Vtyjkyuk

I am trying to determine if the the following is entire $$f(z)= begincases
e^-z^-4 & zneq0 \
0 & z=0\
endcases
$$

My attempt:

Consider $zne 0$. $f(z)=e^-z^-4$ is the composition of two differentiable functions, hence $f$ is differentiable when $zneq 0$.

Next I considered $z=0$. The Cauchy-Riemann equations are satisfied at $z=0$, as $u_x=u_y=v_x=v_y=0$. But this is not sufficient condition for complex differentiability. I wish to show $f$ is continuous/discontinuous at $z=0$.

I tried defining the limit $$f(z)=lim_zto 0 e^-z^-4=e^-infty=0$$

This would imply $f$ is continuous at $z=0$, but the answer I have disagrees.

edit

If I let $z=re^itheta$, then
beginalign
lim_zto 0 e^-z^-4&=lim_rto 0 e^-frac1r^4e^-4itheta \
&=lim_rto 0 e^frac1r^4rightarrow infty textletting left(theta=fracpi4right) \
endalign
Hence, $f$ is not continuous at $z=0implies f$ is not differentible at $z=0$.

So $f$ is not entire.

Hint: consider $z=x+xi$ for real $xto0$.

For each positive interger $n$ lat $c_n$ be a fourth root of $ 2nipi $ and $z_n = frac 1 c_n$Then $|e^-z_n^-4 |=1$ for all $n$ and $z_n to 0$.