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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 $\partial (p(K)).$

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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: ...

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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 ...

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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 ...

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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 ...

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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 $\{f_n\}$ ...

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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 ...

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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 $0\lt||Z||\leq 1$.

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No. Just use the facts that $\sin z$ and $e^z$ are analytic and composition of analytic functions remains analytic.

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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.

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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. $\sum (-1)^{n}\frac{1}{(2n+1)!}\frac{1}{(2n+2)!}$

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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 $w(z)$, ...

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$f(z)= iz$ will satisfie this equation and it is a non constant entire function

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