What is the Laurent series of $\exp(\frac{1}{z})\exp (2z)$?

Question

What is the Laurent series of $\exp(\frac{1}{z})\exp (2z)$ ?

I know how to do the Laurent series of $ \dfrac{1}{z(z+5)}$ (I make use of the geometric series of $\dfrac{1}{1-z}$) but I don't know how to do the above.

Answer 1

Using the power series for the exponential function, which converges to the function for all $\;z\in\Bbb C\;$ :

$$e^{1/z}e^{2z}=e^{\frac{2z^2+1}z}=\sum_{n=0}^\infty \frac{(2z^2+1)^n}{z^nn!}$$

Added on request : Observing carefully the above, we realize $\;z^{-1}\;$ appears only when $\;n\;$ is odd, and then:

$$n=2k-1\implies\frac{(2z^2+1)^{2k-1}}{z^{2k-1}}=\frac1{z^{2k-1}}\sum_{m=0}^{2k-1}\binom{2k-1}m2^mz^{2m}$$

and in the last summand, we'll have

$$2m=2k-2\iff m=k-1\implies\;\text{the coefficient is}\;$$

$$\binom{2k-1}{k-1}\frac{2^{k-1}}{(2k-1)!}=\frac{2^{k-1}}{k!(k-1)!}$$

so the coefficient of $\;z^{-1}\;$ in the above is

$$\sum_{k=1}^\infty\frac{2^{k-1}}{k!(k-1)!}$$