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 $2$.

1.

2.

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 $\lim_{z \to z_0} \, (z-z_0)f(z)$.

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 $f(z)$.

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

Click 1. a removable singularity.

2. a pole.

3. an essential singularity.

4. the residue of $f(z)$ at $z = 0$ is $2$.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.****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.****here**for more problems.

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