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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?

Community
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Santos
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    Do you perhaps mean $\exp(-z^{-4})$ or something similar? (Or, the one who constructed the exercise could have meant so, if you copied it as it was.) – mickep Apr 16 '17 at 14:05
  • No, it's $\exp(-z^4)$ and not $\exp(-z^{-4})$. – Santos Apr 16 '17 at 14:09
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    Then I would assume there is an error, since $\exp(-z^4)$ is holomorphic in $\mathbf C$. – mickep Apr 16 '17 at 14:10
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    The exercise definitely meant $\exp(-z^{-4})$ because 1) $\exp(-z^4)$ is holomporphic in $\mathbb C$ and 2)$\exp(-z^{-4})$ is a classic exercise for complex analysis. – Rebellos Apr 16 '17 at 14:18
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    The exercise is definitely wrong. The function is neither continuous at $0$, nor it has all partial derivatives at $0$. – egreg Apr 16 '17 at 14:22

1 Answers1

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As $-z^4$ is entire then so must be $e^{-z^4}$.

Nitin Uniyal
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