Complex Analysis: Limits
I. Definition via Sequences
Let \( w=f(z) \) where \( f:D \longrightarrow \mathbb{C} \). Let \( z_0 \) be a limit point (l.p.) of \( D \).
The Three Sets
1. The Domain Sequence Set (\( \mathcal{A}_{z_0} \))
$$ \mathcal{A}_{z_0} = \{ \langle z_n \rangle : \lim z_n = z_0 ; \ z_n \in D \setminus \{z_0\} \} $$(An uncountable set of sequences in D converging to \( z_0 \))
2. The Image Sequence Set (\( \mathcal{B}_{z_0} \))
$$ \mathcal{B}_{z_0} = \{ \langle f(z_n) \rangle : \langle z_n \rangle \in \mathcal{A}_{z_0} \} $$3. The Limit Set (\( \mathcal{C}_{z_0} \))
$$ \mathcal{C}_{z_0} = \text{The set of limit points of sequences in } \mathcal{B}_{z_0} $$Case I: Boundedness Condition
If \( f \) is bounded in a neighborhood (nbd) of \( z_0 \):
- Every sequence in \( \mathcal{B}_{z_0} \) is bounded.
- \( \mathcal{C}_{z_0} \) is a non-empty bounded subset of \( \mathbb{C} \).
Conclusion: We say \( \lim_{z \to z_0} f(z) \) exists iff \( \mathcal{C}_{z_0} \) is a singleton.
$$ \text{If } \mathcal{C}_{z_0} = \{l\}, \text{ we write } \lim_{z \to z_0} f(z) = l $$II. Sequential Criterion
If \( f \) is bounded in a nbd of \( z_0 \), then \( \lim_{z \to z_0} f(z) \) exists finitely iff whenever \( \langle z_n \rangle \to z_0 \), the sequence \( \langle f(z_n) \rangle \) is convergent.
Implication: If for any \( \langle z_n \rangle, \langle t_n \rangle \to z_0 \):
$$ \lim f(z_n) = \lim f(t_n) = l $$Then, \( \lim_{z \to z_0} f(z) = l \).
Examples: Polynomials
Ex (i): Real Polynomials
Let \( f(z) = P(x,y) + i\phi(x,y) \) where \( P, \phi \in \mathbb{R}[x,y] \).
For any \( z_0 = a_0 + ib_0 \), the limit exists:
$$ \lim_{z \to z_0} f(z) = P(a_0, b_0) + i\phi(a_0, b_0) $$Reason: Sequences \( f(z_n) \to P(a_0, b_0) + i\phi(a_0, b_0) \).
Ex (ii): Calculation
\( f(z) = x^3 – xy + y^2 + i(2x + 3y) \)
Limit at \( z \to 1+i \):
$$ = 1^3 – (1)(1) + 1^2 + i(2(1) + 3(1)) = 1 + 5i $$Ex 3: Polynomial Ring in \( z, \bar{z} \)
If \( w = f(z, \bar{z}) \in \mathbb{C}[z, \bar{z}] \), then \( \lim_{z \to z_0} f(z) = f(z_0, \bar{z}_0) \).
Example: \( f(z) = 2z^2 + 4z|z|^2 + 5\bar{z} + 1 \)
Using \( |z|^2 = z\bar{z} \):
$$ = 2z^2 + 4z(z\bar{z}) + 5\bar{z} + 1 = 2z^2 + 4z^2\bar{z} + 5\bar{z} + 1 $$Variable Substitution & Non-Existence
Ex 4: Rewriting x and y
\( f(z) = x^2 + y|z|^2 + \bar{z}z^3 \cdot x \)
Substitute \( x = \frac{z+\bar{z}}{2} \) and \( y = \frac{z-\bar{z}}{2i} \):
$$ = \left(\frac{z+\bar{z}}{2}\right)^2 + \left(\frac{z-\bar{z}}{2i}\right)z\bar{z} + z \cdot z^3 \left(\frac{z+\bar{z}}{2}\right) $$Ex 5: General Rule
If \( f \in \mathbb{C}[x, y, z, \bar{z}, |z|^2] \), then \( \lim_{z \to z_0} f(z) = f(z_0) \).
Ex 6(a): Limit Does Not Exist
Evaluate \( \lim_{z \to 0} e^{-1/z^2} \).
- Path 1 (Real): \( z_n = 1/n \). \( f(z_n) = e^{-n^2} \to 0 \).
- Path 2 (Imaginary): \( t_n = i/n \). \( f(t_n) = e^{n^2} \to \infty \).
Since \( 0 \neq \infty \), the limit does not exist.
III. Real & Imaginary Parts
Ex 7: Composite Functions
$$ \lim_{z \to 0} [\sin(e^z + 2\pi i \cos(e^z – 1))]^2 + (3z^2 + z)\sin z $$Direct substitution (since continuous):
$$ = \sin(1 + 2\pi i) $$Theorem: Component-wise Limits
Let \( f(z) = u(x,y) + iv(x,y) \). Then \( \lim_{z \to z_0} f(z) \) exists iff:
$$ \lim_{(x,y) \to (x_0,y_0)} u(x,y) \quad \text{and} \quad \lim_{(x,y) \to (x_0,y_0)} v(x,y) \text{ exist.} $$If they exist:
$$ \lim_{z \to z_0} f(z) = \lim u(x,y) + i \lim v(x,y) $$Counter-Example
\( f(z) = \frac{x^2+y^2}{xy} + i \frac{x^4+y^2}{x^2y} \)
Limit at \( z \to 0 \) does not exist because the multivariable real limits (e.g., \( \lim \frac{x^4+y^2}{x^2y} \)) do not exist.
Advanced Examples & Formal Definition
Example: Squeeze Theorem Application
\( f(z) = x \sin \frac{1}{y} + i \sqrt{|xy|} \quad (y \neq 0) \)
Check limit as \( z \to 0 \):
- Real Part: \( \lim_{(x,y) \to (0,0)} x \sin \frac{1}{y} = 0 \) (Boundedness).
- Imaginary Part: \( \lim_{(x,y) \to (0,0)} \sqrt{|xy|} = 0 \).
Example: Path Dependence (y = mx)
Check \( \lim_{z \to 0} \frac{\bar{z}}{z} \).
Substitute \( z = x + iy \). Along the line \( y = mx \):
$$ \lim_{x \to 0} \frac{x – iy}{x + iy} = \lim_{x \to 0} \frac{x – i(mx)}{x + i(mx)} $$ $$ = \frac{x(1 – im)}{x(1 + im)} = \frac{1 – im}{1 + im} $$The result depends on \( m \). Thus, limit does not exist.
IV. The \( \epsilon – \delta \) Definition
We say \( \lim_{z \to z_0} f(z) = l \) if:
For any \( \epsilon > 0 \), there exists \( \delta > 0 \) such that:
$$ |f(z) – l| < \epsilon $$whenever \( 0 < |z - z_0| < \delta \) in \( D \).