Alternative way to check cauchy riemann equation. I have a function $$f(z)=\begin{cases} 
e^{-z^{-4}} & z\neq0 \\
0 & z=0
\end{cases}$$
I have to show cauchy riemann equation is satisfied everywhere. I have shown that it isn't differentiable at $z=0$. 
Usually I will have to convert it in $$f(z)=u+iv$$ which seems very tedious. Is there some way to do this while keeping it in $f(z)$ form. 
 A: As the question asks to use Cauchy-Riemann equations so either you convert it to get $u$ and $v$  in $x$ and $y$; or use polar coordinates $r$, $\theta$ using $z=re^{i\theta}$, i.e.
$f(re^{i\theta})=e^\frac{-cos\theta}{r^4}cos(\frac{sin\theta}{r^4})+e^\frac{-cos\theta}{r^4}sin(\frac{sin\theta}{r^4})$. 
Then Cauchy Riemann equations in polar form are: $\partial u/\partial r=\partial v/r\partial \theta,\partial v/\partial r=-\partial u/r\partial \theta$
A: This is quite easy. For example, $\frac {f(h+i0)-f(0)} h=\frac {e^{-h^{-4}}} h$ and the limit as  $h \to 0$ through real values is $0$. [ $e^{x^{4}} \to \infty$ faster than any power of $x$ as $x \to \infty$. Put $x=\frac 1 h$].  For partial derivatives w.r.t. $y$ we get the same limit since $i^{4}=1$. 
A: A function $f$ is holomorphic at $z$ if $\lim_{|h| \to 0} \frac{f(z+h)-f(z)}{h}$ exists. This equation is more fundamental than the Cauchy-Riemann equations, which are derived from this.
In the actual case we get
$$
f'(0) 
= \lim_{|h| \to 0} \frac{f(h)-f(0)}{h} 
= \lim_{|h| \to 0} \frac{e^{-h^{-4}}}{h}
= \{ R = 1/h \}
= \lim_{|R| \to \infty} R e^{-R^{4}}
.
$$
When we take partial derivatives, $h$ follows an axis. Then so does $R$ and we have $R^4 = |R|^4.$ This leads to $|R e^{-R^{4}}| = |R| e^{-|R|^{4}} \to 0$ as $|R| \to \infty.$ Thus all partial derivatives are $0$ at $z=0$, which implies that the Cauchy-Riemann equations are satisfied.
