How to verify Picard's theorem if $z$ is an essential singularity If we are given a function such as $\cos(1/z)$, how do we verify that it satisfies Picard's theorem without expanding the Laurent series.
Also if instead we are asked to show $z=0$ is an essential singularity of $\cos(1/z)$ how should it be done? (without expanding Laurent series)
The main issue I have here is how to verify Picard's Theorem here.
 A: Here is a proof that $\cos(1/z)$ has essential singularity at $0$ which does not invoke Laurent series.
Observe that if $f$ has a pole or removable singularity at $0$, then at least one of the functions $f$ and $1/f$ is bounded in a neighborhood of $0$. Neither $e^{1/z}$ nor $e^{-1/z}$ is bounded in any neighborhood of $0$. Hence, $e^{1/z}$ has an essential singularity there. 
Suppose that $\cos (1/z)$   has a pole or removable singularity at $0$. Then the same is true for its derivative $z^{-2}\sin (1/z)$. (This does not require Laurent series: pick $n$ such that $z^nf(z)$ is holomorphic at $0$, and consider $(z^nf(z))'=nz^{n-1}f(z)+z^n f'(z)$.)
Since 
$$e^{i/z}=\cos (1/z)+i\sin(1/z)$$
we have a contradiction: essential singularity on the left but not on the right.
Related: Types of singularities
A: The Riemann principle states that if a singularity is neither removable nor a pole, then it is an essential singularity.
Equivalently, if the Laurent expansion of a function has an infinite number of negative terms, then the singularity is essential.
The easiest way to show this is by assuming that the Taylor expansion of $\cos \frac{1}{z}$ is valid at $z = 0$, then use uniqueness of the Laurent series.
As for verifying Picard's theorem... start with the Laurent series.

$$\cos w = 1 - \frac{w^2}{2} + \frac{w^4}{4!} + \cdots$$
so
$$\cos \frac{1}{z} = 1 - \frac{1}{2}z^{-2}+\frac{1}{4!}z^{-4} + \cdots$$
for all $z \neq 0$.
This obviously has infinitely many negative terms, $a_{-2} = \frac{1}{2},\ldots$, so the singularity must be essential.
