Which one of the following are true? 
Consider the function $e^{-z^{-4}}$  for $z≠0$ and $f(0)=0$. Then,
(A) $f$ is not analytic.
(B)$f$ is not differentiable at $z=0.$
(c)$f$ does not satisfy the C-R(Cauchy-Riemann) equation.
(d)$f$ satisfies the C-R(Cauchy-Riemann) equation and not analytic.

(A) $f$ is not analytic. $(A)$is False.  Laurentz series expansion has principle part.
(B) $$\lim_{h\to 0}\frac{e^{-h^{-4}}}{h}.$$ Let $h=u+iv$,
Path 1: Along real axis. $$\lim_{u\to 0^+}\frac{e^{-u^{-4}}}{u}=0.$$
$$\lim_{u\to 0^-}\frac{e^{-u^{-4}}}{u}=0.$$ 
Path 2: Along the imaginary axis. $$\lim_{v\to 0}\frac{e^{-v^{-4}}}{iv}=0.$$
I am not able to disprove the differentiability. I tried to prove $$ |\frac{f(z)}{z}|^2=|\frac{e^{-z^{-4}}}{z}|^2=\frac{e^{-z^{-4}}}{z}\overline{\frac{e^{-z^{-4}}}{z}}=\frac{e^{-z^{-4}-\overline{z^{-4}}}}{|z|^2}$$. I have no Idea how to prove it.
(C) For C-R equation. I need to check whether $$if_x=f_y.$$ $$f(x,y)=e^{-z^{-4}}$$ $$=e^{-\frac{1}{(x+iy)^4}}$$ $$=e^{-\frac{(x-iy)^4}{(x^2+y^2)^4}}$$
I got $$f_x=e^{-\frac{(x-iy)^4}{(x^2+y^2)^4}}\frac{(x^2+y^2)^44(x-iy)^3-(x-iy)^48x(x^2+y^2)^3}{(x^2+y^2)^8}$$
$$f_y=e^{-\frac{(x-iy)^4}{(x^2+y^2)^4}}\frac{(x^2+y^2)^44(x-iy)^3(-i)-(x-iy)^4 8y(x^2+y^2)^3}{(x^2+y^2)^8}$$. $if_x\neq f_y$. C-R equation won't satisfy. How do I prove $f(z)$ is differentiable at $z=0$?

Consider $\zeta_4$ a $4$-th root of $-1$ and for $\epsilon >0$, $z_\epsilon = \zeta_4 \epsilon$. You have $$
\lim\limits_{\epsilon \to 0}f(z_\epsilon) = \lim\limits_{\epsilon \to 0}e^{-z_\epsilon^{-4}} = \lim\limits_{\epsilon \to 0}e^{1/\epsilon^4} = \infty.$$
Therefore $f$ is not even continuous at $0$.
Hence:


*

*$f$ is not analytic.

*Not differentiable at $0$.

*Satisfy the Cauchy Riemann equation for $z\neq 0$ as $f$ is analytic for $z \neq 0$.

