The function f(z) = $\frac{\cosh(z-3i) -1}{(z-3i)^{5}}$ has one singular point in $\mathbb{C}$.

I understand that the singular point is an isolated singularity at 3i, and I know there are certain theorems about poles and zeroes to calculate the residue of this function at z = 3i. However, the numerator equals 0 at 3i, and hence I cant apply these theorems and must expand $\cosh$ as a series.

It is known that the Maclaurin Series of $\cosh(z)$ is equal to $\sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}$

Therefore, when calculating the series expansion of $\cosh$, is it as simple as stating $\cosh(z-3i) = \sum_{n=0}^{\infty} \frac{(z-3i)^{2n}}{(2n)!}$, which I presume is still a Maclaurin Series, just manipulated, or must I calculate the Taylor Series of $\cosh(z)$ centred around 3i, ie $$\cosh(z-3i) = \sum_{n=0}^{\infty} a_{n}(z-3i)^n$$, where $a_{n} = \frac{f^{(n)}(3i)}{n!}$.

My gut instinct tells me yes, as I assume you would want the Taylor Series centred around the singularity, and obviously my manipulation of the Maclaurin Series of $\cosh(z)$ to $\cosh(z-3i)$ is different to the Taylor series centred around 3i, due to both the fact of where it is centred, and also the $a_{n}$ term.

Any help on this would be appreciated.


1 Answer 1


Your approach is quite ok. $\cosh(z)$ is an entire function and its Taylor series at $z=3i$ is \begin{align*} \cosh(z-3i)=\sum_{n=0}^\infty\frac{(z-3i)^{2n}}{(2n)!} \end{align*}

We therefore obtain \begin{align*} f(z)&=\frac{\cosh(z-3i)-1}{(z-3i)^5}\\ &=(z-3i)^{-5}\sum_{n=1}^\infty\frac{(z-3i)^{2n}}{(2n)!}\\ &=\sum_{n=1}^\infty\frac{(z-3i)^{2n-5}}{(2n)!}\\ &=\sum_{n=0}^\infty\frac{(z-3i)^{2n-3}}{(2n+2)!}\\ &=\frac{1}{2(z-3i)^3}+\frac{1}{24(z-3i)}+\frac{1}{720}(z-3i)\cdots \end{align*} and observe $z=3i$ is a pole of order $3$.

Hint: Note the term Maclaurin Series is reserved for Taylor expansions around $z=0$.


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