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HINT.-$$\frac{c_kx}{(x_k-ix)^2}=\frac{c_k}{x_k-ix}\cdot \frac{x}{x_k-ix}$$ $$\left|\frac{c_kx}{(x_k-ix)^2}\right|=\left|\frac{c_k}{x_k-ix}\right|\cdot \left|\frac{x}{x_k-ix}\right|\le\left|\frac{c_k}{x_k-ix}\right|\cdot\left|\frac{x^2}{x_k^2+x^2}\right|\le \left|\frac{c_k}{x_k-ix}\right|$$

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To show this function is analytic, you need to show that the Cauchy Riemann equations are valid; equivalently, the differential operator below satisfies $$\frac{\partial g}{\partial \bar z} = \frac{1}{2}\bigg( \frac{\partial g}{\partial x} - \frac{1}{i}\frac{\partial g}{\partial y} \bigg) = 0.$$ If you are unfamiliar with this very useful formulation of ...

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Since $b_m\neq 0$, the fraction $$\frac{a_0 z^{m-2}+a_1 z^{m-3}+a_2 z^{m-4}+\cdots +z_n z^{m-n-2}}{b_0 z^m+b_1 z^{m-1}+b_2 z^{m-2}+\cdots +b_m}$$ is defined and continuous at $z=0$ (the denominator does not vanish). This means that the singularity of $$g(z)=\frac{1}{z^2} \cdot \frac{P(1/z)}{Q(1/z)}$$ at $z=0$ is removable, so $g(z)$ can be extended to be ...

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The intersection of $\{\gamma\}$ and $\partial G$ is trivial, since the image of $\gamma$ is contained in $G$. There exists $(x_n), x_n\in \{\gamma\}$ $(y_n\in \partial G)$ such that $lim_nd(x_n,y_n)=d(\{\gamma\},\partial G)$. Since $\{\gamma\}$ is compact, there exists a subsequence $(x_{f(n)}$ of $(x_n)$ which converges towards $x\in\{\gamma\}$, this ...

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For a fixed set, the distance from it is a continuous function. If the set is closed, the distance is positive for each point not contained in it. A continuous function on a compact set has a minimum.

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Because the distance between disjoint sets, one of which is closed and the other compact, is positive. Since the boundary of $G$ is closed, your statement follows.

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