How to Evaluate the Integral of z(z+1)cosh(1/z) Over a Unit Circle?

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SUMMARY

The integral of z(z+1)cosh(1/z) over the unit circle can be evaluated by parameterizing the contour with z = e^(it) for 0 ≤ t ≤ 2π. This leads to the expression ∮_C z(z + 1) cosh(1/z) dz transforming into an integral involving e^(it) and its derivatives. The differential dz is represented as i e^(it) dt, simplifying the evaluation process. Proper substitution and simplification of the resulting integral will yield the final result.

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brunette15
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Hey everyone,

I am trying to evaluate the following integral: \int z(z+1)cosh(1/z) dz with a C of |z| = 1. Can someone please guide me with how to start? I have tried to parametrise the integral in terms of t so that z(t) = e^it however the algebra doesn't seem to work...
 
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brunette15 said:
Hey everyone,

I am trying to evaluate the following integral: \int z(z+1)cosh(1/z) dz with a C of |z| = 1. Can someone please guide me with how to start? I have tried to parametrise the integral in terms of t so that z(t) = e^it however the algebra doesn't seem to work...

Please write your initial attempt.
 
brunette15 said:
Hey everyone,

I am trying to evaluate the following integral: \int z(z+1)cosh(1/z) dz with a C of |z| = 1. Can someone please guide me with how to start? I have tried to parametrise the integral in terms of t so that z(t) = e^it however the algebra doesn't seem to work...

If you parameterise the contour $\displaystyle \begin{align*} \left| z \right| = 1 \end{align*}$ with $\displaystyle \begin{align*} z = \mathrm{e}^{\mathrm{i}\,t} , \, 0 \leq t \leq 2\,\pi \end{align*}$, then $\displaystyle \begin{align*} \mathrm{d}z = \mathrm{i}\,\mathrm{e}^{\mathrm{i}\,t}\,\mathrm{d}t \end{align*}$ and we get the integral

$\displaystyle \begin{align*} \oint_C{ z\,\left( z + 1 \right) \cosh{ \left( \frac{1}{z} \right) } \,\mathrm{d}z } &= \int_0^{2\,\pi}{ \mathrm{e}^{\mathrm{i}\,t}\,\left( \mathrm{e}^{\mathrm{i}\,t} + 1 \right) \cosh{ \left( \mathrm{e}^{-\mathrm{i}\,t} \right) } \, \mathrm{i}\,\mathrm{e}^{\mathrm{i}\,t}\,\mathrm{d}t } \end{align*}$

How do you think you can go from here?
 

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