Question about $f(z)=\exp (-\frac{1}{z^4})$

Question

Let $f(z)=\exp (-\frac{1}{z^4})$ for $z\neq 0$ and $f(0)=0$.

I know this is a famous example and got asked a lot However my question is not about the origin, but what is the best way to actually show that this functions satisfies Cauchy Riemann everywhere? I only read that it fulfills it everywhere but nowhere could I find an actual proof. Because writing $ z = x+iy$ and then expanding the denominator in the exponent and then dividing into real and imaginary part ist really exhausting. Is this the only way or is there some shortcut here?

Answer 1

As I think there is a bit of confusion about complex differentiability vs C-R and there are lots of subtle points, let's note that if we have $f: U \to \mathbb C$ a complex differentiable function in an open (domain) $U \subset \mathbb C$, then by the usual theorems, $f$ is infinitely differentiable, analytic and satisfies the C-R equations which are succintly written as $$\frac{\partial f}{\partial y}=i\frac{\partial f}{\partial x}$$

However, the converse is not true in the sense that if $f$ is defined in $U$, has partial derivatives $\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}$ everywhere in $U$ and said partial derivatives satisfy the C-R above, then it does not follow that $f$ is complex differentiable in $U$ and $f(z)=\exp (-\frac{1}{z^4}), z \ne 0$, while $f(0)=0$ is such a counterexample as clearly $f$ is complex differentiable on $\mathbb C-0$ being a composition of complex differentiable functions there, so by part I above it does satisfy C-R there.

Now it is a straightforward exercise to show that $\frac{\partial f}{\partial y}(0)=\frac{\partial f}{\partial x}(0)=0$ so indeed $f$ has partial derivatives everywhere in the plane, satisfies C-R everywhere too, but it is not complex differentiable everywhere and there are counterexamples where the points where such a function (partial derivatives everywhere, C-R everywhere) is not complex differentiable have non zero Lebesgue measure.

(Note that $f$ above is separately continuous and real differentiable at $0$ in $x$ and $y$ as $f(x+i0)=e^{-1/x^4}$ which is the typical $C^{\infty}$ but not real analytic function, while $f(0+iy)=e^{-1/y^4}$ also)

In usual CA texts, one assumes stronger hypothesis like $f$ real differentiable with continuous differential on an $U$ plus C-R or at least $f$ continuous and the partial derivatives are continuous on $U$ too and deduce that $f$ is complex differentiable on $U$, but there is a theorem of Looman-Menchoff (strengthened by Montel) which is quite difficult and subtle (took a few decades to realize that the original proof of Looman had a gap and fix it) which says that $C-R$ plus continuity on $U$ (or just local boundness - Montel) are enough for complex differentiability as the $f$ here clearly fails continuity or boundness at the origin.

There is a very informative paper by Gray and Morris discussing this and much more:

"When is a Function that Satisfies the Cauchy-Riemann Equations Analytic?"

Answer 2

We rarely work with definitions alone; we use all the theorems available at our disposal. If I asked you to prove that $f(x)=e^{-1/x^4}$ is differentiable on $\Bbb{R}\setminus\{0\}$, you had better tell me that the reason (in full) this is true is that:

• $x\mapsto x$ is differentiable (obvious from $\epsilon$-$\delta$)
• $x\mapsto x^4$ is thus differentiable, being a four-fold product of differentiable functions
• $x\mapsto \frac{1}{x^4}$ is differentiable away from the origin (the quotient of continuous functions is differentiable away from the zeros of the denominator)
• $x\mapsto -\frac{1}{x^4}$ is differentiable away from the origin (product of the constant function $-1$ with a differentiable function)
• $\exp:\Bbb{R}\to\Bbb{R}$ is differentiable (a well-known fact)
• $x\mapsto e^{-1/x^4}$ is differentiable away from the origin (composition of differentiable functions is differentiable).

In the complex case $f:\Bbb{C}\setminus\{0\}\to\Bbb{C}$, $f(z)=e^{-1/z^4}$, the reasoning above works word for work once you replace $\Bbb{R}$ with $\Bbb{C}$. The sum, product, quotient, chain rules (compositions) are all stated and proved exactly as in the real case. So, no Cauchy-Riemann necessary here (and this is often an extremely inefficient way of doing things when you’re given nice simple explicit formulas $f(z)=\dots$, for exactly the reasons you mentioned). Once you know that the function is complex-differentiable, you can then state as a simple corollary that it must therefore satisfy the Cauchy-Riemann equations.