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Steps to Solve

Problem 1: Möbius Transformation

1. Analyze the properties of Möbius transformations:

   Möbius transformations map circles or lines to circles or lines. The set $E_1 = {z \in \mathbb{C} : |z| = 1}$ is a circle (specifically, the unit circle). Therefore, $f(E_1)$ will be either a circle or a line. To determine which, we can check if the image of $z=\infty$ is finite. Since $f(z) = \frac{z+i}{2z-3}$, $f(\infty) = \frac{1}{2}$, which is finite. Since $z= \infty$ is not in $E_1$ we can conclude that $f(E_1)$ will be a circle. We can see that option (C) aligns.

2. Consider $E_2$ :

   $E_2 = {z \in \mathbb{C} : z = \frac{3}{2} + te^{i\pi/4}, t \in \mathbb{R}} - {\frac{3}{2}}$. This represents a straight line passing through $3/2$ with an angle of $\pi/4$ with respect to the real axis, minus the point $3/2$. The Mobius transformation $f(z) = \frac{z+i}{2z-3}$ has a pole at $z=3/2$. As such, this point will be mapped to infinity. So, $f(E_2)$ will be a circle or a line, and since $\frac{3}{2}$ is excluded from $E_2$ its image $f(3/2)$ is excluded from the image $f(E_2)$, which in this case is $f(3/2)= \infty$. Therefore, $f(E_2)$ is a circle or line, minus a point.

To determine whether it's a circle or a line, examine the function $f(z)$. If $3/2$ were included in $E_2$, then $E_2$ would be a line that ends up mapping to a circle $f(E_2 \cup {3/2})$. Now, if we let $E_2$ be the line $E_2 \cup {3/2}$ with ${3/2}$ excluded we have a circle with $f(3/2) = \infty$ excluded. Since infinity is on the image of $E_2$ then the set $f(E_2)$ is actually a line. Since $3/2$ is excluded from $E_2$, $\infty$ is excluded from the line $f(E_2)$. Therefore, $f(E_2)$ is a line minus a point. We can eliminate (B).

3. Consider $E_3$: $E_3 = {z \in \mathbb{C} : 5|z-1| = |z-4|} - {\frac{3}{2}}$. This represents a circle, excluding the point ${3/2}$. Since Möbius transformations map circles to circles or lines, $f(E_3)$ will be a circle or a line, possibly excluding a point. Note that $3/2$ is excluded from $E_3$.

First, determine whether $5|z-1| = |z-4|$ is a circle

$5|z-1| = |z-4|$ $25(z-1)(\overline{z}-1) = (z-4)(\overline{z}-4)$ $25(z\overline{z} -z -\overline{z} + 1) = z\overline{z} -4z-4\overline{z}+16$ $24z\overline{z} -21z -21\overline{z} + 9 = 0$ $z\overline{z} - \frac{7}{8}z - \frac{7}{8}\overline{z} + \frac{3}{8} = 0$ $(z - \frac{7}{8})(\overline{z} - \frac{7}{8}) - (\frac{7}{8})^2 + \frac{3}{8} = 0$ $|z - \frac{7}{8}|^2 = \frac{49}{64} - \frac{24}{64} = \frac{25}{64}$ $|z - \frac{7}{8}|^2 = (\frac{5}{8})^2$ This is, indeed, a circle. Since $3/2$ is excluded from $E_3$, its image under f is not in f($E_3$). $f(3/2) = \infty$, since $z=3/2$ is a pole of $f(z)$. Hence f($E_3$) is a circle minus a point. We can see that option (D) aligns.

Problem 2: Power Series

1. Find the derivative $f'(z)$: Given $f(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}$, the derivative is $f'(z) = \sum_{n=1}^{\infty} \frac{nz^{n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{z^{n-1}}{n}$.

2. Calculate $zf'(z)$: $zf'(z) = z \sum_{n=1}^{\infty} \frac{z^{n-1}}{n} = \sum_{n=1}^{\infty} \frac{z^n}{n}$.

3. Find the derivative of $zf'(z)$: $(zf'(z))' = \sum_{n=1}^{\infty} \frac{nz^{n-1}}{n} = \sum_{n=1}^{\infty} z^{n-1} = \sum_{n=0}^{\infty} z^n = \frac{1}{1-z}$ for $|z| < 1$ using the formula for a geometric series.

4. Compute $(1-z)(zf'(z))'$: $(1-z)(zf'(z))' = (1-z) \cdot \frac{1}{1-z} = 1$.

Problem 3: Definite Integral

1. Use complex analysis: Consider the integral $\int_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2} dx$. We evaluate $\int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^2} dx$ and take the real part.

2. Apply Residue Theorem: Let $f(z) = \frac{e^{iz}}{1+z^2}$. The poles are at $z = \pm i$. The pole in the upper half-plane is $z=i$. The residue at $z=i$ is $$Res(f,i) = \lim_{z \to i} (z-i) \frac{e^{iz}}{(z-i)(z+i)} = \frac{e^{i(i)}}{i+i} = \frac{e^{-1}}{2i}$$

3. Evaluate the integral: By the Residue Theorem, $\int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^2} dx = 2\pi i \cdot Res(f,i) = 2\pi i \cdot \frac{e^{-1}}{2i} = \frac{\pi}{e}$.

4. Take the real part: Since $\int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^2} dx = \int_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2} dx + i \int_{-\infty}^{\infty} \frac{\sin(x)}{1+x^2} dx$. The real part is $\int_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2} dx = \frac{\pi}{e}$.