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So I was reading about Laurent series and $\mathrm{e}^{-1/x^2}$ was used as an example. We define the function $f(x) = \mathrm{e}^{-1/x^2}$ for $x \neq 0$ and $f(0)=0$. Then it was stated that, as a real function, $f(x)$ is infinitely differentiable everywhere, but as a complex function it is not differentiable at $x=0$. Why is this?

We know that $f$ is real differentiable if the derivative exists at every point of its domain. We have $$f'(x) = -\frac{f(x)}{x^2},$$ so that $f'(0)=0/0$?

It is complex differentiable if $$\lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}$$ exists. I guess we have an essential singularity in the argument of the exponential function at $z_0=0$ which is why it is not complex differentiable?

Numbersandsoon
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  • $\lim_{x \to 0} \dfrac{e^{-1/x^2}}{x} = 0$, but $\lim_{y \to 0} \dfrac{e^{-1/(iy)^2}}{iy} = \infty$. – njguliyev Oct 18 '13 at 20:07
  • $f$ is $C^\infty$ and $f^{(k)}(0) = 0$ for all $k$, but $f$ not identically zero. Any complex function that is differentiable in a ball around zero and has $f^{(k)}(0) = 0$ for all $k$ must be identically zero. – copper.hat Oct 18 '13 at 20:17
  • Right! I guess I misread the condition for the complex function as implying that $z=0$ instead of just the real part. I thought of this as a possibility but got $$\frac{ \mathrm{e}^{-1/(x+iy)}-\mathrm{e}^{-1/0} }{ x+iy }.$$ How did you account for that limit? I see that letting $y=0$ and taking the limit and $x=0$ and taking the limit results in different limits. This is what I tried to do. Thanks! But how about that argument of the exponential? – Numbersandsoon Oct 18 '13 at 20:20
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    To avoid some of your confusion, define $f$ more carefully. As you have done for $x \ne 0$, but add $f(0)=0$. – GEdgar Oct 18 '13 at 20:31
  • Ah, right. Got it. – Numbersandsoon Oct 18 '13 at 20:32
  • If you're asking why it has derivatives of all orders at $0$ as a function of a real variable, that's the "hard" part. If you're only asking why it's NOT differentiable at $0$ as a function of a complex variable, that's much easier: just think about what happens to $f(z)$ as $z$ approaches $0$ along the imaginary axis. – Michael Hardy Oct 18 '13 at 21:51

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