MTH2003 Complex Analysis Lecture 3

• Previous Lecture: MTH2003 Complex Analysis Lecture 2.
• Pre-Lecture: MTH2003 Complex Analysis Pre-Lecture 3.
• Flashcards: MTH2003 Complex Analysis Flashcards.
• Next Lecture: MTH2003 Complex Analysis Lecture 4.

How are limits defined for complex functions?

What is the formal $ϵ-\delta$ definition?

A limit ${\lim }_{z\to z_0}f(z)=l$ exists if:

$$\forall ϵ>0,\exists \delta >0:0<|z-z_0|<\delta ⟹|f(z)-l|<ϵ.$$

This extends real analysis limits to two-dimensional approaches in $\mathbb{C}$, requiring $f(z)$ to approach $l$ regardless of the path taken toward $z_0$.

Why is this definition stricter than real limits?

In $\mathbb{R}$, limits only consider left/right approaches. In $\mathbb{C}$, $z$ can spiral or approach $z_0$ along infinitely many paths (e.g., lines, curves). The limit must hold for all possible approaches.

How do we compute complex limits using real and imaginary parts?

What decomposition lemma applies?

If $f(z)=u(x,y)+iv(x,y)$ and $l=a+ib$, then:

$$\underset{z\to z_0}{\lim }f(z)=l⟺\{\begin{array}{l}{\lim }_{(x,y)\to (x_0,y_0)}u(x,y)=a\\ {\lim }_{(x,y)\to (x_0,y_0)}v(x,y)=b\end{array}$$

This reduces complex limits to coupled real limits, which can be evaluated using multivariable calculus techniques.

Example: Compute ${\lim }_{z\to i}{z}^2$.

1. Express $z^2$ in real/imaginary parts: $z=x+iy⟹z^2=(x^2-y^2)+i(2xy)$.
2. Evaluate real and imaginary limits at $(0,1)$:
   • Real part: ${\lim }_{(x,y)\to (0,1)}{x}^2-y^2=-1$.
   • Imaginary part: ${\lim }_{(x,y)\to (0,1)}2xy=0$.
3. Combine: ${\lim }_{z\to i}{z}^2=-1+i0=-1$.

Key Insight: Both limits must agree independent of approach order (e.g., $x\to 0$ first vs. $y\to 1$ first). Contradictions imply no limit exists.

What techniques simplify calculating complex limits?

How do polynomial/rational functions behave?

Polynomials $P(z)$ are continuous everywhere, so:

$$\underset{z\to z_0}{\lim }P(z)=P(z_0).$$

For rational functions $f(z)=\frac{P(z)}{Q(z)}$, the limit exists at $z_0$ if $Q(z_0)\ne 0$. Discontinuities occur at roots of $Q(z)$.

What algebraic tricks help resolve indeterminate forms?

• Factor cancellation: Simplify $\frac{z^3+1}{z+1}$ by factoring numerator.
• Conjugate multiplication: Rationalize denominators like $\frac{1}{z-i}$.

What defines continuity in complex functions?

What is the formal criterion?

A function $f:D\to \mathbb{C}$ is continuous at $z_0\in D$ if:

1. $f(z_0)$ exists.
2. ${\lim }_{z\to z_0}f(z)$ exists.
3. ${\lim }_{z\to z_0}f(z)=f(z_0)$.

Why is continuity critical in complex analysis?

Continuous functions preserve limits under operations (addition, multiplication, composition). This underpins key results like the Intermediate Value Theorem and integration.

How do we determine continuity for complex rational functions?

Example: Is $f(z)=\frac{z^3+2z+1}{z^3+1}$ continuous?

1. Identify discontinuities: Solve $z^3+1=0⟹z=e^{i\pi /3},-1,e^{i5\pi /3}$ (cube roots of $-1$).
2. Domain: $D=\mathbb{C}\setminus \{e^{i\pi /3},-1,e^{i5\pi /3}\}$.
3. Continuity on $D$: For all $z_0\in D$, ${\lim }_{z\to z_0}f(z)=f(z_0)$ holds since numerator/denominator are polynomials and $z_0$ isn’t a root of the denominator.

Conclusion: $f(z)$ is continuous on $D$ but has removable discontinuities at the excluded points (if redefined, it could be continuous there too).

Key Takeaways:

• Complex limits require agreement across all approach paths.
• Decomposing into real/imaginary parts simplifies verification.
• Continuity in $\mathbb{C}$ mirrors $\mathbb{R}$ but with richer geometric implications.