Prove $ \lim_{z \rightarrow z_o} f(z)$ is finite if $\lim_{z \rightarrow z_o} (z-z_0) f(z) = 0$

Question

Context: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3 (asked about here)

Question

From (here): How do we have that $$\lim_{z \rightarrow z_o} f(z) \in \mathbb{C} \ \text{for removable singularity} \ z_0 \tag{1}$$ under the characterisation that $f$ has a removable singularity if $$\lim_{z \rightarrow z_o} (z-z_0) f(z) = 0 \tag{2}$$?

Update: Attempt is now moved to answer.

---

Some intuition:

$(2) \Leftarrow (1)$

• If $L \in \mathbb{C}$, then $\lim_{z \rightarrow z_o} (z-z_0) f(z) = \lim_{z \rightarrow z_o} (z-z_0) \lim_{z \rightarrow z_o} f(z) = (0)(L) = 0$

$(2) \Rightarrow (1)$

• If $L = \infty$, then $\lim_{z \rightarrow z_o} (z-z_0) f(z) = 0 \cdot \infty$, indeterminate.

• If $L$ dne, then $\lim_{z \rightarrow z_o} (z-z_0) f(z)$ dne.

Upon reflection, this actually seems like a simple complex extension of a simple elementary real analysis proof.

Answer 1

Pf $(2) \iff (1)$:

Firstly, let $\lim_{z \rightarrow z_o} f(z) =: L$, a letter which could be a finite complex number, $\infty$ or 'dne'.

Now for the forward implication:

$(2) \Leftarrow (1)$:

• Given: $\forall \varepsilon_1 > 0, \exists \delta_1 > 0: |f(z)-L|<\varepsilon_1 \ \text{whenever} \ 0<|z-z_0|<\delta_1$

• TS: $\forall \varepsilon > 0, \exists \delta > 0: |(z-z_0)f(z)|<\varepsilon \ \text{whenever} \ 0<|z-z_0|<\delta$

• Pf: Let $\varepsilon > 0$. $$|(z-z_0)f(z)| = |(z-z_0)[f(z)-L+L]|$$

$$= |(z-z_0)| |f(z)-L+L| \le |z-z_0| (|f(z)-L|+|L|)$$

$$\stackrel{(*)}{\le} |z-z_0| \varepsilon_1+|L| < \varepsilon \ \text{whenever} \ 0 < |z-z_0| < \delta := \min\{\frac{\varepsilon}{\varepsilon_1+|L|},\delta_1\}$$

(*) $\text{whenever} \ 0 < |z-z_0| < \delta_1$

QED $(2) \Leftarrow (1)$

Now for the backward implication:

$(2) \Rightarrow (1)$

• Given: $\forall \varepsilon > 0, \exists \delta > 0: |(z-z_0)f(z)|< \varepsilon \ \text{whenever} \ 0<|z-z_0|<\delta$

• TS: $\forall \varepsilon_1 > 0, \exists \delta_1 > 0: |f(z)-L|< \varepsilon_1 \ \text{whenever} \ 0<|z-z_0|<\delta_1$

• Pf: Let $\varepsilon_1 > 0$. $$|f(z)-L| = |\frac{(z-z_0)[f(z)]}{z-z_0}-L| \le |\frac{(z-z_0)[f(z)]}{z-z_0}| + |L|$$

$$\stackrel{(**)}{\le} |\frac{\varepsilon}{z-z_0}| + |L| \le \varepsilon_1 \ \text{whenever} \ 0<|z-z_0|<\delta_1 := \min\{\delta,\frac{\varepsilon}{\varepsilon_1-|L|}\}$$

(**) $\text{whenever} \ 0 < |z-z_0| < \delta$

QED $(2) \Rightarrow (1)$

Now both the forward and backward implications have been proven.

QED $(2) \iff (1)$