The Central Idea
Let \(C\) be an oriented curve in the complex plane and let \(f\) be a complex-valued function defined along that curve.
The complex line integral is written
\[ \boxed{ \int_C f(z)\,dz. } \]The symbol \(dz\) records an infinitesimal complex displacement along the contour.
To compute the integral, we parametrize the curve and convert the problem into an ordinary real-variable integral.
What Is a Contour?
A contour is a directed curve in the complex plane.
A convenient parametrization is
\[ \boxed{ z=z(t), \qquad a\le t\le b. } \]As \(t\) increases from \(a\) to \(b\), the point \(z(t)\) traces the contour in a specified direction.
Rectangular Parametrization
A contour may be written
\[ z(t)=x(t)+iy(t). \]Then
\[ \boxed{ z'(t)=x'(t)+iy'(t). } \]The derivative \(z'(t)\) is the complex tangent vector to the curve.
The Differential \(dz\)
If
\[ z=z(t), \]then
\[ \boxed{ dz=z'(t)\,dt. } \]This is the substitution that converts a complex contour integral into an ordinary integral with respect to the real parameter \(t\).
Definition of the Complex Line Integral
If
\[ C:z=z(t), \qquad a\le t\le b, \]then
\[ \boxed{ \int_C f(z)\,dz = \int_a^b f(z(t))z'(t)\,dt. } \]This formula is the basic computational definition of contour integration.
Example: Integrate Along a Line Segment
Let \(C\) be the line segment from \(0\) to \(1+i\).
Parametrize it by
\[ z(t)=t(1+i), \qquad 0\le t\le1. \]Then
\[ z'(t)=1+i. \]Evaluate
\[ \int_C z\,dz. \]Substitute:
\[ \int_0^1 t(1+i)(1+i)\,dt. \]Since
\[ (1+i)^2=2i, \]we obtain
\[ 2i\int_0^1 t\,dt. \]Therefore,
\[ \boxed{ \int_C z\,dz=i. } \]A General Line Segment
The line segment from \(z_0\) to \(z_1\) can be parametrized by
\[ \boxed{ z(t)=z_0+t(z_1-z_0), \qquad 0\le t\le1. } \]Then
\[ \boxed{ z'(t)=z_1-z_0. } \]This is one of the most useful standard contour parametrizations.
Parametrizing a Circle
The circle centered at \(z_0\) with radius \(R\) can be parametrized counterclockwise by
\[ \boxed{ z(t)=z_0+Re^{it}, \qquad 0\le t\le2\pi. } \]Then
\[ \boxed{ z'(t)=iRe^{it}. } \]Orientation Matters
A contour has direction.
If \(C^{-}\) denotes the same geometric curve traversed in the opposite direction, then
\[ \boxed{ \int_{C^{-}}f(z)\,dz = -\int_C f(z)\,dz. } \]Reversing orientation changes the sign of the integral.
Positive Orientation
For a simple closed contour in the plane, the standard positive orientation is counterclockwise.
Clockwise traversal is the negative orientation.
This convention becomes especially important in Cauchy's theorem and the residue theorem.
Closed Contours
A contour is closed if it starts and ends at the same point.
If
\[ z(a)=z(b), \]then the contour is closed.
Integrals around closed contours are often written
\[ \boxed{ \oint_C f(z)\,dz. } \]Simple Contours
A contour is simple if it does not cross itself, except possibly where a closed contour returns to its starting point.
Simple closed contours divide the plane into an interior and an exterior.
This geometric distinction becomes important when singularities lie inside or outside a contour.
Piecewise Smooth Contours
Many contours are made from several smooth pieces.
A polygonal path, for example, has corners but each side is smooth.
Such a curve is called piecewise smooth.
Complex line integrals are commonly defined over piecewise smooth contours.
Additivity Across Contour Pieces
If a contour \(C\) is formed by traversing \(C_1\) and then \(C_2\), then
\[ \boxed{ \int_C f(z)\,dz = \int_{C_1}f(z)\,dz + \int_{C_2}f(z)\,dz. } \]This allows a complicated contour to be broken into manageable pieces.
Linearity
For complex constants \(a\) and \(b\),
\[ \boxed{ \int_C [af(z)+bg(z)]\,dz = a\int_C f(z)\,dz + b\int_C g(z)\,dz. } \]Complex contour integration is linear just like ordinary integration.
Separating Real and Imaginary Parts
Write
\[ f(z)=u(x,y)+iv(x,y) \]and
\[ dz=dx+i\,dy. \]Then
\[ f(z)\,dz = (u+iv)(dx+i\,dy). \]Expanding,
\[ \boxed{ f(z)\,dz = (u\,dx-v\,dy) + i(v\,dx+u\,dy). } \]A Complex Integral as Two Real Line Integrals
Therefore,
\[ \boxed{ \int_C f(z)\,dz = \int_C (u\,dx-v\,dy) + i\int_C(v\,dx+u\,dy). } \]This formula connects complex integration directly with vector calculus and real line integrals.
Example: Integrate by Components
Suppose
\[ f(z)=z=x+iy \]and \(C\) is any smooth path from \(0\) to \(1+i\).
Since
\[ \frac{d}{dz}\left(\frac{z^2}{2}\right)=z, \]an antiderivative exists.
Therefore,
\[ \int_C z\,dz = \frac{(1+i)^2}{2}-0. \]Since
\[ (1+i)^2=2i, \]we obtain
\[ \boxed{ \int_C z\,dz=i. } \]Complex Antiderivatives
A function \(F\) is an antiderivative of \(f\) on a domain if
\[ \boxed{ F'(z)=f(z). } \]When an antiderivative exists throughout a region containing the contour, contour integration becomes especially simple.
The Fundamental Theorem for Complex Line Integrals
If \(F'(z)=f(z)\) on a domain containing a contour \(C\) from \(z_0\) to \(z_1\), then
\[ \boxed{ \int_C f(z)\,dz = F(z_1)-F(z_0). } \]The integral then depends only on the endpoints, not on the particular path.
Path Independence
An integral is path independent when every contour with the same starting and ending points gives the same value.
If \(f\) has an antiderivative on the domain, then
\[ \boxed{ \int_C f(z)\,dz } \]is path independent.
Antiderivatives and Closed Contours
If \(f\) has an antiderivative \(F\) and \(C\) is closed, then its endpoints coincide.
Therefore,
\[ \int_C f(z)\,dz = F(z_0)-F(z_0). \]Hence,
\[ \boxed{ \oint_C f(z)\,dz=0. } \]Example: Polynomial Around a Closed Contour
Let
\[ f(z)=3z^2+2z+1. \]An antiderivative is
\[ F(z)=z^3+z^2+z. \]Therefore, for every closed contour \(C\),
\[ \boxed{ \oint_C(3z^2+2z+1)\,dz=0. } \]Example: Direct Integration Around a Circle
Evaluate
\[ \oint_C z\,dz \]where \(C\) is the unit circle traversed counterclockwise.
Parametrize:
\[ z(t)=e^{it}, \qquad 0\le t\le2\pi. \]Then
\[ dz=ie^{it}\,dt. \]Therefore,
\[ \oint_C z\,dz = \int_0^{2\pi} e^{it}ie^{it}\,dt. \] \[ = i\int_0^{2\pi}e^{2it}\,dt. \]The integral is zero over a complete period, so
\[ \boxed{ \oint_C z\,dz=0. } \]A Crucial Example: \(\displaystyle \oint_C \frac{1}{z}\,dz\)
Let \(C\) be the positively oriented unit circle:
\[ z(t)=e^{it}, \qquad 0\le t\le2\pi. \]Then
\[ dz=ie^{it}\,dt. \]Therefore,
\[ \oint_C\frac1z\,dz = \int_0^{2\pi} \frac1{e^{it}} ie^{it}\,dt. \]The factors cancel:
\[ =i\int_0^{2\pi}dt. \]Hence,
\[ \boxed{ \oint_C\frac1z\,dz=2\pi i. } \]Why \(\displaystyle \oint_C \frac1z\,dz\) Is So Important
The function
\[ \frac1z \]is analytic everywhere except at the origin.
The contour surrounds that singular point.
The integral is therefore not zero:
\[ \boxed{2\pi i}. \]This example foreshadows the central role of enclosed singularities in Cauchy's integral formula and the residue theorem.
General Circle Around a Point
Let
\[ C:|z-z_0|=R \]be positively oriented.
Parametrize:
\[ z=z_0+Re^{it}. \]Then
\[ dz=iRe^{it}\,dt. \]Therefore,
\[ \boxed{ \oint_C\frac{1}{z-z_0}\,dz = 2\pi i. } \]Powers Around a Circle
Consider
\[ \oint_C(z-z_0)^n\,dz \]around a positively oriented circle centered at \(z_0\).
The result is
\[ \boxed{ \oint_C(z-z_0)^n\,dz = \begin{cases} 2\pi i, & n=-1,\\ 0, & n\neq-1. \end{cases} } \]The exponent \(-1\) is exceptional.
Length of a Contour
If
\[ C:z=z(t), \qquad a\le t\le b, \]then its length is
\[ \boxed{ L(C) = \int_a^b|z'(t)|\,dt. } \]This is the ordinary arc-length formula written in complex notation.
The \(ML\) Estimate
Suppose
\[ |f(z)|\le M \]for every point on a contour \(C\), and suppose the contour has length
\[ L. \]Then
\[ \boxed{ \left| \int_C f(z)\,dz \right| \le ML. } \]This is one of the most useful estimation tools in complex integration.
Why the \(ML\) Estimate Works
Parametrize \(C\) by \(z(t)\).
Then
\[ \left| \int_C f(z)\,dz \right| = \left| \int_a^b f(z(t))z'(t)\,dt \right|. \]By the triangle inequality for integrals,
\[ \le \int_a^b |f(z(t))||z'(t)|\,dt. \]If \(|f(z)|\le M\), then
\[ \le M\int_a^b|z'(t)|\,dt. \]Therefore,
\[ \boxed{ \left| \int_C f(z)\,dz \right| \le ML. } \]Example: Estimating an Integral
Suppose \(C\) is the circle
\[ |z|=2 \]and
\[ f(z)=\frac1{z-5}. \]On the circle,
\[ |z-5| \ge 5-|z| = 3. \]Therefore,
\[ |f(z)| \le \frac13. \]The contour length is
\[ L=2\pi(2)=4\pi. \]Hence,
\[ \boxed{ \left| \oint_C\frac1{z-5}\,dz \right| \le \frac{4\pi}{3}. } \]Independence from Parametrization
Different smooth parametrizations can trace the same oriented contour.
As long as they preserve the same orientation and satisfy the appropriate regularity conditions, they produce the same contour integral.
Thus the integral depends on the oriented geometric path, not on the particular parameter used to describe it.
When Path Dependence Can Occur
If a function does not possess an antiderivative on the relevant domain, two paths with the same endpoints may produce different integral values.
A classic source of path dependence is a domain containing a hole or singularity that prevents a global antiderivative.
Why \(1/z\) Has No Global Antiderivative on \(\mathbb{C}\setminus\{0\}\)
If \(1/z\) had an antiderivative throughout
\[ \mathbb{C}\setminus\{0\}, \]then every closed contour integral would be zero.
But for the unit circle,
\[ \oint_C\frac1z\,dz=2\pi i. \]Therefore,
\[ \boxed{ \frac1z \text{ has no single-valued global antiderivative on } \mathbb{C}\setminus\{0\}. } \]Local Antiderivatives and the Logarithm
Although \(1/z\) has no global antiderivative on the punctured plane, it does possess local antiderivatives on suitable regions where a branch of the logarithm can be chosen.
On such a region,
\[ \boxed{ \frac{d}{dz}\operatorname{Log}z = \frac1z. } \]This is another example of how domain topology matters in complex analysis.
A Preview of Simply Connected Domains
Informally, a domain is simply connected if it has no holes that obstruct continuous contraction of closed curves.
In simply connected regions, analytic functions have especially strong integral properties.
This idea will appear in Cauchy's theorem.
A Preview of Contour Deformation
If a function is analytic throughout the region between two contours, one contour can often be continuously deformed into the other without changing the integral.
Singularities obstruct such deformations.
This principle becomes a major computational tool later in the theory.
Analytic Functions and Closed-Contour Integrals
We have already seen that functions with antiderivatives integrate to zero around closed contours.
Cauchy's theorem will show that analytic functions satisfy an even more powerful closed-contour result under appropriate domain conditions.
This is one of the central turning points of complex analysis.
Example: Integrate Along a Circular Arc
Let \(C\) be the quarter-circle
\[ |z|=2 \]from \(2\) to \(2i\), traversed counterclockwise.
Parametrize:
\[ z(t)=2e^{it}, \qquad 0\le t\le\frac{\pi}{2}. \]Then
\[ dz=2ie^{it}\,dt. \]Evaluate
\[ \int_C z^2\,dz. \]Substitute:
\[ \int_0^{\pi/2} (2e^{it})^2 (2ie^{it})\,dt. \] \[ = 8i \int_0^{\pi/2} e^{3it}\,dt. \]Integrating,
\[ = \frac{8}{3} \left[ e^{3it} \right]_0^{\pi/2}. \]Since
\[ e^{3i\pi/2}=-i, \]we obtain
\[ \boxed{ \int_C z^2\,dz = \frac{8}{3}(-1-i). } \]Checking the Circular-Arc Example with an Antiderivative
Since
\[ \frac{d}{dz}\left(\frac{z^3}{3}\right)=z^2, \]we can instead write
\[ \int_C z^2\,dz = \frac{(2i)^3}{3} - \frac{2^3}{3}. \]Because
\[ (2i)^3=-8i, \]we obtain
\[ \boxed{ \frac{-8i-8}{3} = \frac{8}{3}(-1-i). } \]The two methods agree.
Common Error: Forgetting the \(z'(t)\) Factor
After parametrizing
\[ z=z(t), \]the differential becomes
\[ dz=z'(t)\,dt. \]Therefore,
\[ \int_C f(z)\,dz \]becomes
\[ \boxed{ \int_a^b f(z(t))z'(t)\,dt, } \]not merely \(\int_a^b f(z(t))\,dt\).
Common Error: Ignoring Orientation
The same geometric curve traversed in opposite directions gives integrals with opposite signs.
\[ \boxed{ \int_{-C}f(z)\,dz = -\int_C f(z)\,dz. } \]Always identify the starting point, ending point, and direction.
Common Error: Assuming Every Closed-Contour Integral Is Zero
Closed-contour integrals vanish under specific analytic or antiderivative conditions.
The example
\[ \oint_C\frac1z\,dz=2\pi i \]shows that a closed contour can produce a nonzero value when a singularity affects the region.
Common Error: Using an Antiderivative Outside Its Valid Domain
A branch of
\[ \log z \]may serve as a local antiderivative of \(1/z\), but no single-valued branch works on the entire punctured plane.
Domain restrictions must be respected.
Common Error: Using a Parametrization That Traces the Wrong Curve
A valid parametrization must reproduce:
- the correct geometric path;
- the correct starting point;
- the correct ending point;
- the correct orientation.
Always test the parameter endpoints before integrating.
Common Error: Using the \(ML\) Estimate as an Equality
The estimate
\[ \left| \int_C f(z)\,dz \right| \le ML \]provides an upper bound.
It does not normally give the exact integral.
A Reliable Procedure for Direct Contour Integration
- Identify the contour and its orientation.
- Choose a convenient parametrization \(z=z(t)\).
- Verify the parameter endpoints.
- Compute \(z'(t)\).
- Substitute \(z(t)\) into \(f(z)\).
- Replace \(dz\) by \(z'(t)\,dt\).
- Evaluate the resulting real-variable integral.
- Simplify the final complex value.
A Reliable Procedure When an Antiderivative Exists
- Identify a function \(F\) satisfying \(F'=f\).
- Confirm that \(F\) is valid on a domain containing the contour.
- Identify the starting point \(z_0\).
- Identify the ending point \(z_1\).
- Compute \(F(z_1)-F(z_0)\).
- Use path independence to avoid unnecessary parametrization.
A Reliable Procedure for an \(ML\) Estimate
- Find or bound \(|f(z)|\) on the contour.
- Choose a valid maximum bound \(M\).
- Compute or bound the contour length \(L\).
- Apply \[ \left|\int_C f(z)\,dz\right|\le ML. \]
- State clearly that the result is an upper bound.
Worked Example: A Line Segment
Evaluate
\[ \int_C \overline{z}\,dz \]where \(C\) is the line segment from \(0\) to \(1+i\).
Parametrize:
\[ z(t)=t(1+i), \qquad 0\le t\le1. \]Then
\[ \overline{z(t)} = t(1-i) \]and
\[ dz=(1+i)\,dt. \]Therefore,
\[ \int_C\overline{z}\,dz = \int_0^1 t(1-i)(1+i)\,dt. \]Since
\[ (1-i)(1+i)=2, \]we obtain
\[ 2\int_0^1t\,dt. \]Therefore,
\[ \boxed{ \int_C\overline{z}\,dz=1. } \]Worked Example: A Piecewise Linear Contour
Let \(C\) travel from \(0\) to \(1\), then from \(1\) to \(1+i\).
Evaluate
\[ \int_C z\,dz. \]Since \(z\) has antiderivative
\[ F(z)=\frac{z^2}{2}, \]path independence gives
\[ \int_C z\,dz = \frac{(1+i)^2}{2}-0. \]Therefore,
\[ \boxed{i}. \]No separate piecewise parametrization is needed.
Worked Example: A Singular Integrand Around a Circle
Evaluate
\[ \oint_{|z-2|=3} \frac{1}{z-2}\,dz \]counterclockwise.
Parametrize:
\[ z=2+3e^{it}, \qquad 0\le t\le2\pi. \]Then
\[ dz=3ie^{it}\,dt. \]Also,
\[ z-2=3e^{it}. \]Therefore,
\[ \oint \frac1{z-2}\,dz = \int_0^{2\pi} \frac{3ie^{it}}{3e^{it}}\,dt. \] \[ = i\int_0^{2\pi}dt. \]Hence,
\[ \boxed{2\pi i}. \]Practice 1: Parametrize a Segment
Give a parametrization of the line segment from
\[ z_0=1-i \]to
\[ z_1=3+2i. \]Practice 2: Direct Integral
Evaluate
\[ \int_C z\,dz \]where \(C\) is the line segment from \(0\) to \(2\).
Practice 3: Antiderivative
Evaluate
\[ \int_C z^3\,dz \]along any contour from \(1\) to \(i\).
Practice 4: Closed Polynomial Integral
Evaluate
\[ \oint_C(z^4+2z)\,dz \]for any closed contour \(C\).
Practice 5: Unit Circle
Evaluate
\[ \oint_{|z|=1}\frac1z\,dz \]counterclockwise.
Practice 6: Reverse Orientation
What is
\[ \oint_{|z|=1}\frac1z\,dz \]if the unit circle is traversed clockwise?
Practice 7: Circle Parametrization
Parametrize the circle
\[ |z-(1+i)|=4 \]counterclockwise.
Practice 8: Contour Length
Find the length of the circle
\[ |z|=3. \]Practice 9: \(ML\) Estimate
Suppose
\[ |f(z)|\le5 \]on a contour of length \(4\).
Give an upper bound for
\[ \left| \int_C f(z)\,dz \right|. \]Practice 10: Path Independence
Explain why
\[ \int_C e^z\,dz \]depends only on the endpoints of \(C\).
Practice Answers
Practice 1
Use the line-segment formula:
\[ z(t) = z_0+t(z_1-z_0). \]Here,
\[ z_1-z_0 = (3+2i)-(1-i) = 2+3i. \]Therefore,
\[ \boxed{ z(t)=1-i+t(2+3i), \qquad 0\le t\le1. } \]Practice 2
An antiderivative of \(z\) is
\[ \frac{z^2}{2}. \]Therefore,
\[ \int_Cz\,dz = \frac{2^2}{2}-0. \] \[ \boxed{2}. \]Practice 3
An antiderivative of \(z^3\) is
\[ \frac{z^4}{4}. \]Therefore,
\[ \int_Cz^3\,dz = \frac{i^4}{4} - \frac{1^4}{4}. \]Since
\[ i^4=1, \]we get
\[ \boxed{0}. \]Practice 4
The integrand is a polynomial and therefore has an antiderivative everywhere.
Hence,
\[ \boxed{ \oint_C(z^4+2z)\,dz=0. } \]Practice 5
\[ \boxed{ \oint_{|z|=1}\frac1z\,dz=2\pi i. } \]Practice 6
Reversing orientation changes the sign:
\[ \boxed{-2\pi i}. \]Practice 7
\[ \boxed{ z(t) = 1+i+4e^{it}, \qquad 0\le t\le2\pi. } \]Practice 8
The circumference is
\[ 2\pi r. \]Therefore,
\[ \boxed{6\pi}. \]Practice 9
Using
\[ \left| \int_Cf(z)\,dz \right| \le ML, \]we obtain
\[ \boxed{20}. \]Practice 10
Since
\[ \frac{d}{dz}e^z=e^z, \]the function \(e^z\) has the global antiderivative \(e^z\).
Therefore,
\[ \boxed{ \int_Ce^z\,dz = e^{z_1}-e^{z_0}, } \]so the integral depends only on the endpoints.
Checking Strategies
- Test the parameter endpoints before integrating.
- Confirm the contour orientation.
- Always include \(z'(t)\) after parametrization.
- Look for an antiderivative before doing a long direct computation.
- For a closed contour, ask whether singularities lie inside or affect the region.
- Use the \(ML\) estimate when an exact value is unnecessary or difficult.
- Check results against orientation reversal or antiderivative methods when possible.
Connections to Earlier Mathematics
Complex integration combines several earlier ideas:
- Parametric curves: contours are traced by \(z(t)\).
- Vector calculus: complex integrals split into real line integrals.
- Complex differentiation: antiderivatives create path independence.
- Geometry: orientation and contour length affect the integral.
- Topology: holes and singularities can obstruct global antiderivatives.
- Analysis: estimates control integral magnitudes without exact evaluation.
What to Remember
If
\[ C:z=z(t), \qquad a\le t\le b, \]then
\[ \boxed{ \int_Cf(z)\,dz = \int_a^b f(z(t))z'(t)\,dt. } \]Reversing orientation changes the sign:
\[ \boxed{ \int_{-C}f(z)\,dz = -\int_Cf(z)\,dz. } \]If \(F'=f\), then
\[ \boxed{ \int_Cf(z)\,dz = F(z_1)-F(z_0). } \]For a closed contour and a valid antiderivative,
\[ \boxed{ \oint_Cf(z)\,dz=0. } \]But
\[ \boxed{ \oint_C\frac1z\,dz=2\pi i } \]when \(C\) circles the origin once counterclockwise.
The central lesson is: complex integrals depend not only on the integrand, but also on the geometry, orientation, domain, and singularities associated with the contour.
Where We Go Next
We now know how to integrate directly along contours, how antiderivatives create path independence, and why the integral of \(1/z\) around the origin behaves differently from polynomial integrals.
The next lesson, Cauchy's Theorem and Integral Formula, explains the deeper reason analytic functions have vanishing contour integrals under suitable conditions and develops the remarkable formula \[ f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z-z_0}\,dz, \] which allows the value of an analytic function inside a contour to be recovered entirely from values along the boundary.