I am working on a complex analysis exercice. I need to find the Laurent series of the function:
$$f(z) = \frac{e^z}{z + 1}$$
about $z = −1$.
I know that the result is
$$\sum_{n=-1}^{\infty}\frac{(z+1)^n}{e\cdot(n+1)!}$$
for $|z+1| > 0$, but I cannot see how one can get this result.
On
Put $z+1=:t\ $; in other words: Consider the auxiliary function $$g(t):=f(-1+t)={e^{-1+t}\over t}={1\over e\ t}\sum_{k\geq0}^\infty{t^k\over k!} ={1\over e}\left({1\over t}+1+{t\over 2}+{t^2\over 6}+{t^3\over24}+\ldots\right)\ .$$ It follows that $$f(z)=g(z+1)={1\over e}\left({1\over z+1}+1+{z+1\over 2}+{(z+1)^2\over 6}+{(z+1)^3\over24}+\ldots\right)\qquad(z\ne-1)\ .$$
You can develop the Laurent series (which is in fact a Taylor series) of $e^z$ about $z=-1$:
$$e^z = \sum_{n=0}^\infty \frac{e^{-1}}{n!}(z+1)^n = \sum_{n=0}^\infty \frac{1}{e\cdot n!}(z+1)^n$$
because $\frac{d^n}{dz^n}e^z=e^z$. Now just divide:
$$\frac{e^z}{z+1} = \sum_{n=0}^\infty \frac{1}{e\cdot n!}(z+1)^{n-1} = \sum_{n=-1}^\infty \frac{1}{e\cdot (n+1)!}(z+1)^{n}$$