### CSIR JUNE 2011 PART C QUESTION 83 (Singularities of $f(z) = \frac{e^z + 1}{e^z -1}$)

The function $f(z) = \frac{e^z + 1}{e^z -1}$ has what type of singularity at $z=0$?
1. a removable singularity.
2. a pole.
3. an essential singularity.
4. the residue of $f(z)$ at $z = 0$ is $2$.
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Solution:
Results:
1. If $z_0$ is a removable singularity of $f(z)$ if $\lim_{z \to z_0} (z - z_0)f(z) = 0$.
2.  If $z_0$ is a pole of $f(z)$ if $\lim_{z \to z_0} (z - z_0)f(z)$ exists and non-zero.
3. The residue of the function $f(z)$ at $z = z_0$ is given by the coefficient of $\frac{1}{z}$ in the Laurent series expansion of $f(z)$. Which can be calculated by $\lim_{z \to z_0} \, (z-z_0)f(z)$.
Remark: From the above results, we observe that the limit used to check whether a singularity $z_0$ is removable or pole of $f(z)$ is same the limit used to find the residue of $f(z)$ at $z_0$. Indeed, in both the cases the limit is $\lim_{z \to z_0}(z-z_0)f(z)$. We conclude that,
1. The residue of $f(z)$ at $z = z_0$ is zero if and only if $z_0$ is a removable singularity of f(z). If particular there is no $\frac{1}{z}$ term in the Laurent series expansion of $f(z)$.
2. The residue of $f(z)$ at $z = z_0$ is non-zero if and only if $z_0$ is a pole of f(z). If particular, the coefficient of $\frac{1}{z}$ term in the Laurent series expansion of $f(z)$ is equal to the residue.

We have $e^z - 1 = z + \frac{z^2}{2} + \frac{z^3}{3!} + \cdots$. Then $\frac{e^z -1}{z} = 1 + \frac{z}{2} + \frac{z^2}{3!} + \cdots$. We observe that the series expansion of the function $h(z) = \frac{e^z - 1}{z}$ has constant term $1$ and hence $h(0) = 1$. This shows that the function $g(z) = \frac{e^z+1}{h(z)}$ is analytic at $z = 0$. In particular, $\lim z \to 0 \,g(z)$ exists. We observe that $$f(z) = \frac{g(z)}{z}$$ and $\lim_{z \to 0} (z-0) f(z) = \lim_{z \to 0} g(z) = \lim_{z \to 0} \frac{z(e^z+1)}{e^z - 1} = 2 \ne 0$. This shows that, from the results given at the beginning, $0$ is a simple pole for $f(z)$.
So option 2 is true.

Now, the residue of $f(z)$ at $z=0$ is equal to $\lim_{z \to 0} \, (z-0)f(z) = \lim_{z \to 0}g(z) = 2$. Hence option 4 is True.

### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
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