I need to find the coefficient $a_{-1}$ in the Laurent series of $f(z)=z^{3}\cdot\cos(\frac{1}{z})\cdot e^{\frac{1}{z^{2}}}$.
I tried two methods:
- $a_{-1}=\frac{1}{2\pi i}\oint_{C_{r}}f(z)dz$
I chose $r=1$ and wrote $a_{-1}=\frac{1}{2\pi i}\oint_{C_{r}}f(z)dz=\frac{1}{2\pi i}\int_0 ^{2\pi}ie^{it}f(e^{it})dt$, which didn't lead me anywhere.
- Trying to compute directly, I know these Laurent series about $z=0$:
$\cos(\frac{1}{z})=1-\frac{z^{-2}}{2!}+\frac{z^{-4}}{4!}+...$
$e^{\frac{1}{z^{2}}}=1+z^{-2}+\frac{z^{-4}}{2!}+\frac{z^{-6}}{3!}+...$
So $f(z)=(1-\frac{z^{-2}}{2!}+\frac{z^{-4}}{4!}+...)(z^3+z+\frac{z^{-1}}{2!}+\frac{z^{-3}}{3!}+...)$ and $a_{-1}$ is the coefficient of $(\frac 1 z)$, so $a_{-1}=\frac {1}{2!}-\frac {1}{2!}+\frac{1}{4!}=\frac 1 {24}$.
Is my work okay? Did I miss any theoretical point? When is it advised to use each method? And what if I needed to calculate the whole series, not just one coefficient?
Any sharing of your experience will be greatly appreciated.
Looks fine to me. You could also consider \begin{align} e^{w^2}\cos w &=\frac{e^{w^2+iw}+e^{w^2-iw}}{2} \\ &=\frac{1}{2}\left(1+(w^2+iw)+\frac{(w^2+iw)^2}{2}+\frac{(w^2+iw)^3}{6}+\frac{(w^2+iw)^4}{24}\right)+\\ &\phantom{{}={}} \frac{1}{2}\left(1+(w^2-iw)+\frac{(w^2-iw)^2}{2}+\frac{(w^2-iw)^3}{6}+\frac{(w^2+iw)^4}{24}\right)+o(w^4)\\ &=1+w^2+w^4-w^2-w^4+\frac{1}{24}w^4+o(w^4)\\ &=1+\frac{1}{24}w^4+o(w^4) \end{align} Thus $$ z^3e^{1/z^2}\cos\frac{1}{z}=z^3+\frac{1}{24}z^{-1}+o(z^{-1}) $$