 |z|^2 = c_1(x^2+y^2) \le 2c_1y^2 \le 3c_1 y^2 - c_1 x^2, $$and so by the assumption we have$$|f(z)|\le O(e^{c_1|z|^2}) = O(\exp(- c_1 x^2 + 3c_1 y^2)).$$By symmetry we may consider ... 0 Note that if x\in U, then x\in U_{i_0} for some i_0. This implies |f(x)| \le \|f\|_{U_{i_0}} \le \sum_{i=1}^{l} \sup_{U_i}|f|. Therefore$$\|f_n\|_{U} \le \sum_{i=1}^{l}\|f_n\|_{U_i}.$$So we have$$\infty >\sum_{i=1}^{l}\sum_{n=0}^{\infty} \|f_n\|_{U_i} = \sum_{n=0}^{\infty} \sum_{i=1}^{l}\|f_n\|_{U_i} \ge \sum_{n=0}^{\infty} \|f_n\|_{U}.$$1 For a C^1 counterexample, consider a path \gamma: [-1,1] \to \mathbb C as follows. For -1 \le t \le 0, \gamma(t) = -t^2. For positive integers n, from t = (n+1)^{-1/2} to t=n^{-1/2} the path goes from in straight line segments from 1/(n+1) to e^{i/(n+1)}/n to e^{-i/(n+1)}/(n+1) to 1/n. Note that the length of these three segments is ... 2 No the answer will be only harmonic, as example f(z)=Re(z) the real part of z, we know using Cauchy-Riemann equation that a real valued holomorphic function will be constant. but for all a=\alpha+i\beta we have$$ \alpha=f(a)=\frac{1}{2\pi}\int_{0}^{2\pi} (\alpha+R\cos(\theta))d\theta=\frac{1}{2\pi}\int_{0}^{2\pi} f(a+Re^{i\theta})d\theta=\frac{1}{2\...

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Well...yes, because then you would have an expression for the nth derivative of f, making f infinitely differentiable/analytic in U and thus holomorphic. https://en.wikipedia.org/wiki/Analytic_function

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If you parametrize your integral, you see that your equation is equivalent to $$f(a) = \dfrac{1}{2\pi} \int_0^{2\pi} f(a + r e^{i\theta})\; d\theta$$ and this is true for harmonic functions, not just holomorphic ones.

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For an open-interval domain, i believe the answer is "no." The gist is this: Take the graph of $x \mapsto \sin(\frac{1}{x})$ for $0 < x < 1$. Connect the $x = 1$ end to the top of a vertical line from $(0, 1)$ to $(0, -1)$. Connect the bottom of that line to a similar sine-curve on the left side of the axis. At the origin, every disk about the ...

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Suppose $p_n$ are holomorhic polynomials such that $p_n(z) \to 1/z$ uniformly on the unit circle. Then by uniform convergence and Cauchy's theorem, $$0 = \int_{|z|=1}p_n(z)\,dz \to \int_{|z|=1}(1/z)\,dz = 2\pi i,$$ contradiction. So there is no such sequence.

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I think i found answer to that. The sequence of polynomial consists of entire functions, so we can apply the cauchy formula at $z_0 = 0$. We now have $$p_n(0) = \int_{C_r(0)} \frac{p_n(\zeta)}{\zeta} d\zeta \quad \forall r>0$$ In my lecture we have shown that for an compact converging sequence $fn \to f$ it is: $F_n \to_{comp} F$ where $$F_n:= \int_{\... 1 Yes, the sequence of polynomials would then converge too in 0. If (p_n)_{n\in\mathbb N} is the given sequence then p_n(0) = \frac{1}{2\pi}\int_0^{2\pi}p_n(e^{it})\mathrm dt. By uniform convergence on S^1 and Lebesgue's theorem the integrals converge to \frac{1}{2\pi}\int_0^{2\pi}e^{-it}\mathrm dt = 0 1 Hint: The real part and imaginary part of a holomorphic function are both harmonic. 0 Hint: |f(z) - 2| < 1 for |z|<1 by the maximum modulus theorem. 0 Fix a point w in the open unit disk \mathbb{D}, and define$$ f_w(z)=\frac{w-z}{1-\overline{w}z}$$for z\in \mathbb{D}. One can show that f_w is a biholomorphic map \mathbb{D}\to \mathbb{D} such that f_w(0)=w and f_w(w)=0. Therefore given two points w_1,w_2\in \mathbb{D}, the function f(z)=(f_{w_2}\circ f_{w_1})(z) is a biholomorphic map \... 1 Hint: the homothetic map whose ratio is R_1/R_2 is a conformal map. 2 Start with the series expansion$$e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdotsand use the triangle and reverse triangle inequalities to get the result. By the triangle inequality and the condition \lvert z\rvert \le 1, we have\begin{align}\lvert e^z - 1\rvert &\le \lvert z \rvert + \frac{\lvert z\rvert^2}{2!} + \frac{\lvert z\rvert^... 2 The same objection as before holds. If we consider f(z)=1-a_2 z+a_4 z^2-a_6 z^3 +\ldots $$the fact that \{a_{2n}\}_{n\geq 1} is a positive decreasing sequence do not give that Newton's inequalities are fulfilled. If Newton's inequalities are not fulfilled, f(z) cannot have only real roots and the same applies to your original function. For instance, ... 1 If we assume r\geq 2 and we consider the integral over a large circle centered at the origin with radius R, such integral \phantom{}^{(A)} is bounded by a quantity tends to zero as R\to +\infty, so the sum of the residues (at z=-1,z=-2,\ldots,z=-r) is zero, just like the integral over |z|=2r. The case r=1 is trivial. (A): For instance, in ... 0 Yes. We have (U \cap D)^\complement = U^\complement \cup D^\complement. Then C is a connected component of this set, as it is a compact connected component of U^\complement, and it is disjoint from the closed set D^\complement. 3 There is no holomorphic function (on any neighborhood of 0) with f^{(n)}(0) = (n!)^2, but you can find a C^\infty smooth function with arbitrarily prescribed partial derivatives (this is a theorem of Borel). In the complex setup, you can for example ask for a smooth function satisfying$$ \frac{\partial^n f}{\partial z^n}(0) = (n!)^2 \qquad \frac{\...

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