Complex Analysis: Continuity
1. Definition of Continuity
Let \( f \) be defined on \( D \subset \mathbb{C} \), i.e., \( f: D \longrightarrow \mathbb{C} \).
Let \( z_0 \in D \cap D’ \) (Limit point belonging to the domain).
Definition: \( f \) is continuous at \( z = z_0 \) iff
$$ \lim_{z \to z_0} f(z) = f(z_0) $$Note: Isolated Points
If \( z_0 \in D \) but \( z_0 \notin D’ \) (an isolated point), then \( f \) is continuous at \( z_0 \) by definition.
Basic Examples
- Polynomials: \( w = P(x,y) + i\phi(x,y) \) is continuous at every complex number.
- Removable Discontinuity Fixed: $$ f(z) = \begin{cases} \frac{\sin z}{z} & z \neq 0 \\ 1 & z = 0 \end{cases} $$ $$ \lim_{z \to 0} \frac{\sin z}{z} = 1 = f(0) $$ Thus, \( f \) is continuous at \( z=0 \).
2. Equivalent Definitions of Continuity
Let \( w = f(z) = u(x,y) + iv(x,y) \) and \( z_0 = a + ib \).
I. Component-wise Definition
\( f \) is continuous at \( z_0 \) iff:
$$ u(x,y) \text{ and } v(x,y) \text{ are continuous at } (a,b). $$II. \( \epsilon – \delta \) Definition
\( f \) is continuous at \( z = z_0 \) if for any \( \epsilon > 0 \), there exists \( \delta > 0 \) such that:
$$ |f(z) – f(z_0)| < \epsilon $$whenever \( |z – z_0| < \delta \) in \( D \).
III. Sequential Definition
\( f \) is continuous at \( z = z_0 \) if:
$$ \langle z_n \rangle \longrightarrow z_0 \implies \langle f(z_n) \rangle \longrightarrow f(z_0) $$3. Algebra of Continuity & Counter-Examples
Properties
- (i) Basic functions \( f_i \) (from our list) are continuous everywhere in \( \mathbb{C} \).
- (ii) Quotient: \( f(z) = \frac{f_i(z)}{f_j(z)} \) is continuous on its natural domain.
- (iii) Composition: \( K(z) = f_i(H(z)) \) is continuous on its natural domain. (e.g., \( \sin(1/z) \)).
Ex (iv): Discontinuity at Origin
Consider:
$$ f(z) = \begin{cases} z \sin \frac{1}{z} & z \neq 0 \\ 0 & z = 0 \end{cases} $$Claim: \( f \) is NOT continuous at \( z=0 \) because the limit doesn’t exist.
Proof via Sequences:
- Sequence 1: \( a_n = \frac{1}{n\pi} \to 0 \). $$ f(a_n) = \frac{1}{n\pi} \sin(n\pi) = 0 \longrightarrow 0 $$
- Sequence 2: \( b_n = \frac{i}{n} \to 0 \). $$ f(b_n) = \frac{i}{n} \sin\left(\frac{1}{i/n}\right) = \frac{i}{n} \sin(-in) $$ Using \( \sin(-ix) = -i \sinh(x) \): $$ = \frac{i}{n} (-i \sinh n) = \frac{-i^2}{n} \sinh n = \frac{1}{n} \sinh n $$ $$ = \frac{e^n – e^{-n}}{2n} \approx \frac{e^n}{2n} \longrightarrow \infty $$
Since \( 0 \neq \infty \), the limit does not exist.
4. More Discontinuities & Key Lemma
Ex (v): Exponential Singularity
$$ f(z) = \begin{cases} e^{-1/z^n} & z \neq 0, n \in \mathbb{N} \\ 0 & z = 0 \end{cases} $$Evaluate limit along paths \( z = r e^{i\theta} \) as \( r \to 0 \):
- Along \( \theta = 0 \) (Real Axis): $$ \lim_{r \to 0} e^{-1/r^n} = 0 $$
- Along \( \theta = \frac{\pi}{n} \):
Here \( z^n = r^n e^{i\pi} = -r^n \). $$ \lim_{r \to 0} e^{-1/(-r^n)} = \lim_{r \to 0} e^{1/r^n} = \infty $$
Thus, \( \lim_{z \to 0} f(z) \) does not exist.
Double Star: Unboundedness Lemma
Let \( f: D \longrightarrow \mathbb{C} \).
Theorem: If \( f \) is unbounded on \( D \), then for any \( n \in \mathbb{N} \), there exists \( z_n \in D \) such that:
$$ |f(z_n)| > n $$In other words, there exists a sequence \( \langle z_n \rangle \) in \( D \) such that the sequence of images \( \langle |f(z_n)| \rangle \) is divergent.