I have to classify the singularity at the origin of the function $f(z)=z \cos(1/z)$. How do I get this answer without passing through the Laurent serie directly (in using (*)). Is anyone could help me at this point?
I think this is an essential singularities because of the behavior of $\frac{1}{z}$ in the Taylor serie of $cos$ in $f$.
(*) Definition : An isolated singularity is essential if the Laurent serie contain an infinite number of term $(z-z_0)^k$, $k < 0$.
$$ \cos z^{-1} = \sum_{j=0}^\infty \frac{(-1)^j z^{-2j}}{(2j)!} = \sum_{j=-\infty}^0 \frac{(-1)^j z^{2j}}{(-2j)!} $$ so $z \cos z^{-1}$ has a essential singularity at $0$.
You can also use Riemann's theorem to prove it.