Finding the Laurent Series for $e^{1/z}\cos(1/z)$

Question

Finding the Laurent Series for $e^{1/z}\cos(1/z)$.

First, I begin by finding the expansion for $e^{1/z}$. Recall that $$e^{z}=\sum_{j=0}^{\infty}\frac{z^{j}}{j!}, \forall z\in \mathbb{C}$$

Then, I substitute in $1/z$ for $z$ to find the expansion for $e^{1/z}$, giving: $$e^{1/z}=\sum_{j=0}^{\infty}\frac{1}{z^{j}j!}, \forall |z|>0$$

Similarly for $\cos(1/z)$: $$\cos(1/z)=\sum_{j=0}^{\infty}\frac{(-1)^{j}}{z^{2j}(2j)!}, \forall |z|>0$$.

Here, is where I am confused. I know I am supposed to use Cauchy Product to produce an answer that is "1 term," but I am confused as to what to do after using the formula. I find: $$e^{1/z}\cos(1/z)=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{(-1)^{l}}{z^{k+l}(2l)!(k-l)!}, \forall |z|>0$$.

Is this my final answer? Can I simplify this further? Perhaps is there a better way to approach this problem altogether?

Answer 1

Actually, start with the Taylor series of $e^z\cos(z)$, utilizing DanielFischer's hint

$$f(1/z)=e^z\cos(z)=e^z\left(\frac{e^{iz}+e^{-iz}}2\right)=\frac12(e^{(1+i)z}+e^{(1-i)z})$$

$$f(1/z)=\frac12\sum_{n=0}^\infty\frac{[(1+i)z]^n+[(1-i)z]^n}{n!}=\frac12\sum_{n=0}^\infty\frac{(1+i)^n+(1-i)^n}{n!}z^n$$

Finally,

$$f(z)=\frac12\sum_{n=0}^\infty\frac{(1+i)^n+(1-i)^n}{n!}z^{-n}$$