Ways to see the result of $\lim_{z\to 0}f(z)$

Question

On my notes, it's written that a complex function is continuous iff writing $f(x,y)=u(x,y)+i\cdot v(x,y)$, where $u(x,y)$ is the real part of $f(z)$ and $v(x,y)$ the imaginary part, we have that $u$ and $v$ are continuous.

Suppose that I have the function $$f(z)=\exp(-1/z^4)\quad if\quad z\neq 0$$

and $0$ if $z=0$.

Imagine that we don't want to expand that $z^4=(x+yi)^4$ to separate this function. Is there a way to see that this function is not continuous on $z_0=0$ without having to separate it?

Thanks for your time.

Answer 1

Since every complex function has a real part and an imaginary parte, under your definition every complex function is continuous. Perhaps that you forgot to add that $u$ and $v$ are continuous.

If $x\in\mathbb R$, then $\exp(-1/(x+xi)^4)=\exp(1/(4x^4))$, and therefore the limit$$\lim_{x\to0}\exp(-1/(x+xi)^4)$$doesn't exist (in $\mathbb C$). In particular, the limit is not $f(0)$ and so $f$ is not continuous at $0$.

Answer 2

Twisted proof: suppose $$\lim_{z\to 0}\exp(-1/z^4)$$ exists. Then $f$ has a removable singularity at $z = 0$ and (I will use the same name) $$f(z) = \exp(-1/z^4), z\ne0; \qquad f(0) = \lim_{z\to 0}\exp(-1/z^4)$$ is holomorphic in the whole plane. But $|z|\to\infty\implies f(z)\to 1$ and $f$ would be nonconstant and bounded.