[Math] Complex power series expansion of $\frac{e^z}{1+z}$

complex-analysispower series

I'm trying to find complex power series expansion of $\frac{e^z}{1+z}$ centered at $z=0$ and its radius of convergence. Here is my attempt:

Since $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$, we can divide both terms by $1+z$ to get $e^z = \sum_{n=0}^\infty \frac{z^n}{(1+z)n!}$. Then I can get radius of convergence using the usual Cauchy-Hadamard formula.

Is this correct?
Thanks for any help!

Best Answer

In order to get the power series expansion we could multiply the series $e^z$ with $\frac{1}{1+z}$ using the Cauchy product

We obtain \begin{align*} \frac{e^z}{1+z}&=\left(\sum_{k=0}^{\infty}\frac{z^k}{k!}\right)\left(\sum_{l=0}^\infty (-z)^l\right)\tag{1}\\ &=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{(-1)^l}{k!}\right)z^n\tag{2}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\right)z^n \end{align*}

Comment:

  • In (1) we use the series expansion for the exponential function and the geometric series expansion

  • In (2) we multiply the series using the Cauchy product formula

We observe the function \begin{align*} \frac{1}{1+z} \end{align*} has a simple pole at $z=-1$. We also know that the exponential function is an entire function, i.e. analytic in $\mathbb{C}$. Since the radius of convergence is the distance from the center $z=0$ to the nearest singularity, we conclude the radius is $1$.

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