Let $\varepsilon>0$. I want to find the image of $\{z\in\mathbb{C}:0<|z|<\varepsilon\}$ through of $f(z)=\cos\left(\frac{1}{z}\right)$. I know that $f\left(\{z\in\mathbb{C}:0<|z|<\varepsilon\}\right)=\mathbb{C}$ using Picard's theorem but, How can I show this directly?
I have shown direcly that the image of $\{z\in\mathbb{C}:0<|z|<\varepsilon\}$ throught of $g(z)=e^\frac{1}{z}$ is $\mathbb{C}\setminus\{0\}$. Is it useful?
Thank you.
Since $\{|z|>R\}$ is a base of (punctured, if you work in $\mathbb{C}$) neighbourhood of $\infty$, any neighbourhood of $\infty$ contains a strip $\{z\in\mathbb{C}:\Re z\in [a,a+2\pi]\}$.
The desired result now follows from periodicity and surjectivity of $\cos\colon\mathbb{C}\to\mathbb{C}$.