Complex Analysis – Limit of f(z)=1/z as z Approaches 0

Question

I know that the answer is infinity, but doesn't this contradict the theorem stating that $\displaystyle\lim_{z\to0} \Im(f(z)) = \Im(\displaystyle\lim_{z\to0} f(z))$? I mean, the LHS doesn't exist because $\Im(f(z))$ approaches $-\infty$ and $\infty$ as $z$ approaches $0$ along the imaginary axis from the positive and negative directions respectively. I know I'm misunderstanding something fundamental, but what is it?

Answer 1

The key is to remember that $\infty$ is not a complex number! In reality, in complex analysis, when we say that $$\lim_{z \to z_0} f(z) = \infty,$$ what we really mean is $$\lim_{z \to z_0} |f(z)| = \infty.$$ So we bring it back to real numbers. And of course, this statement doesn't exactly make sense, since $\infty$ is not a real number, either. What we really mean is that for all (large) $M>0$, we can find a (small) $\delta > 0$ such that for all $z$ where $|z - z_0| < \delta$, we have $|f(z)| > M$. And this is definitely true for $1/z$ as $z \to 0$.

Now, in real analysis, we can often talk about $-\infty$ and $\infty$, while in complex analysis, we can perhaps imagine $\infty e^{i \theta}$ for $\theta \in [0, 2\pi)$. But, it turns out that one of the best ways to think of large complex numbers is that they are actually somewhat close to each other; if we have complex numbers $z$ that we know are large, say $|z|>M$, then we can look at $1/z$ and note that $\left| \frac{1}{z} \right| < 1/M$. Therefore, we know that (as we defined a limit of complex numbers going to $\infty$) $$\lim_{z \to z_0} f(z) = \infty$$ if and only if $$\lim_{z \to z_0} \frac{1}{f(z)} = 0.$$

This notion is formalized with the Riemann sphere. The Riemann sphere allows us to think of meromorphic functions as functions where we can "fill-in" the poles by mapping them to $\infty$, and to think of rational functions as mappings from the Riemann sphere to itself. But note that even the Riemann sphere isn't "enough" to handle essential singularities.