The Central Idea
Suppose
\[ f:\mathbb{C}\to\mathbb{C}. \]We write
\[ \boxed{ \lim_{z\to z_0}f(z)=L } \]when the values of \(f(z)\) approach the complex number \(L\) whenever \(z\) approaches \(z_0\), regardless of the direction or path of approach.
The phrase regardless of the path is the central feature of complex limits.
Real Limits Versus Complex Limits
On the real line, a point can be approached primarily from two directions:
\[ x\to x_0^- \]and
\[ x\to x_0^+. \]In the complex plane, the point
\[ z_0=x_0+iy_0 \]may be approached along infinitely many lines, curves, spirals, or other paths.
Therefore, agreement along only two paths is not enough to establish a complex limit.
Approach Is Measured by Modulus
The distance between two complex numbers \(z\) and \(z_0\) is
\[ |z-z_0|. \]Thus the statement
\[ z\to z_0 \]means
\[ \boxed{|z-z_0|\to0}. \]Similarly, saying
\[ f(z)\to L \]means
\[ \boxed{|f(z)-L|\to0}. \]The Formal \(\varepsilon\)-\(\delta\) Definition
The statement
\[ \lim_{z\to z_0}f(z)=L \]means that for every
\[ \varepsilon>0, \]there exists
\[ \delta>0 \]such that
\[ 0<|z-z_0|<\delta \]implies
\[ |f(z)-L|<\varepsilon. \]Symbolically,
\[ \boxed{ \forall\varepsilon>0\; \exists\delta>0: 0<|z-z_0|<\delta \Rightarrow |f(z)-L|<\varepsilon }. \]Geometric Meaning of \(\varepsilon\) and \(\delta\)
The inequality
\[ |z-z_0|<\delta \]describes a disk of radius \(\delta\) around the input point \(z_0\).
The inequality
\[ |f(z)-L|<\varepsilon \]describes a disk of radius \(\varepsilon\) around the output value \(L\).
A limit exists if inputs sufficiently close to \(z_0\) are forced into an arbitrarily small output disk around \(L\).
The Punctured Neighborhood
In a limit,
\[ 0<|z-z_0|<\delta, \]the condition
\[ |z-z_0|>0 \]excludes the point \(z=z_0\) itself.
Therefore, a limit depends on values near \(z_0\), not necessarily on the value of \(f(z_0)\).
The set
\[ 0<|z-z_0|<\delta \]is called a punctured disk or punctured neighborhood.
Example: A Simple Polynomial Limit
Evaluate
\[ \lim_{z\to1+i}(z^2+2z). \]Polynomial functions are continuous everywhere in \(\mathbb{C}\), so direct substitution is valid:
\[ (1+i)^2+2(1+i). \]Since
\[ (1+i)^2=2i, \]we obtain
\[ 2i+2+2i. \]Therefore,
\[ \boxed{ \lim_{z\to1+i}(z^2+2z)=2+4i }. \]Limits Are Unique
If a complex limit exists, it can have only one value.
It is impossible for both
\[ \lim_{z\to z_0}f(z)=L_1 \]and
\[ \lim_{z\to z_0}f(z)=L_2 \]to hold with
\[ L_1\neq L_2. \]This is the same uniqueness principle used in real analysis.
Testing a Limit Along Paths
If a complex limit exists, every path approaching \(z_0\) must produce the same limiting value.
Therefore, to prove that a limit does not exist, it is enough to find two paths that produce different values.
This gives an efficient nonexistence test.
Example: A Limit That Does Not Exist
Consider
\[ f(z)=\frac{\overline{z}}{z}, \qquad z\neq0. \]Examine
\[ \lim_{z\to0}\frac{\overline{z}}{z}. \]Approach along the real axis:
\[ z=x, \qquad \overline{z}=x. \]Then
\[ \frac{\overline{z}}{z}=1. \]Approach along the imaginary axis:
\[ z=iy, \qquad \overline{z}=-iy. \]Then
\[ \frac{\overline{z}}{z}=-1. \]Because the two paths produce different values,
\[ \boxed{ \lim_{z\to0}\frac{\overline{z}}{z} \text{ does not exist}. } \]The Same Example in Polar Form
Write
\[ z=re^{i\theta}. \]Then
\[ \overline{z}=re^{-i\theta}. \]Therefore,
\[ \frac{\overline{z}}{z} = e^{-2i\theta}. \]As \(r\to0\), the expression still depends on \(\theta\).
Since the output changes with direction,
\[ \boxed{ \lim_{z\to0}\frac{\overline{z}}{z} \text{ does not exist}. } \]Matching a Few Paths Does Not Prove Existence
Testing paths is powerful for disproving a limit.
But checking only a few paths cannot prove a limit exists, because infinitely many other paths remain.
To prove existence, use:
- limit laws;
- continuity;
- estimates involving modulus;
- the \(\varepsilon\)-\(\delta\) definition;
- component-function analysis.
Limits Through Real and Imaginary Components
Let
\[ z=x+iy \]and
\[ f(z)=u(x,y)+iv(x,y). \]Suppose
\[ z_0=x_0+iy_0 \]and
\[ L=A+iB. \]Then
\[ \boxed{ \lim_{z\to z_0}f(z)=L } \]exactly when
\[ \boxed{ \lim_{(x,y)\to(x_0,y_0)}u(x,y)=A } \]and
\[ \boxed{ \lim_{(x,y)\to(x_0,y_0)}v(x,y)=B. } \]Example: Evaluate a Limit by Components
Let
\[ f(z)=z^2. \]With
\[ z=x+iy, \]we have
\[ f(z)=x^2-y^2+i(2xy). \]As
\[ z\to1+i, \]we have
\[ (x,y)\to(1,1). \]Therefore,
\[ x^2-y^2\to0 \]and
\[ 2xy\to2. \]Hence,
\[ \boxed{ \lim_{z\to1+i}z^2=2i }. \]Basic Limit Laws
Suppose
\[ \lim_{z\to z_0}f(z)=L \]and
\[ \lim_{z\to z_0}g(z)=M. \]Then:
\[ \boxed{ \lim_{z\to z_0}[f(z)+g(z)]=L+M } \] \[ \boxed{ \lim_{z\to z_0}[f(z)-g(z)]=L-M } \] \[ \boxed{ \lim_{z\to z_0}f(z)g(z)=LM } \]and, if \(M\neq0\),
\[ \boxed{ \lim_{z\to z_0}\frac{f(z)}{g(z)} = \frac{L}{M}. } \]Constant Multiples
If \(c\in\mathbb{C}\) is constant and
\[ \lim_{z\to z_0}f(z)=L, \]then
\[ \boxed{ \lim_{z\to z_0}cf(z)=cL. } \]Complex constants behave under limits just as real constants do.
Powers and Polynomial Limits
Since
\[ \lim_{z\to z_0}z=z_0, \]repeated multiplication gives
\[ \boxed{ \lim_{z\to z_0}z^n=z_0^n } \]for every positive integer \(n\).
Therefore, every polynomial
\[ P(z)=a_nz^n+\cdots+a_0 \]satisfies
\[ \boxed{ \lim_{z\to z_0}P(z)=P(z_0). } \]Rational Function Limits
If
\[ f(z)=\frac{P(z)}{Q(z)} \]and
\[ Q(z_0)\neq0, \]then
\[ \boxed{ \lim_{z\to z_0}\frac{P(z)}{Q(z)} = \frac{P(z_0)}{Q(z_0)}. } \]Direct substitution is valid whenever the denominator remains nonzero.
Example: Removing an Algebraic Obstruction
Evaluate
\[ \lim_{z\to1}\frac{z^2-1}{z-1}. \]Direct substitution gives \(0/0\), so factor:
\[ z^2-1=(z-1)(z+1). \]For \(z\neq1\),
\[ \frac{z^2-1}{z-1}=z+1. \]Therefore,
\[ \boxed{ \lim_{z\to1}\frac{z^2-1}{z-1}=2. } \]The Limit Need Not Equal the Function Value
A function may fail to be defined at \(z_0\) and still have a finite limit there.
In the previous example,
\[ \frac{z^2-1}{z-1} \]is undefined at
\[ z=1, \]but the limit exists and equals \(2\).
This distinction leads directly to continuity.
Continuity at a Point
A complex function \(f\) is continuous at \(z_0\) if:
- \(f(z_0)\) is defined;
- \(\lim_{z\to z_0}f(z)\) exists;
- the limit equals the function value.
Symbolically,
\[ \boxed{ \lim_{z\to z_0}f(z)=f(z_0). } \]The \(\varepsilon\)-\(\delta\) Definition of Continuity
A function \(f\) is continuous at \(z_0\) if for every
\[ \varepsilon>0 \]there exists
\[ \delta>0 \]such that
\[ |z-z_0|<\delta \]implies
\[ |f(z)-f(z_0)|<\varepsilon. \]Unlike the limit definition, the point \(z=z_0\) does not need to be excluded.
Continuity Through Component Functions
If
\[ f(z)=u(x,y)+iv(x,y), \]then \(f\) is continuous at
\[ z_0=x_0+iy_0 \]exactly when both
\[ u(x,y) \]and
\[ v(x,y) \]are continuous at
\[ (x_0,y_0). \]Polynomial Functions Are Continuous Everywhere
Every complex polynomial
\[ P(z) = a_nz^n+\cdots+a_1z+a_0 \]is continuous on all of
\[ \boxed{\mathbb{C}}. \]This follows directly from the limit laws.
Rational Functions Are Continuous on Their Domains
A rational function
\[ f(z)=\frac{P(z)}{Q(z)} \]is continuous at every point where
\[ Q(z)\neq0. \]Its possible discontinuities occur at the zeros of the denominator.
The Complex Exponential Is Continuous
Recall
\[ e^{x+iy} = e^x\cos y + i e^x\sin y. \]The component functions
\[ u(x,y)=e^x\cos y \]and
\[ v(x,y)=e^x\sin y \]are continuous everywhere in \(\mathbb{R}^2\).
Therefore,
\[ \boxed{e^z\text{ is continuous for every }z\in\mathbb{C}.} \]The Conjugation Function Is Continuous
Consider
\[ f(z)=\overline{z}. \]If
\[ z=x+iy, \]then
\[ f(z)=x-iy. \]Its component functions are
\[ u(x,y)=x, \qquad v(x,y)=-y. \]Both are continuous everywhere, so
\[ \boxed{ \overline{z} \text{ is continuous on }\mathbb{C}. } \]The Modulus Function Is Continuous
The modulus function
\[ f(z)=|z| \]is continuous everywhere.
One useful estimate is the reverse triangle inequality:
\[ \boxed{ \bigl||z|-|w|\bigr| \le |z-w|. } \]Thus, if \(z\to w\), then
\[ |z|\to|w|. \]The Reverse Triangle Inequality
The ordinary triangle inequality gives
\[ |z| \le |z-w|+|w|. \]Therefore,
\[ |z|-|w| \le |z-w|. \]Interchanging \(z\) and \(w\) gives
\[ |w|-|z| \le |z-w|. \]Combining the two,
\[ \boxed{ \bigl||z|-|w|\bigr| \le |z-w|. } \]This inequality is frequently useful in complex limit proofs.
Example: An \(\varepsilon\)-\(\delta\) Proof
Prove
\[ \lim_{z\to z_0}(2z+3)=2z_0+3. \]Consider
\[ |(2z+3)-(2z_0+3)|. \]Simplify:
\[ |2(z-z_0)| = 2|z-z_0|. \]To make this less than \(\varepsilon\), it is enough to require
\[ |z-z_0|<\frac{\varepsilon}{2}. \]Therefore, choose
\[ \boxed{\delta=\frac{\varepsilon}{2}}. \]Then
\[ |z-z_0|<\delta \]implies
\[ |(2z+3)-(2z_0+3)|<\varepsilon. \]Example: Proving \(\lim_{z\to0}z^2=0\)
We want
\[ |z^2|<\varepsilon. \]Since
\[ |z^2|=|z|^2, \]it is enough to require
\[ |z|<\sqrt{\varepsilon}. \]Therefore choose
\[ \boxed{\delta=\sqrt{\varepsilon}}. \]Then
\[ 0<|z|<\delta \]implies
\[ |z^2|<\varepsilon. \]Estimating with Modulus
Complex limits are often proved by bounding the modulus of an expression.
If
\[ |f(z)| \le g(z) \]and
\[ g(z)\to0, \]then
\[ f(z)\to0. \]This is the complex analogue of squeeze reasoning.
Example: A Limit by Estimation
Evaluate
\[ \lim_{z\to0}z\overline{z}. \]Since
\[ z\overline{z}=|z|^2, \]we immediately obtain
\[ |z|^2\to0. \]Therefore,
\[ \boxed{ \lim_{z\to0}z\overline{z}=0. } \]Example: Path Dependence in Two Real Variables
Let
\[ f(z) = \frac{x^2-y^2}{x^2+y^2}, \qquad z=x+iy\neq0. \]Along the real axis,
\[ y=0, \]so
\[ f(z)=1. \]Along the imaginary axis,
\[ x=0, \]so
\[ f(z)=-1. \]Hence,
\[ \boxed{ \lim_{z\to0}f(z) \text{ does not exist}. } \]Approaching Along Lines
One common family of paths toward the origin is
\[ y=mx. \]Substituting this relationship into a function can reveal whether the limit depends on the slope \(m\).
If different values of \(m\) produce different limits, the complex limit does not exist.
Curved Paths Can Matter
Agreement along every straight line still does not guarantee a multivariable or complex limit.
A function can behave differently along a curve such as
\[ y=x^2 \]or
\[ y=x^3. \]Therefore, line testing is a diagnostic tool, not a complete proof of existence.
Operations Preserve Continuity
If \(f\) and \(g\) are continuous at \(z_0\), then so are:
- \(f+g\);
- \(f-g\);
- \(fg\);
- \(cf\) for any constant \(c\in\mathbb{C}\);
- \(f/g\), provided \(g(z_0)\neq0\).
This allows complicated continuous functions to be built from simpler ones.
Composition Preserves Continuity
If \(g\) is continuous at \(z_0\) and \(f\) is continuous at
\[ g(z_0), \]then
\[ f\circ g \]is continuous at \(z_0\).
Symbolically,
\[ \boxed{ \lim_{z\to z_0}f(g(z)) = f(g(z_0)). } \]Removable Discontinuities
Suppose \(f\) is undefined at \(z_0\), but
\[ \lim_{z\to z_0}f(z)=L \]exists and is finite.
Then the discontinuity can be removed by defining
\[ \boxed{f(z_0)=L}. \]This creates a continuous extension of the function at \(z_0\).
Example: Creating a Continuous Extension
Let
\[ f(z)=\frac{z^2-1}{z-1}, \qquad z\neq1. \]For \(z\neq1\),
\[ f(z)=z+1. \]Since
\[ \lim_{z\to1}f(z)=2, \]defining
\[ \boxed{f(1)=2} \]makes the extended function continuous at \(z=1\).
A Discontinuity That Is Not Removable
Consider
\[ f(z)=\frac1z. \]As \(z\to0\),
\[ |f(z)|=\frac1{|z|}. \]The modulus becomes arbitrarily large.
There is no finite complex number \(L\) such that
\[ \lim_{z\to0}\frac1z=L. \]This type of behavior will later be associated with a pole.
A Preview of Infinite Limits
Although \(\infty\) is not an ordinary complex number, it is useful to write
\[ f(z)\to\infty \]when
\[ |f(z)|\to\infty. \]For example,
\[ \frac1z\to\infty \qquad \text{as }z\to0 \]in this extended sense.
Later, the extended complex plane will make this idea more precise.
Limits as \(z\to\infty\)
The statement
\[ z\to\infty \]means
\[ |z|\to\infty. \]For example,
\[ \lim_{z\to\infty}\frac1z=0 \]because
\[ \left|\frac1z\right| = \frac1{|z|} \to0. \]Thus,
\[ \boxed{ \lim_{z\to\infty}\frac1z=0. } \]Limits and Sequences
A useful equivalent viewpoint uses sequences.
The limit
\[ \lim_{z\to z_0}f(z)=L \]exists exactly when every sequence
\[ z_n\to z_0 \]with
\[ z_n\neq z_0 \]satisfies
\[ \boxed{f(z_n)\to L}. \]This gives another way to reason about path independence.
Continuity and Sequences
A function \(f\) is continuous at \(z_0\) exactly when every sequence
\[ z_n\to z_0 \]satisfies
\[ \boxed{ f(z_n)\to f(z_0). } \]This sequential formulation is often convenient in proofs.
Neighborhoods and Open Sets
A disk centered at \(z_0\) with radius \(r>0\) is
\[ D(z_0,r) = \{z:|z-z_0|Open sets provide the natural domains on which complex differentiability and analyticity are studied.
Interior Points
A point \(z_0\) is an interior point of a set \(S\) if there exists
\[ r>0 \]such that
\[ D(z_0,r)\subseteq S. \]This means a small disk around \(z_0\) remains entirely inside the domain.
Such neighborhoods are important because complex derivatives require nearby points in every direction.
Common Error: Checking Only the Real and Imaginary Axes
If two paths give different values, the limit does not exist.
But if the real-axis and imaginary-axis limits agree, the complex limit may still fail along another line or curve.
Agreement along a few paths is evidence, not proof.
Common Error: Substituting Before Checking the Domain
Direct substitution is valid for continuous functions at points in their domain.
If substitution produces
\[ \frac00, \]the expression is indeterminate, not automatically zero and not proof that the limit fails.
Simplify or analyze the expression further.
Common Error: Treating \(0/0\) as a Number
The expression
\[ \frac00 \]is undefined.
In a limit calculation, it signals that additional analysis is needed.
Possible next steps include:
- factoring;
- canceling a common factor away from the limiting point;
- using an estimate;
- testing paths;
- rewriting in rectangular or polar form.
Common Error: Confusing Existence of a Limit with Continuity
A limit can exist even if \(f(z_0)\) is undefined.
Continuity requires all three:
\[ f(z_0)\text{ exists}, \] \[ \lim_{z\to z_0}f(z)\text{ exists}, \]and
\[ \lim_{z\to z_0}f(z)=f(z_0). \]Do not merge these separate conditions.
Common Error: Thinking “Approach” Means Only Straight-Line Motion
The notation
\[ z\to z_0 \]does not specify a line.
It includes every possible way for the distance
\[ |z-z_0| \]to approach zero.
A Reliable Procedure for Evaluating a Complex Limit
- Identify the limiting point \(z_0\).
- Check whether the function is already continuous there.
- If direct substitution works, use it.
- If an indeterminate form appears, simplify algebraically.
- Use rectangular components when helpful.
- Use polar form when radial and angular behavior matter.
- For suspected nonexistence, compare different paths.
- For a proof of existence, use limit laws, estimates, or \(\varepsilon\)-\(\delta\) reasoning.
A Reliable Procedure for Testing Continuity
- Confirm that \(f(z_0)\) is defined.
- Compute or determine \(\lim_{z\to z_0}f(z)\).
- Check whether the limit exists independently of path.
- Compare the limit with \(f(z_0)\).
- If they agree, the function is continuous at \(z_0\).
- If the limit exists but the value is missing or different, test for a removable discontinuity.
Worked Example: Rational Limit
Evaluate
\[ \lim_{z\to2+i} \frac{z+1}{z-1}. \]The denominator at \(z=2+i\) is
\[ 1+i\neq0. \]Therefore direct substitution is valid:
\[ \frac{3+i}{1+i}. \]Multiply numerator and denominator by \(1-i\):
\[ \frac{(3+i)(1-i)}{2}. \]The numerator is
\[ 3-3i+i-i^2 = 4-2i. \]Therefore,
\[ \boxed{ \lim_{z\to2+i} \frac{z+1}{z-1} = 2-i. } \]Worked Example: Continuity by Components
Consider
\[ f(z)=z\overline{z}. \]If
\[ z=x+iy, \]then
\[ z\overline{z}=x^2+y^2. \]Thus
\[ u(x,y)=x^2+y^2 \]and
\[ v(x,y)=0. \]Both component functions are continuous everywhere, so
\[ \boxed{ f(z)=z\overline{z} \text{ is continuous on }\mathbb{C}. } \]Worked Example: Detecting Path Dependence
Consider
\[ f(z)=\frac{x^2}{x^2+y^2}, \qquad z=x+iy\neq0. \]Along \(y=0\),
\[ f(z)=1. \]Along \(x=0\),
\[ f(z)=0. \]Since the two path limits differ,
\[ \boxed{ \lim_{z\to0}f(z) \text{ does not exist}. } \]Worked Example: Removable Discontinuity
Define
\[ f(z) = \frac{z^2+z-2}{z-1}, \qquad z\neq1. \]Factor the numerator:
\[ z^2+z-2=(z-1)(z+2). \]For \(z\neq1\),
\[ f(z)=z+2. \]Therefore,
\[ \lim_{z\to1}f(z)=3. \]Defining
\[ \boxed{f(1)=3} \]removes the discontinuity.
Practice 1: Direct Substitution
Evaluate
\[ \lim_{z\to2-i}(z^2+3z-1). \]Practice 2: Rational Limit
Evaluate
\[ \lim_{z\to i}\frac{z+2}{z+1}. \]Practice 3: Factor First
Evaluate
\[ \lim_{z\to3}\frac{z^2-9}{z-3}. \]Practice 4: Test Path Dependence
Determine whether
\[ \lim_{z\to0}\frac{x^2-y^2}{x^2+y^2} \]exists.
Practice 5: Conjugate Limit
Evaluate
\[ \lim_{z\to2+3i}\overline{z}. \]Practice 6: Modulus Limit
Evaluate
\[ \lim_{z\to-1+2i}|z|. \]Practice 7: Continuity
Is
\[ f(z)=\frac1{z-4} \]continuous at
\[ z=2+i? \]Practice 8: Removable Discontinuity
Find the value that should be assigned to \(f(2)\) to make
\[ f(z)=\frac{z^2-4}{z-2} \]continuous at \(z=2\).
Practice 9: Limit at Infinity
Evaluate
\[ \lim_{z\to\infty}\frac{3}{z}. \]Practice 10: Component Continuity
Determine whether
\[ f(z)=x^2-y^2+i(x+y) \]is continuous on \(\mathbb{C}\).
Practice Answers
Practice 1
Substitute
\[ z=2-i. \]Since
\[ (2-i)^2=3-4i, \]we get
\[ 3-4i+6-3i-1. \]Therefore,
\[ \boxed{8-7i}. \]Practice 2
Since \(i+1\neq0\), direct substitution gives
\[ \frac{i+2}{i+1}. \]Multiply by \(1-i\):
\[ \frac{(2+i)(1-i)}{2}. \]The numerator is
\[ 3-i. \]Therefore,
\[ \boxed{ \frac32-\frac12 i }. \]Practice 3
\[ z^2-9=(z-3)(z+3). \]Therefore,
\[ \lim_{z\to3}\frac{z^2-9}{z-3} = \lim_{z\to3}(z+3) = \boxed{6}. \]Practice 4
Along \(y=0\), the expression equals \(1\).
Along \(x=0\), it equals \(-1\).
Therefore,
\[ \boxed{ \text{the limit does not exist}. } \]Practice 5
Conjugation is continuous, so
\[ \lim_{z\to2+3i}\overline{z} = \overline{2+3i}. \]Therefore,
\[ \boxed{2-3i}. \]Practice 6
Modulus is continuous, so
\[ \lim_{z\to-1+2i}|z| = |-1+2i|. \] \[ = \sqrt{1+4}. \]Therefore,
\[ \boxed{\sqrt5}. \]Practice 7
The only excluded point is
\[ z=4. \]Since
\[ 2+i\neq4, \]the function is continuous there.
\[ \boxed{\text{Yes.}} \]Practice 8
\[ z^2-4=(z-2)(z+2). \]Therefore,
\[ \lim_{z\to2}\frac{z^2-4}{z-2} = \lim_{z\to2}(z+2) = 4. \]Define
\[ \boxed{f(2)=4}. \]Practice 9
Since
\[ \left|\frac3z\right| = \frac3{|z|} \to0, \]we have
\[ \boxed{ \lim_{z\to\infty}\frac3z=0. } \]Practice 10
The component functions are
\[ u(x,y)=x^2-y^2 \]and
\[ v(x,y)=x+y. \]Both are polynomials in \(x\) and \(y\), so both are continuous everywhere.
Therefore,
\[ \boxed{ f\text{ is continuous on }\mathbb{C}. } \]Checking Strategies
- If direct substitution works, verify that the denominator is nonzero.
- If a proposed limit is finite, test whether the modulus of the difference from that limit tends to zero.
- For suspected nonexistence, compare at least two strategically chosen paths.
- When a function is written as \(u+iv\), verify both component limits.
- For continuity, separately check the function value, the limit, and their equality.
- If a discontinuity appears algebraic, factor before concluding that it is nonremovable.
Connections to Earlier Mathematics
Complex limits and continuity unify several earlier ideas:
- Real analysis: \(\varepsilon\)-\(\delta\) reasoning and limit laws.
- Multivariable calculus: approach through the plane and component-function limits.
- Geometry: neighborhoods are disks measured by modulus.
- Algebra: factoring and simplification resolve removable singular behavior.
- Complex functions: continuity describes stable mappings of nearby points.
Why This Matters for Complex Differentiation
The complex derivative is defined by the limit
\[ f'(z_0) = \lim_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0}. \]Because \(z\) may approach \(z_0\) from any direction, this quotient must converge to the same value along every possible path.
This is why complex differentiability is substantially more restrictive than ordinary real differentiability.
What to Remember
A complex limit
\[ \boxed{ \lim_{z\to z_0}f(z)=L } \]requires the same output limit along every path toward \(z_0\).
The formal condition is
\[ \boxed{ 0<|z-z_0|<\delta \Rightarrow |f(z)-L|<\varepsilon. } \]Continuity at \(z_0\) means
\[ \boxed{ \lim_{z\to z_0}f(z)=f(z_0). } \]For
\[ f(z)=u(x,y)+iv(x,y), \]limits and continuity can be studied through the two real component functions \(u\) and \(v\).
The most important diagnostic principle is: different path limits prove nonexistence.
Where We Go Next
Complex limits now give us the machinery needed to define a derivative in the complex plane.
The next lesson, Complex Differentiation, studies the limit \[ f'(z) = \lim_{h\to0} \frac{f(z+h)-f(z)}{h}, \] explains why the direction of \(h\) matters, and shows why differentiability imposes powerful relationships on the real and imaginary component functions.