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Since $$\cosh(z)=0\iff e^z=-e^{-z}\iff e^{2z}=-1\iff z=\left(k+\tfrac12\right)\pi i,k\in\mathbb{Z}$$ the singularities of $\frac{\cosh(az)}{\cosh(z)}$ are at $z=\left(k+\tfrac12\right)\pi i$ for $k\in\mathbb{Z}$. The residue of $\frac1{\cosh(z)}$ at $z=\left(k+\tfrac12\right)\pi i$ is $$\frac1{\sinh\left(\left(k+\tfrac12\right)\pi i\right)} ... 2 The point is that Louville's Theorem can be used to provide a proof of the Fundamental Theorem of Algebra. Suppose there exists a polynomial P(z) of degree at least 1 with no complex roots. Then f(z) = 1/P(z) is holomorphic on the entire complex plane (entire). Once you show that \vert P(z) \vert is unbounded on \mathbb{C}, you can conclude that ... 1$$\int_C x dz=\int_C \frac{z+\overline z}{2} dz\\ \left(z=e^{i\theta}, dz=izd\theta\right)\\ =\int_0^{2\pi}\frac{e^{i\theta}+e^{-i\theta}}{2}ie^{i\theta}d\theta =\int_0^{2\pi}i\frac{e^{2i\theta}+1}{2}d\theta\\ =\left[\frac{e^{2i\theta}}{4}+i\frac{\theta}{2}\right]_0^{2\pi}=\pi i $$1 If I understand your question correctly, it is the following: Let \epsilon > 0 be given. For what \alpha, \beta \in \mathbb C does the following hold:$$ \left|\frac{\alpha+z_0}{\beta z_1+z_0}\right|<\epsilon \quad \text{ for all } z_0, z_1 \in \mathbb C \text{ with } |z_0|<\epsilon \text{ and } |z_1|<\epsilon \, . $$First let ... 1 You have developed the partial fractions into a power series or Laurent series at z_0 =0 and z_0 = \infty, which unfortunately does not help to get the Laurent series at z_0 = i. Instead you have proceed as follows: For |z-i| < 2 you have$$ \frac{1}{z+i} = \frac{1}{(z-i) + 2i} = \frac{1}{2i(\frac{z-i}{2i} + 1)} = -\frac i2 \frac{1}{1 - \frac ...

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The absolute value of $i$ is $1$ and so it is removed (using $|ab| = |a||b|$). Regarding your work, note that in the second line, you replace $|\zeta - z|^2$ in the denominator of the integrand with $R^2$. If $z = 0$, then indeed $|\zeta - z|^2 = R^2$ but if $z \neq 0$, you need to estimate $|\zeta- z|^2$ and it is not true with your assumptions that ...

1

$$\left|\frac{az+b}{cz+d}\right|=\left|\frac{a(z+\tfrac{b}{a})}{c(z+\tfrac{d}{c})}\right|=\left|\frac{a}{c}\cdot\frac{z+\tfrac{b}{a}}{z+\tfrac{d}{c}}\right|=\left|\frac{a}{c}\right|\cdot\left|\frac{z+\tfrac{b}{a}}{z+\tfrac{d}{c}}\right|=\frac{|a|}{|c|}\cdot\frac{|z+\tfrac{b}{a}|}{|z+\tfrac{d}{c}|}.$$

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We consider the integral of analytic function $$F(z)=\frac{\cosh(az)}{\cosh(z)}$$ on the rectangle contour of $C=C_1\cup C_2\cup C_3\cup C_4$, which are $y=0, \:x=R,\:y=\pi,\:x=-R$ respectively. By Cauchy's residue theorem, we have $$\int_{-R}^RF(x)dx+\int_{C_2}F(z)dz+\int_{C_3}F(z)dz+\int_{C_4}F(z)dz=2\pi iRes\left(F,\frac{\pi i}{2}\right)\tag1$$ Note ...

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That is not correct because $| e^{iz} | = 1$ holds only for real numbers $z$: $$| e^{iz} | = | e^{i(x+iy)} | = | e^{ix - y} | = | e^{ix} | |e^{-y}| = e^{-y} \, .$$ And since $$|f(-iy)| = \frac{y e^y}{y^4+1} \to \infty \text{ for } y \to +\infty$$ such an $R(\epsilon)$ can not exist. If $z$ is restricted to the upper halfplane then $| e^{iz} | < 1$ ...

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I give some more or less useful criteria, but mostly without proofs, so these should be regarded as exercises. Some conditions use popular modular forms and functions. Other conditions are elementary, using only divisibility or requiring some expressions to evaluate to integers. You do not need modforms for those, so if you are only interested in elementary ...

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