Is the complex function $f(z)=e^{-z^{-4}}$ analytic at $z=0$? Is the complex function $$f(z)=\begin{cases}
e^{-z^{-4}} & z\ne0\\
0 & z=0
\end{cases}$$ analytic at $z=0$ ?
I am able to show that the Cauchy-Riemann equations are satisfied.
 A: The other answers are using thermonuclear weapons to conclude...
The function is not even continuous at zero. For example, there exists in $\mathbb C$ a sequence $(w_n)_{n\geq0}$ such that $w_n\to\infty$ and $\exp w_n^4=1$ for all $n\geq0$. It follows that $1/w_n\to0$ as $n\to\infty$ and $f(1/w_n)=\exp(-(1/w_n)^{-4})=1$ for all $n$, and this together with $f(0)=0$ imply that $f$ is not continuous at $0$.
A: The function $f(z)$ thus defined has a Laurent series expansion
$$ f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!} z^{-4n} $$
around $z=0$ with an infinite principal part (i.e. with infinitely many non-zero coefficients in front of the negative powers of $z$). 
Thus $z=0$ is an essential singularity of $f(z)$, and by the Casoratti-Weierstraß theorem, there is no way to define $f(0)$ so that $f(z)$ be continuous at $z=0$. In fact, we have 
$$ |f(z)| = e^{-\Re(z^{-4})} = \exp\left(\frac{-x^4+6x^2y^2-y^4}{(x^2+y^2)^4}\right) \quad \text{for} \quad z = x +iy \neq 0.$$
Restricting $z$ to the line $y = tx$ with $t\in\mathbb{R}$, we observe that
$$ \lim_{x\to0}|f(x+itx)| = \lim_{x\to 0}\,\exp\left(\frac{-1+6t^2-t^4}{x^4(1+t^2)^4}\right)\tag{1}\label{limit}$$
is not independent of $t \in \mathbb{R}$. Indeed, we can factor the biquadratic polynomial $-1+6t^2-t^4$ as
$$-\big((t^2-3)^2 - 8\big) = -\big(t^2-(\sqrt{2}-1)^2\big)\big(t^2-(\sqrt{2}+1)^2\big) = -(t+\sqrt{2}+1)(t+\sqrt{2}-1)(t-\sqrt{2}+1)(t-\sqrt{2}-1),$$
whence we deduce that the limit \eqref{limit} equals either $1$ (for $t = \pm\sqrt{2}\pm1$), or $\infty$ (whenever $|t+\sqrt{2}| < 1$ or $|t-\sqrt{2}|<1$), or $0$ (otherwise). Therefore, we conclude that $\lim_{z\to0}|f(z)|$ does not exist. Hence, the limit $\lim_{z\to0}f(z)$ does not exist either, and thus, $f(z)$ cannot be analytic at $z=0$.
Edit: I would like to thank Abhishek Verma for pointing out an error in the original solution: the denominator appearing in the expression for $|f(z)|$ was incorrectly calculated as $(x^2+y^2)^2$, instead of $(x^2+y^2)^4$. 
A: Since you already know the function is continuous, we can conclude directly that it is analytic from Riemann's theorem (about removable singularity).
A: I would say no, because every derivative is $0$ at $0$, (due to expontential growth), see:
http://en.wikipedia.org/wiki/Non-analytic_smooth_function
