# Tag Info

0

As you say, by the ratio test the radius of convergence as a power series about $0$ is indeed $R=1$, so the series must converge in $D=\{z \in \mathbb{C} : |z|<1 \}$. Lets see what happens if $z \in \partial D=\{ z: |z|=1\}$. Consider the set $$E=\{ e^{2\pi i \cdot r} : r \in \mathbb{Q} \} \subset \partial D$$ For every $z \in E$, we have $$... 0 What I'd do: We have |P(z)/Q(z)| on the order of 1/|z|^2 as |z| \to \infty. By Cauchy,$$\int_C \frac{P(z)}{Q(z)}\,dz = \int_{|z|=R} \frac{P(z)}{Q(z)}\,dz$$for large R. Use the "M-L" estimate on the latter to see its absolute value is no more than a constant times (1/R^2)\cdot2\pi R. The latter \to 0 as R\to \infty. Conclusion: \int_C ... 0 Part of the solution: Let c>0. Suppose there is no such \epsilon. Then there exists a sequence f_n(z)=\sum_{k=1}^{\infty}a_k(n)z^k, with \sum_k |a_k(n)|\le c for each n, such that$$\sup_{[1/2,3/4]}|1-f_n(x)| \to 0as n\to \infty. Note that I use [1/2,3/4]; I think the full interval [1/2,1] is a bit of a red herring. Now the condition ... 0 After working on it I believe this is a counterexample. Let f(z)=z=x+iy=Re(f)+iIm(f). Then f(z) is entire. Furthermore for u=x we have u_{xx}+u_{yy}=0. Thus u is harmonic. In particular it is harmonic on D(0,1). Take \gamma=e^{i\theta} for 0\leq\theta\leq 2\pi. Then \begin{align*}\int_{\gamma}u=\int_\gamma Re(f) ... 0 Take the closed rectangular contour \Gamma  counterclockwise oriented in the complex plane from -T  to T , a vertical segment from T  to T+ia , a segment from T+ia  to -T+ia  and the last segment from -T+ia  to -T . Take f\left(z\right)=e^{-z^{2}} . Then by Cauchy theorem we have ... 1 f(z) anti-holomorphic over \mathbb{C} if and only if f(z) = g(\bar{z}) for some holomorphic function g(z). Notice on the unit circle C(0,1), \bar{z} = \frac{1}{z}, we have \frac{1}{2\pi i}\oint_{C(0,1)} \frac{f(z)}{z-z_0} dz = \frac{1}{2\pi i}\oint_{C(0,1)} \frac{g(\bar{z})}{z-z_0} dz = \frac{1}{2\pi i}\oint_{C(0,1)} ...

0

As Travis pointed out, there are very few conformal maps in dimension higher than $2$. The natural generalization of Riemann's mapping theorem to higher dimension is to ask, not for a conformal bijection, but for a biholomorphic map, i.e. a bijective holomorphic map $\Omega \to \Omega'$ with a holomorphic inverse. Such maps do not preserve all angles, but ...

2

Anything can happen, pretty much. Using $R=1$ as an example: $\displaystyle \sum_{k=1}^\infty \frac1{k^2}z^k$ converges absolutely for all $|z|=1$; $\displaystyle \sum_{k=1}^\infty \frac1{\sqrt k}z^k$ converges conditionally for most $|z|=1$ but diverges at $z=1$; $\displaystyle \sum_{k=1}^\infty kz^k$ diverges for all $|z|=1$.

1

Taylor expanding $\cos z$ about $z=\frac{\pi}{2}$ one finds that: $$\cos z=-\left(z-\frac{\pi}{2}\right)+\frac{1}{6}\left(z-\frac{\pi}{2}\right)^3+\ldots$$ Similarly, Taylor expanding $\frac{1}{z+\frac{\pi}{2}}$ about $z=\frac{\pi}{2}$ one finds that: ...

9

I believe you mean $|e^{i\lambda z}| = |e^{i\lambda x}||e^{-\lambda y}|$. This equality follows from the fact that $z = x + iy$: $$e^{i\lambda z} = e^{i\lambda(x+iy)} = e^{i\lambda x - \lambda y} = e^{i\lambda x}e^{-\lambda y}.$$ Then, as $\lambda, x \in \mathbb{R}$, $|e^{i\lambda x}| = 1$. To see this, note that $e^{i\lambda x} = \cos(\lambda x) + ... 0 Let$U$be the larger domain. Certainly$f$is analytic in$U\setminus \{1\}.$But$f$is also analytic in$D(1,1).$Why? Because it equals a power series centered at$1$there. Thus for any point in$U,f'(z)$exists. Hence$f$is analytic in$U.$3 Phragmén-Lindelöf is not the right tool here. The person to turn to is Paul Montel. Since$f$is bounded, the family$\mathscr{F} = \bigl\{ f_n \colon z \mapsto f(2^{-n}\cdot z) \,\;\big\vert\;\, n\in \mathbb{N}\bigr\}$is normal. Since$(f_n)$converges (locally uniformly) to$0$on$(0,+\infty)$, it follows by normality that$f_n \to 0$locally uniformly ... 1 If such a function$g$existed, its integral over any closed curve would be zero by Cauchy's theorem. However,$\oint_{|z|=2}g(z)dz$is obviously nonzero. 2 Suppose$\;g(z)\;$is entire (i.e. analytic in the whole complex plane) and$\;g(z)=\dfrac1z\;$on$\;|z|>1\;$. Since$\;\dfrac1z\;$is analytic on$\;0<|z|\le1\;$as well,$\;g(z)$continues analitically$\;\dfrac1z\;$on the punctured unit disk. But since for any$\;z_k\to 0\;$have that$\;g(z_k)\to g(0)\;$and$\;g(0)\;$is well defined, we get ... 0 We need case 2, since if$f$is meromorphic on$D_r(0) \setminus \{0\}$and has a sequence of poles converging to$0$, then$0$is not an isolated singularity of$f$, and thus Picard's theorem doesn't apply. It may be that Picard's theorem applies to a simple transformation of$f$, consider e.g. $$f(z) = \frac{1}{\sin \frac{1}{z}},$$ where$1/f$has an ... 1 Yes, a continuous function with the "discrete mean value property" is harmonic. Such a function satisfies the maximum (and minimum) principle in the form If$u$has a local maximum (minimum) at$z_0$, then$u$is constant on a neighbourhood of$z_0$. Postponing the proof of this, we use it to deduce the harmonicity: Fix an arbitrary$z_0\in ...

1

Putting $z=x+iy$ I get $$|f|^2=(x^2-y^2-y+3)^2+(2xy+x-1)^2$$ Now differentiating, and applying the condition $x^2+y^2=1$ I get the critical $x$ values to be roots of the equation $$2x^3-17x+3=0$$ The roots are $-3$ and $\frac 32 \pm \frac {\sqrt7}{2}$ and of these only $\frac 32-\frac {\sqrt7}{2}$ satisfies the constraint. It is then a matter of ...

1

$|f|^2$ for the value $e^{it}$ is $$6 - 2 \sin t - \sin 2t - \cos t + 3 \cos 2 t$$ Some plotting suggests that the maximum is achieved at for $t= \pi$, value 10. Checking: \begin{eqnarray} 10 -(6 - 2 \sin t - \sin 2t - \cos t + 3 \cos 2 t) = 4 + 2 \sin t + \sin 2t + \cos t - 3 \cos 2 t = \\ = 2 \cos^2 (t/2)\cdot ( \cos^2 (t/2) + 13 \sin^2 (t/2)\, ) \ge 0 ...

1

One (not elegant solution) is to substitute $z=a+bi$ with $a^2+b^2=1$, than use Lagrange's multipliers to find $|f|^2$ extremum (as function of $a$ and $b$) on unit circle.

0

Yes, if none of them are in the circle the answer is just $0$. Cauchy's theorem says that if the function is holomorphic in the inside of $\Gamma$, and the inside is simply connected, then the integral is $0$. THis is precisely your case. The problem with curves that go around $0$ with functions like $\frac 1z$ is that the function is holomorphic in ...

0

About the easy-to-see part in @N.S.'s answer on this same page. On the unit circle $1/z = \cos(t)-i\sin(t)$, integrating which over $0\leq t\leq 2\pi$ is clearly zero. The confusion that might arise is the implied integrand $\mathrm{d}z$. $$\int_{C} \frac{1}{z} \; \mathrm{d}z$$ When integrating in $\mathbb{R}^2$, $\mathrm{d}x$ is a function that reflects ...

1

No, this is only true for $\Bbb C^1 \leftrightarrow \Bbb R^2$. By (one of the many significant results called) Liouville's Theorem (in particular, not the well-known theorem from complex analysis that bears that name), for $n > 2$ any conformal isometry from one open subset of $\Bbb R^n$ to another is actually a restriction of a global isometry from the ...

1

$f$ is continuous. Then $$|f(-r)|\le\ln r+\pi$$ for $r>0$.

1

One can also prove the theorem using Nevanlinna theory. See here, for example.

1

You're right that the Schwarz reflection principle is applicable here. We start with $f$ analytic on the strip $\{-1<\operatorname{Re} z<1\}$ and then starting with the right side, we "flip" $f$ over about the line $\operatorname{Re}z=1$. That is, we define $f$ analytically in $\{1\leq\operatorname{Re}z\leq3\}$ by $f(z)=\overline{f(\overline{1-z}+1)}$ ...

0

I am not sure if I understood well your question. So forgive me if the following answer does not satisfy you. Anyway, if you have an equation like $$x^2-6x+10=0$$ not having real solutions, you can still plot the function $$f(x)=x^2-6x+10.$$ See the blue curve below. (The coefficient belonging to $x^2$ will be $1$ during this argumentation.) The fact that ...

1

Step 0: A proper holomorphic map is not constant. For if $f(z) \equiv w_0$, then $f^{-1}(\{w_0\}) = D_1$ is not compact, although the singleton set $\{w_0\}$ is compact. Step 1: For each $w\in D_2$, the fibre $f^{-1}(\{w\})$ is finite. Since $f$ is not constant, the fibre is discrete. A discrete compact space is finite. Step 2: The function $N\colon D_2 ... 0 Given$y = mx + b$, find the perpendicular to it from the origin. That line is given by$y = -\frac{1}{m}x$. The complex number$1 + \frac{1}{m}i$will rotate and scale the first line so that it's vertical. To find the real part, compute the real part of$(1+\frac{1}{m}i)bi$, which is$\frac{-b}{m}$. So that gives$\Re((1+\frac{1}{m}i)z) = \frac{-b}{m}$1 $$f(z)=\dfrac{1}{z_0\left(1+\dfrac{z-z_0}{z_0}\right)}=\dfrac{1}{z_0}\cdot\dfrac{1}{1+\dfrac{z-z_0}{z_0}}=\dfrac{1}{z_0}\cdot\sum_{k=0}^\infty (-1)^k\left(\dfrac{z-z_0}{z_0}\right)^k$$ $$\therefore f(z)=\dfrac{1}{z_0}\left[1-\left(\dfrac{z-z_0}{z_0}\right)+\left(\dfrac{z-z_0}{z_0}\right)^2-\left(\dfrac{z-z_0}{z_0}\right)^3+\ldots\right]$$ 0 All lines in$\mathbb C$can be written as $$\beta z + \bar\beta \ \bar z + \gamma = 0$$ where$\beta \in \mathbb C$,$\gamma \in \mathbb R$. Now write$z = x + iy$and simplify the expression; impose that it should be equal to$y = mx + b$and you'll find$\beta$0 Check the book for notation, but from the above, it's clear that$\mathbb{T} = \partial\mathbb{D}$If$p$is a polynomial in$n $variables, then$q(z_1) = p(z_1,0,0,\ldots,0)$is a polynomial in$z_1$, so if$w\in \bar{\mathbb{D}}$, then $$|q(w)| \le \max_{\mathbb{T}} |q| = \max_{\mathbb{T}\times(0,\ldots,0)} |p|$$ Hence$\bar{\mathbb{D}}\times(0,\ldots,0) ...

0

You do not need any particular expansion to find the residues of: $$f(z)=\frac{\pi}{(2z+1)^3\sin(\pi z)}.$$ The zeroes of $\sin(\pi z)$ occurs at $z\in\mathbb{Z}$ and they are all simple zeroes, hence $f(z)$ has simple poles at $z\in\mathbb{Z}$ and a double pole at $z=-\frac{1}{2}$. We trivially have: $$\forall n\in\mathbb{Z},\quad\text{Res}(f(z),z=n) = ... 1 HINT: It's all due to the Euler's formula. Recall that for a complex number z = x+ iy we have$$ e^{i\theta} = \cos \theta + i \sin \theta \iff z = x+ i y = |z| e^{\operatorname{arg}z} = \sqrt{x^2 + y^2}\,e^{\,\operatorname{atan}\!\frac{y}{x}} $$The line x=c  can be written as  z_1 = c + i\cdot0. The z_1 gets mapped into z_2 by  w(z), ... 1 No. Just use the facts that \sin z and e^z are analytic and composition of analytic functions remains analytic. 0$$z^4=-1=e^{\pi i\left(1+2k\right)}\;,\;\;k=0,1,2,3\implies z_k=e^{\frac{\pi i}4(1+2k)}$$thus in the first quadrant we have only one (simple) pole:$$z_0=e^{\pi i/4}=\frac1{\sqrt2}(1+i)$$The residue here is$$\lim_{z\to z_0}(z-z_0)\frac1{z^4+1}\stackrel{l'Hospital}=\lim_{z\to z_0}\frac1{4z^3}=\frac1{4e^{3\pi ...

1

I know this post is old, but maybe this answer helps someone. then why do they only use $(1−e^{2iz})$? This change makes it easier to solve the integral because of limit properties, while having the same zeroes as the sine. The post you linked in short: Using Cauchy's theorem, the contour integral of a holomorphic function (at least inside and ...

1

Looks like you only took the $z$ term from the sine power series and the $\frac{1}{2z^2}$ term from the series of the exponent. What you should take is an infinite sum of terms over all powers that sum up to -1. $\sum (-1)^{n}\frac{1}{(2n+1)!}\frac{1}{(2n+2)!}$

0

$$\int_{W} \frac{e^{\frac{1}{z}}}{(z-3)^3} dz$$ $$=2\pi i[Res(f,0)+Res(f,3)]=-2\pi iRes(f,\infty)=0$$where , $f(z)=\frac{e^{\frac{1}{z}}}{(z-3)^3}$. Since, sum of residues at finite poles and the residue at $\infty$ is equal to zero. Since, $Res(f,\infty)=-\lim_{z\to \infty}zf(z)$ , as $f(\infty)=0$ $=-\lim_{z\to \infty}\frac{z}{(z-3)^3}e^{1/z}=0$.

0

Easy Cauchy theorem: Assume $U$ is open and convex, and $f\in H(U).$ If $\gamma$ is a closed contour in $U,$ then $\int_\gamma f(z)\,dz =0.$ Now apply this with $U=\{\operatorname {Re}z > -2\}, f(z) =\operatorname {Log} (z+2),$ and $\gamma = C.$

2

You can actually prove this using Fourier series: consider the Fourier series of $\cos{\alpha x}$ on $[-\pi,\pi]$. The function is continuous, and differentiable except at the endpoints, so the Fourier series converges to the function everywhere. Some easy integration will show you that the Fourier coefficients are such that the series is $$\cos{\alpha x} = ... 1 Some ideas. There's quite a few things to justify, among them taking the Principal Value of the series after the logarithmic differentiation. Take the infinite product for the sine function:$$\sin \pi z=\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)=\pi z\prod_{0\neq n=-\infty}^\infty\left(1+\frac zn\right)\implies\log\sin\pi z=\log\pi ...

1

The concepts are not equivalent: The open subset $\Omega=\mathbb C^2\setminus\{0\}\subset \mathbb C^2$ is Runge but not polynomially convex: indeed for $K$ the unit sphere $||Z||=1$ centered at the origin $\hat K$ is the non compact set $0\lt||Z||\leq 1$.

1

This is not true. Consider $f=\dfrac{1}{z-2}$ with a singularity at $z=2\in \bar{\Omega}$. Obviously, $f\in H(\Omega)$. Suppose there exists $\{f_n\}\in H(A)$ with $\bar{\Omega}\subset A$, s.t. $f_n\to f$ uniformly in $\Omega$. Then we pick up the sequence of points $z_m=2+\dfrac{i}{m}\in \Omega$ which converges to $z=2$. Since the converge of $\{f_n\}$ ...

1

If we take $\operatorname{Log}(z)$ along its principal branch, wherein $\operatorname{Log}(z) = \ln|z| + i \operatorname{Arg}(z)$, for $- \pi < \operatorname{Arg}(z) \leq \pi$, then we know that $\operatorname{Log}(z)$ is analytic everywhere except the real axis where $z \leq 0$. It follows that $\operatorname{Log}(z+2)$ is analytic everywhere except ...

1

Use the Cauchy-Riemann equations for $f(x+iy) = x^2$, with $u(x,y) = x^2$ and $v(x,y) = 0$ in the usual notation. Then: $$\begin{cases} u_x = v_y \\ u_y = -v_x\end{cases} \implies \begin{cases} 2x = 0 \\ 0 = 0\end{cases}$$ The equations are only satisfied for $x = 0$ (and any $y$). The function is differentiable only in the imaginary axis.

0

There is no singularity at $1$. It is canceled out by the double zero of $\sin(z-1)^2$.

2

Yes. the pole's are $\sqrt{2},-\sqrt{2}$ and $0$. All of them are simple pole's. Because: $$f(z)=\frac{\frac{\mathbb e^z\sin(3z)}{(z-\sqrt2)z^2}}{z+\sqrt2}=\frac{\frac{\mathbb e^z\sin(3z)}{(z+\sqrt2)z^2}}{z-\sqrt2}=\frac{\mathbb e^z\frac{\sin(3z)}{z(z^2-2)}}{z}$$ $0$ is simple, because the function $\frac{\sin(3z)}{z}$ is analytic. Since: ...

1

Well, it looks like $\pm \sqrt{2}$ are your two obvious simple poles (Multiplicity 1). However, to analyze the $z^2$ term, we should expand $e^z$ and $\sin(3z)$ into their Taylor series representations to see what happens. $$e^z = 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + ...$$ $$\sin(3z) = 3z + \frac{27z^3}{6} + \frac{243z^5}{120} + ...$$ By doing this we ...

2

No, consider $p(re^{it}) = (\sin r )e^{it}.$ Then $p$ maps $K = \overline {D(0,3\pi /4)}$ onto $\overline {D(0,1)}$ but $p(\partial K)$ doesn't even intersect $\partial (p(K)).$

0

Proof: Since $K$ is compact, we may impose the following assumptions on $K$: $K$ consists of a finite union of closed balls. The poles of $K$ are contained in the interior $K^\circ$. Let $z_1,z_2,\ldots,z_k$ be the poles of $f$ in $K$ with multiplicities $m_1,m_2,\ldots,m_k$. Define $q(z)=(z-z_1)^{m_1}\cdots(z-z_k)^{m_k}$, and define $g(z)=f(z)/q(z)$, ...

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