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Show that there is a complex number $z$ such that:$$\left|\cos{\left(\frac{1}{2z^4+3z^2+1}\right)}+100\tan^2{z}+e^{-z^2}\right|<1$$
It's easy to see that $z=i$ is a simple pole of $\frac{1}{2z^4+3z^2+1}$, but I want to know how to conclude that $z=i$ is an essential singularity of $\cos{\left(\frac{1}{2z^4+3z^2+1}\right)}$ so that I can use the Casorati-Weierstrass theorem.
If you have access to the Open Mapping Theorem at this point, then there's an easier way then trying to directly show that $\displaystyle\cos\left(\frac{1}{2z^4+3z^2+1}\right)$ has an essential singularity at $z = i$.
In any neighborhood of $z = i$ we can make $100 \tan^2(z) + e^{-z^2}$ arbitrarily close to $A = 100 \tan^2(i) + e^{-i^2}$. Find a $\delta$ neighborhood around $z = i$ so that inside it you get $| 100 \tan^2(z) + e^{-z^2} - A| < \frac{1}{2}$. The image $U$ of this neighborhood under $w = 2z^4+3z^2+1$ is (using Open Mapping) an open set containing $w = 0$, since $z = i$ is a root of $2z^4+3z^2+1$.
Note that $\cos(\frac{1}{w})$ has an essential singularity at $w = 0$.
Now using Casoratti-Weierstrass find $w \in U$ so that $|\cos(\frac{1}{w}) + A| \lt \frac{1}{2}$. $w = 2z^4+3z^2+1$ for some $z$ in the $\delta$ neighborhood of $z = i$, so you get
$$ \Big| \cos\left(\frac{1}{2z^4+3z^2+1}\right) + 100 \tan^2(z) + e^{-z^2} \Big| \le \Big| \cos\left(\frac{1}{w}\right) + A\Big| + \Big| 100 \tan^2(z) + e^{-z^2} - A \Big| < 1 $$
Denote $$f(z) = \cos\left(\frac{1}{2z^4+3z^2+1}\right)$$
To prove $z=i$ is an essential singularity of $f(z)$, just find two complex sequences of $\{z_n\}$ and $\{w_n\}$ so that $z_n, w_n\rightarrow i$ but $f(z_n) = 1$, $f(w_n)=-1$.
This means both $\lim_{z->i} f(z)$ and $\lim_{z->i} 1/f(z)$ do not exist. By definition, $z=i$ is an essential singularity.
