Let $f(z) = \exp{(-z^{4})}$, $z \neq 0$, $f(0) = 0$.
Show that $f$ satisfies Cauchy-Riemann equations, but $f$ is not holomorphic.
My attempt
For $t\in \mathbb{R}$, we have:
$$\lim_{t\to0} \exp(-t^4) = 1 \neq f(0)$$
Then $f$ isn't continuous and thus $f$ isn't holomorphic for $z=0$.
I couldn't prove that $f$ satisfies Cauchy-Riemann equations at $z=0$.
I noticed that, for $t \in \mathbb{R}$:
$$\lim_{t \to 0^+} \frac{f(t)-f(0)}{t-0} = u_x(0,0)+iv_x(0,0) = \lim_{t \to 0^+} \frac{\exp(-t^4)}{t} = \infty$$
$$\lim_{t \to 0^+} \frac{f(it)-f(0)}{it-0} = -i u_y(0,0)+ v_y(0,0) = \lim_{t \to 0^+} \frac{\exp(-(it)^4)}{it} = \ \infty$$
Then $u_x(0,0)$, $u_y(0,0)$, $v_x(0,0)$ and $v_y(0,0)$ don't exist.
Is this correct?
If so, is this exercise wrong?
Best Answer
As $-z^4$ is entire then so must be $e^{-z^4}$.