Evaluating the integral $\oint \coth(\frac{1}{2}z) dz$ taken over the contour $C$ and $|z-\frac{\pi}{2}i|^{2} = 1$

Question

I would highly appreciate your valuable feedback on the consistency of my demonstration:

The integral $\oint \coth\left( \frac{1}{2} \cdot z \right)\, \operatorname{d}z$ is taken over the contour $C$, where $C$ is the circle $\left| z - \frac{\pi}{2} \cdot i \right|^{2} = 1$, traversed in a clockwise direction. Now, let's determine whether Cauchy's theorem applies:

1. Yes because the points $\pm 4 \cdot \pi \cdot i$ at which $\sinh\left( \frac{1}{2} \cdot z \right) = 0$ all lie outside the contour of integration, so that Cauchy's theorem applies.
2. Yes because the points $\pm 2 \cdot n \cdot \pi \cdot i$ at which $\cosh\left( \frac{1}{2} \right) \cdot z = π$ all lie outside the contour of integration, so that Cauchy's theorem applies.
3. No because the points $+/-4 \cdot \pi \cdot i$ at which $\sinh\left( \frac{1}{2} \cdot z \right) = 0$ all lie inside the contour of integration, so that Cauchy's theorem does not apply.
4. No because the points ($\pm 2 \cdot n \cdot \pi \cdot i$) at which cosh $\cosh\left( \frac{1}{2} \cdot z \right) = \pi$ all lie outside the contour of integration, so that Cauchy's theorem does not apply.

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Hello everyone, I have tried to solve this problem and I got the correct answer $1$. To find the singularities, we set the denominator of $\coth\left( \frac{1}{2} \cdot z \right)$ equal to zero:

1. $\frac{1}{2} \cdot z = \left( 2 \cdot n + 1 \right) \cdot \pi \cdot i$.

2. Now, we solve for $z$: $z = \frac{2}{2 \cdot n + 1} \cdot \pi \cdot i$. So, the singularities of $\coth\left( \frac{1}{2} \cdot z \right)$ are at $z = \pm 4 \cdot \pi \cdot i,\, \pm 2 \cdot \pi \cdot i,\, \pm 2 \cdot i,\, \pm \frac{2}{3} \cdot \pi \cdot i,\, \pm \frac{2}{5} \cdot \pi \cdot i,\, \dots$, and so on.

3. Next, let's check whether these singularities lie outside the contour $C$, which is the circle $\left| z - \frac{\pi}{2} \cdot i \right|^{2} = 1$:

   3.1 Let's first consider $z = \pm 4 \cdot \pi \cdot i$: $\left| \pm 4 \cdot \pi \cdot i - \frac{1}{2} \cdot \pi \cdot i \right|^{2} = \left| \pm 3.5 \cdot \pi \cdot i \right|^{2} = \left( 3.5 \cdot \pi \right)^{2} \ne 1$. Since the absolute value of $z$ at $\pm \cdot \pm \cdot i$ is greater than $1$, these singularities lie outside the contour $C$.

   3.2 Now, let's consider $z = \pm 2 \cdot \pi \cdot i$: $\left| \pm 2 \cdot \pi \cdot i - \frac{1}{2} \cdot \pi \cdot i \right|^{2} = \left| \pm 1.5 \cdot \pi \cdot i \right|^{2} = \left( 1.5 \cdot \pi \right)^{2} \ne 1$. Similarly, since the absolute value of z at $\pm 2 \cdot \pi \cdot i$ is greater than $1$, these singularities also lie outside the contour $C$.

   3.3 Continuing this analysis, we find that all the singularities of $\coth\left( \frac{1}{2} \cdot z \right)$ lie outside the contour $C$.

   Conclusion: Therefore, since all the singularities lie outside the contour $C$, Cauchy's theorem applies, and we can evaluate the integral $\oint \coth\left( \frac{1}{2} \cdot z \right)\, \operatorname{d}z$ over the contour $C$ so the correct answer should be $1$.