# Laurent series expansion of $\cosh z$

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 $$\sum_{n=0}^{\infty}\frac{z^{2n}}{(2n)!}$$

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

## 1 Answer

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}

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