# Counter-example: Cauchy Riemann equations does not imply differentiability

I need help with this exercise:

Let f(z) = \left\{ \begin{align} &e^{-\frac{1}{z^4}} &\hspace{1mm} \mbox{if} \hspace{1mm} z \neq 0 \\ &0 &\hspace{1mm} \mbox{if} \hspace{1mm} z = 0 \\ \end{align} \right. Show that $f$ satisfies Cauchy-Riemann equations, but $f$ is not differentiable in $z=0$.

I have to compute $u_x(x,y)$, $u_y(x,y)$, $v_x(x,y)$ and $v_y(x,y)$, so I have to find $u(x,y)$ and $v(x,y)$ explicitly. My attempts: if $z = x+iy$, then $z^4 = (x+iy)^4$, doing the math, I found

$$z^4 = (x^4 – 6x^2y^2 + y^4)+i(4x^3y – 4xy^3)$$

I don’t know how to find $-\dfrac{1}{z^4}$. My second attempt was this: (trying to find $f(z)$ in polar coordinates) let be $z = |z|e^{i\theta}$, then $z^4 = |z|^4e^{i(4\theta)}$, thus

$$e^{-\frac{1}{z^4}} = e^{-|z|^4}e^{e^{i\theta}}$$

I appreciate any idea you have.

#### Solutions Collecting From Web of "Counter-example: Cauchy Riemann equations does not imply differentiability"

The function $f$ is analytic off the origin, hence satisfies the CR equations at all $z\ne0$. For $z=0$ look at
$$f_x(0,0)=\lim_{x\to0}{e^{-1/x^4}\over x}=0\ ,$$
and similarly
$$f_y(0,0)=\lim_{y\to0}{e^{-1/(iy)^4}\over y}=0\ .$$
It follows that $u_x(0,0)=v_x(0,0)=u_y(0,0)=v_y(0,0)=0$; so $f$ satisfies the CR equations also at $z=0$.

But $f$ is not even continuous at $z=0$: Consider the points $z(t):=(1+i)t$ for real $t$ near $0$. I leave the details to you.