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How can I find the Laurent series expansion of $\cosh z$ in the immediate neighborhood of $z=i \pi$.

The neighborhood part confused me. Since Laurent expansion of $\cosh z$ is $\sum_{n=0}^{\infty}\frac{z^{2n}}{(2n)!}$

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    $\begingroup$ No, that is the expansion near $z=0$, not near $z=i\pi$ (i.e., you want $\cosh z=\sum_n a_n(z-i\pi)^n$). $\endgroup$ Apr 1, 2021 at 11:11
  • $\begingroup$ so what do i suppose to do? $\endgroup$
    – Aegean
    Apr 1, 2021 at 11:44
  • $\begingroup$ $coshz= \sum a_n (z-i \pi)^n$ and $a_n=\frac{1}{2 \pi i} \oint \frac{f(z)=coshz}{(z-\pi i)^{n+1} }dz$ is it true? $\endgroup$
    – Aegean
    Apr 1, 2021 at 11:51

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The function $\cosh$ is holomorphic on $\Bbb{C}$ so sure, it has a Laurent expansion, but even better, it has a Taylor expansion about any point, and the series has infinite radius of convergence. From elementary calculus, you should recall that the Taylor expansion is given as \begin{align} \cosh(z)&=\sum_{n=0}^{\infty}\frac{\cosh^{(n)}(i\pi)}{n!}(z-i\pi)^n \end{align} Now, recall that $\cosh'=\sinh$ and $\sinh'=\cosh$, so all we have to do is evaluate $\cosh(i\pi)$ and $\sinh(i\pi)$. But from the definitions it is immediate that \begin{align} \cosh(i\pi)=\cos(\pi)=-1 \qquad\text{and}\qquad \sinh(i\pi)=i\sin(\pi)=0. \end{align} So, in the Taylor expansion, only the even powers remain. So, we have \begin{align} \cosh(z)&=-\sum_{k=0}^{\infty}\frac{(z-i\pi)^{2k}}{(2k)!}. \end{align}


Alternatively, you can observe that due to how $\cosh$ is defined in terms of the exponential function, we have \begin{align} \cosh(z)=-\cosh(z-i\pi)=-\sum_{n=0}^{\infty}\frac{(z-i\pi)^{2n}}{(2n)!}. \end{align}

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  • $\begingroup$ does the minus sign come from even terms? $\endgroup$
    – Aegean
    Apr 1, 2021 at 12:16
  • $\begingroup$ @KonstantinNovoselov yes. $\cosh(i\pi)=\cosh^{(2)}(i\pi)=\cosh^{(4)}(i\pi)=\dots = -1$, whereas $\cosh'(i\pi)=\cosh^{(3)}(i\pi)=\cosh^{(5)}(i\pi)=\dots = 0$. $\endgroup$
    – peek-a-boo
    Apr 1, 2021 at 12:17
  • $\begingroup$ thanks for guiding. taylor series looks so brilliant and easier $\endgroup$
    – Aegean
    Apr 1, 2021 at 12:34

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