What conditions are needed for $\int_{\gamma} f(z)\,dz$ to be well-defined in this setup?

The curve must be continuously differentiable (so $\gamma$ and its real/imaginary parts have continuous derivatives), and the curve’s image must lie inside the domain $U\subseteq\mathbb{C}$ where $f$ is defined. The function $f$ is assumed continuous so the integrand $f(\gamma(t))\gamma'(t)$ behaves well enough for the integral to exist.

Complex Analysis 18 | Complex Contour Integral [dark version]

Q: Why does the definition of a contour integral include the derivative $\gamma'(t)$?

Because $\gamma'(t)$ represents how the parameterized curve moves locally in the complex plane. When the curve is reparameterized (same geometric path, different speed), $\gamma'(t)$ changes in a way that compensates, keeping the integral consistent. In the quarter-circle example, $\gamma_1(t)=e^{it}$ and $\gamma_2(t)=e^{i(\pi/2)t}$ trace the same arc from $1$ to $i$ but at different rates; the $\gamma'(t)$ factor ensures both integrals still evaluate to $-1$.

Q: How does the quarter-circle example compute $-1$ when $f(z)=z$?

With $\gamma_1(t)=e^{it}$, the integral becomes $\int_0^{\pi/2} f(\gamma_1(t))\gamma_1'(t)\,dt = \int_0^{\pi/2} (e^{it})(i e^{it})\,dt = \int_0^{\pi/2} i e^{2it}\,dt$. Evaluating the antiderivative gives $-1$ after plugging in the endpoints $t=0$ and $t=\pi/2$.

Q: Why is getting $-1$ again for the straight-line path not a general rule?

Because path-independence depends on the function $f$. For $f(z)=z$, the straight-line parameterization $\gamma_3(t)=1+(i-1)t$ leads to an integral that still simplifies to $-1$. But the discussion explicitly warns that in general the integral can depend on the chosen curve, even if the start and end points match.

TL;DR

A contour integral along a parameterized curve $γ$ is defined as $\int_{a}^{b} f (γ (t)) γ^{'} (t) d t$ .

Briefing Cornell Notes

Briefing

Complex contour integrals are defined by adding up the values of a complex function along a parameterized path, with each contribution weighted by how fast the path moves in the complex plane. The key move is to replace “walking along a curve” with a precise parameterization: a continuously differentiable map $γ : [a, b] \to C$ . As $t$ runs from $a$ to $b$ , $γ (t)$ traces the curve, while $γ^{'} (t)$ captures the curve’s local velocity (the rate of change of both real and imaginary parts). That velocity factor is what makes the contour integral behave like an ordinary integral under change of variables.

Formally, for a complex-valued function $f$ and a curve $γ$ lying inside a domain $U \subseteq C$ , the contour integral is written as $\int_{γ} f (z) d z$ and defined by $\int_{a}^{b} f (γ (t)) γ^{'} (t) d t .$ The integral is well-defined when $f$ is continuous (and the usual conditions for existence hold), and the curve’s image must stay within $U$ . This definition is designed to generalize the idea of integrating along a path in the complex plane while remaining consistent with the real-variable integral concept.

To make the definition concrete, the discussion works through a quarter-circle example using $f (z) = z$ . One parameterization $γ_{1} (t) = e^{i t}$ traces the arc from $1$ to $i$ as $t$ runs from $0$ to $π /2$ . Plugging into the definition yields an integral of $e^{2 i t}$ times $i$ , which evaluates to $- 1$ . A second parameterization $γ_{2} (t) = e^{i (π /2) t}$ runs through the same geometric arc but at a different speed (now $t$ goes from $0$ to $1$ ). The result stays $- 1$ , illustrating why the $γ^{'} (t)$ factor matters: it compensates for reparameterizing the same path.

Next comes a more surprising comparison. A third curve $γ_{3}$ is the straight line from $1$ to $i$ , given by $γ_{3} (t) = 1 + (i - 1) t$ for $t \in [0, 1]$ . Even though the path is different, the integral of $f (z) = z$ still comes out $- 1$ because the function is simple and the calculation reduces to integrating a linear expression. The takeaway is careful: path-independence is not guaranteed in general, but it can happen for certain functions.

Finally, the integral is re-justified through a “weighted curve” picture. The curve is approximated by many small line segments using a partition of the parameter interval. Each segment contributes roughly $f$ evaluated near the segment times the segment’s complex displacement $γ (t_{i + 1}) - γ (t_{i})$ . As the partition gets finer, this Riemann-sum-like construction converges to the contour integral, with the segment displacement playing the role of $γ^{'} (t) d t$ . That geometric weighting is the intuition behind the formal definition.

Cornell Notes

A contour integral $\int_{γ} f (z) d z$ is defined for a complex function $f$ along a parameterized curve $γ : [a, b] \to C$ by $\int_{a}^{b} f (γ (t)) γ^{'} (t) d t$ . The factor $γ^{'} (t)$ is essential because it encodes the curve’s local direction and speed in the complex plane, making the integral consistent under reparameterization. Using $f (z) = z$ and a quarter-circle from $1$ to $i$ , two different parameterizations of the same arc both produce $- 1$ , showing that changing the speed along the same path doesn’t change the value. A straight-line path from $1$ to $i$ also yields $- 1$ for this particular $f$ , but that kind of path-independence is not universal. A Riemann-sum approximation with small line segments motivates why the definition matches the “weighted path” intuition.

Why does the definition of a contour integral include the derivative $γ^{'} (t)$ ?

Because

γ^{'} (t)

represents how the parameterized curve moves locally in the complex plane. When the curve is reparameterized (same geometric path, different speed),

γ^{'} (t)

changes in a way that compensates, keeping the integral consistent. In the quarter-circle example,

γ_{1} (t) = e^{i t}

and

γ_{2} (t) = e^{i (π /2) t}

trace the same arc from

1

i

but at different rates; the

γ^{'} (t)

factor ensures both integrals still evaluate to

- 1

What conditions are needed for $\int_{γ} f (z) d z$ to be well-defined in this setup?

The curve must be continuously differentiable (so

γ

and its real/imaginary parts have continuous derivatives), and the curve’s image must lie inside the domain

U \subseteq C

where

f

is defined. The function

f

is assumed continuous so the integrand

f (γ (t)) γ^{'} (t)

behaves well enough for the integral to exist.

How does the quarter-circle example compute $- 1$ when $f (z) = z$ ?

With

γ_{1} (t) = e^{i t}

, the integral becomes

\int_{0}^{π /2} f (γ_{1} (t)) γ_{1}^{'} (t) d t = \int_{0}^{π /2} (e^{i t}) (i e^{i t}) d t = \int_{0}^{π /2} i e^{2 i t} d t

. Evaluating the antiderivative gives

- 1

after plugging in the endpoints

t = 0

and

t = π /2

What does the second quarter-circle parameterization demonstrate?

It demonstrates reparameterization invariance for the same geometric path. Using

γ_{2} (t) = e^{i (π /2) t}

[0, 1]

changes

γ^{'} (t)

and the limits, but the integral still evaluates to

- 1

. The derivative factor

γ^{'} (t)

effectively accounts for the different traversal speed.

Why is getting $- 1$ again for the straight-line path not a general rule?

Because path-independence depends on the function

f

. For

f (z) = z

, the straight-line parameterization

γ_{3} (t) = 1 + (i - 1) t

leads to an integral that still simplifies to

- 1

. But the discussion explicitly warns that in general the integral can depend on the chosen curve, even if the start and end points match.

How does the “weighted curve” / Riemann-sum viewpoint connect to the formal definition?

The curve is approximated by many small line segments between points

γ (t_{i})

and

γ (t_{i + 1})

. Each segment contributes approximately

f

evaluated near the segment times the complex displacement

γ (t_{i + 1}) - γ (t_{i})

. As the partition gets finer (

n \to \infty

), this sum converges to the contour integral, with the displacement behaving like

γ^{'} (t) d t

, matching

\int_{a}^{b} f (γ (t)) γ^{'} (t) d t

Review Questions

Given $γ (t)$ and $f (z)$ , how would you set up $\int_{γ} f (z) d z$ using the parameter $t$ ?
In the quarter-circle example, what role does $γ^{'} (t)$ play when switching from $γ_{1} (t) = e^{i t}$ to $γ_{2} (t) = e^{i (π /2) t}$ ?
Does equal start and end points guarantee equal contour integrals? Use the examples to justify your answer.

Key Points

1
A contour integral along a parameterized curve $γ$ is defined as $\int_{a}^{b} f (γ (t)) γ^{'} (t) d t$ .
2
The derivative $γ^{'} (t)$ is the local velocity factor, encoding both direction and speed of traversal in the complex plane.
3
The curve must be continuously differentiable, and its image must lie within the domain where $f$ is defined.
4
Continuity of $f$ (and standard integrability conditions) ensures the contour integral exists.
5
Reparameterizing the same geometric path changes $γ (t)$ and $γ^{'} (t)$ , but the integral can remain unchanged because $γ^{'} (t)$ compensates.
6
Path-independence is not universal; different curves between the same endpoints can yield different integrals for other functions.
7
A Riemann-sum approximation using small line segments motivates the definition and shows how $γ^{'} (t) d t$ arises from segment displacements.

Highlights

The contour integral is built from a parameterization: ∫γ​f(z)dz=∫ab​f(γ(t))γ′(t)dt.

Two different parameterizations of the same quarter-circle from 1 to i both give −1 for f(z)=z.

Even with the same endpoints, changing the curve can still preserve the value for some functions (like f(z)=z)—but that behavior isn’t guaranteed in general.

Approximating the curve by line segments turns the contour integral into a Riemann-sum-like limit, with γ(ti+1​)−γ(ti​) playing the role of γ′(t)dt.

Topics

Contour Integral
Parameterization
Quarter Circle
Riemann Sums
Complex Functions