Consider $f(z)=e^{-z^{-4}}$ for $z≠0$ and $f(0)=0$. Show that the Cauchy Riemann equations are satisfied for $z=0$ 
Consider $$f(z)=e^{-z^{-4}}$$ for $z≠0$ and $f(0)=0$. Show that the
  Cauchy Riemann equations are satisfied for $z=0$, and show that $f$ is
  not complex differentiable.

Any smart ideas here ? I hope there is some nicer method than just exanpding $(x+iy)^{-4}$. 
 A: Given that$$f(z)=u+iv=e^{-z^{-4}}=e^{-(x+iy)^{-4}}=e^{-(x-iy)^4/(x^2+y^2)^4}=e^{-\frac{1}{(x^2+y^2)^4}\left[(x^4+y^4-6x^2y^2)-4ixy(x^2-y^2)\right]}$$
$$\implies u=e^{-\frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}}\cdot \cos\left[\dfrac{4xy(x^2-y^2)}{(x^2+y^2)^4}\right]$$
$$~~~~~~~~~~~~~v=e^{-\frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}}\cdot \sin\left[\dfrac{4xy(x^2-y^2)}{(x^2+y^2)^4}\right]$$
At $~z=0~,$ $$\dfrac{\partial u}{\partial x}=\lim_{x\to 0}\dfrac{u(x,0)-u(0,0)}{x}=\lim_{x\to 0}\dfrac{e^{-x^{-4}}}{x}=\lim_{x\to 0}\left[\dfrac{1}{x\cdot e^{1/x^{4}}}\right]=0$$
$$\dfrac{\partial u}{\partial y}=\lim_{y\to 0}\dfrac{u(0,y)-u(0,0)}{y}=\lim_{y\to 0}\dfrac{e^{-y^{-4}}}{y}=\lim_{y\to 0}\left[\dfrac{1}{y\cdot e^{1/y^{4}}}\right]=0$$
Similarly,
$$\dfrac{\partial v}{\partial x}=\lim_{x\to 0}\dfrac{v(x,0)-v(0,0)}{x}=0$$
$$\dfrac{\partial v}{\partial y}=\lim_{y\to 0}\dfrac{v(0,y)-v(0,0)}{y}=0$$
Hence Cauchy-Riemann equations are satisfied at $~z=0~.$
But $$f'(0)=\lim_{z\to 0}\dfrac{f(z)-f(0)}{z}=\lim_{z\to 0}\dfrac{1}{ze^{1/z^{4}}}=\lim_{r\to 0}\dfrac{1}{re^{i(\pi/4)}}\cdot\dfrac{1}{e^{-1/r^4}}=\infty$$taking $~z=re^{i(\pi/4)}~.$
Which shows that $~f'(z)~$ does not exist at $~z=0~$ and hence $~f(z)~$ is not analytic at $~z=0~.$
