I am trying to use the Cauchy integral formula and the residue theorem to show that

$$ \oint_{|z-i| = 3} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{z^2 + 9} dz= 2\pi i\cosh(1) - \pi i \cosh\left( \dfrac{9+6i}{13}\right) $$

In preparation to use the Cauchy integral formula, my first attempt was to rewrite the integral:

$$ \oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{(z-3i)(z+3i)}dz = \oint_{C} \dfrac{\dfrac{z\cosh{\left( \frac{z}{z-2}\right)}}{z+3i}} {z-3i}dz $$

However, the function $g(z) = \frac{z\cosh{\left( \frac{z}{z-2}\right)}}{z+3i}$ fails to be analytic at $z = 2$ which is an interior point of $C = \{z \in \mathbb C : |z-i| = 3 \}$. Therefore, Cauchy's integral theorem does not directly apply. Next, I decomposed one of the multiples into partial fractions:

$$ \dfrac{z}{\left(z - 3 i \right) \left(z + 3 i\right)}=\dfrac{1}{6 i + 2 z} - \dfrac{1}{6 i - 2 z} = \dfrac12\left[ \dfrac{1}{z + 3 i } - \dfrac{1}{3 i - z} \right] $$

Which gives

$$ \oint_{C} \cosh{\left( \frac{z}{z-2}\right)} \frac{z}{(z-3i)(z+3i)}dz = \dfrac12\left[ \oint_C \dfrac{\cosh{\left( \frac{z}{z-2}\right)}}{3 i + z} - \oint_C \dfrac{\cosh{\left( \frac{z}{z-2}\right)}}{3 i - z} \right] $$

This, unfortunately, does not help with the singularity at $z=2$. In addition, the point $z=3i$ is not even in the region of integration. Thus, I decided to try and solve this using the residue theorem. There are three singular points for the initial function: $\{3i, -3i, 2\}$. Since $-3i \notin C$, we only need the residues at the remaining two points.

Residue at $z=3i$:

$$ \begin{align} \lim_{z \to 3i} (z-3i)f(z) &= \lim_{z \to 3i} \dfrac{z^2 - 3zi}{z^2 + 9} \cosh\left(\frac{z}{z-2}\right)\\ &= \cosh\left(\frac{3i}{3i-2}\right)\lim_{z \to 3i} \dfrac{z^2 - 3zi}{z^2 + 9}\\ &=\cosh\left(\frac{3i}{3i-2}\right) \cdot \dfrac12 \end{align} $$

I evaluated the last limit using L'Hopital's rule.

Residue at $z=2$:

The problem is that the limit $$\lim_{z \to 2} (z-2)f(z)$$ does not exist since the left-hand limit is $-\infty$ whereas the right-hand limit is $+\infty$. Therefore, I cannot use the residue theorem. I am unsure what to try next. How do I deal with the singularity at $z=2$ and what other approaches can I try?

  • $\begingroup$ I am not sure why you are attempting to evaluate that limit when you are calculating the residue at $2$, it only applies to simple poles. You should just change the variable to $\zeta=z-2$ in the function, expand everything by binomial theorem or series for $\cosh$ and pick out the coefficient of $\zeta^{-1}$ - I don't think it's too hard. $\endgroup$ Jan 1, 2021 at 10:45
  • $\begingroup$ It is surely an essential singularity, not a pole? $\endgroup$ Jan 1, 2021 at 11:48
  • 1
    $\begingroup$ In fact, given your comments on Don Antonio's solution, I suggest that you revise the definitions of pole, singularity, residue - you seem hung up on simple calculational rules, but these only apply in special cases. $\endgroup$ Jan 1, 2021 at 11:51
  • $\begingroup$ Alright, I'll refer to a different textbook; perhaps the one I'm currently using is poorly worded. $\endgroup$
    – E.Nole
    Jan 1, 2021 at 12:49
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    $\begingroup$ Verify that $z = 2$ is an essential singularity. Find the residues at $z = -3 i$ and $z = \infty$ instead. The expansion around infinity has the form $$(\cosh 1 + O(|z|^{-1})) (z^{-1} + O(|z|^{-2})).$$ $\endgroup$
    – Maxim
    Jan 1, 2021 at 14:42

1 Answer 1


Some ideas for you to evaluate the residue at $\;z=2\;$ of $\;\cosh\frac z{z-2}\;$ (Check carefully the calculations, though the algorithm is the important thing here: sometimes we need, and some even would say must, calculate Laurent series) :

Since $\;\cosh z=\cos(iz)\;$, we get that

$$\cosh z=\sum_{n=0}^\infty \frac{ z^{2n}}{(2n)!}\implies \cosh\frac z{z-2}=\sum_{n=0}^\infty \frac{ \left(\frac z{z-2}\right)^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{ z^{2n}}{(2n)!}\cdot\frac1{(z-2)^n}$$


$$z=2+(z-2)=2\left(1+\frac{z-2}2\right)\implies\cosh z=\sum_{n=0}^\infty \frac{2^n\left(1+\frac{z-2}2\right)^n}{(2n)!}\cdot\frac1{(z-2)^n}$$

Let us try to calculate the coefficient $\;c_{-1}\;$ of $\;(z-2)^{-1}\;$ , which is the residue of $\;\cosh\frac z{z-2}\;$ at $\;z=2\;$ . First we write

$$\left(1+\frac{z-2}2\right)^n\cdot\frac1{(z-2)^n}=\left(\frac1{z-2}+\frac12\right)^n=\sum_{k=0}^n\binom nk(z-2)^{-k}\cdot 2^{-n+k}$$

Thus, for each $\;n=1,2,....\;$, the coefficient of $\;(z-2)^{-1}\;$ is given when $\;k=1\;$ (not for $\;n=0\;$...!), and thus it is given by

$$\sum_{n=1}^\infty\binom n1\frac1{2^{n-1}}=\sum_{n=1}^\infty\frac{n}{2^{n-1}}=4\;$$

For the last series' value: differentiate $\;\cfrac1{1-z}\;$ and substitute the correct value of $\;z$

  • $\begingroup$ What do you mean by last series' value? And I'm unclear how exactly to use this in the original problem. That is, the residue at $z=2$ of $\cosh \frac{z}{z-2}$ won't equal that of $\cosh{\left( \frac{z}{z-2}\right)} \frac{z}{z^2 + 9}$ since the terms in the Laurent series will each get multiplied. So, is there any particular reason for calculating the residue at $z=2$ of just part of the function? $\endgroup$
    – E.Nole
    Jan 1, 2021 at 11:41
  • $\begingroup$ You're right: I only gave a way to evaluate the residue fo the hyperbolic cosine, which is the function that was giving you problems. You can know learn from this how to do it with the exact integrand function you have...and do it by yourself. And the last series' value is that $\;4\;$ and that line that appears there almost at the end of my answer. $\endgroup$
    – DonAntonio
    Jan 1, 2021 at 12:37
  • $\begingroup$ Oh, I get it now, thanks. You meant to use $$\sum_{n=1}^{\infty}z^n=\frac1{1-z}$$ $\endgroup$
    – E.Nole
    Jan 1, 2021 at 13:11
  • $\begingroup$ @E.Nole Yup...bu the sum begin with $\;n=0\;$ and you should try to differentiate both sides (you can do it termwise inside the convergence interval, which is $\;(-1,1)\;$ $\endgroup$
    – DonAntonio
    Jan 1, 2021 at 13:19
  • $\begingroup$ When you made the substitution $z=2\left(1+\frac{z-2}2\right)$, shouldn't the exponent in the sum be $2n$? That is $$ \sum_{n=0}^\infty \frac{2^\color{red}{2n}\left(1+\frac{z-2}2\right)^\color{red}{2n}}{(2n)!}\cdot\frac1{(z-2)^n} $$ $\endgroup$
    – E.Nole
    Jan 1, 2021 at 14:07

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