Complex Line Integrals

Integrating Along a Curve

A complex function assigns a value to each point in a region, but an integral along a curve also depends on how the curve is traversed. The curve’s parametrization supplies both the points visited and the direction and speed of motion. This makes a complex line integral an ordinary integral in a real parameter, with the complex differential $dz$ represented by the curve’s derivative.

We will define the integral for piecewise continuously differentiable curves. This class includes line segments and circles, as well as paths formed by joining finitely many such pieces. As with power series in the Analytic Functions tutorial, complex-valued expressions are handled through their real and imaginary parts.

Definition (Piecewise $C^1$ Curve): Let $U\subseteq\mathbb{C}$ be open. A curve in $U$ is a continuous function $\gamma:[a,b]\to U$. It is piecewise $C^1$ if there is a finite partition $a=t_0<t_1<\cdots<t_m=b$ such that $\gamma$ is continuously differentiable on each subinterval $[t_{j-1},t_j]$, with one-sided derivatives allowed at the partition points. The curve is closed if $\gamma(a)=\gamma(b)$.

Definition (Complex Line Integral): Let $f:U\to\mathbb{C}$ be continuous, and let $\gamma:[a,b]\to U$ be piecewise $C^1$. The line integral of $f$ along $\gamma$, with respect to $dz$, is $$ \int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\gamma'(t)\,dt. $$ The integral on the right is a complex-valued real-parameter integral, defined by integrating its real and imaginary parts. For a piecewise $C^1$ curve, the integral is taken separately on the finitely many smooth subintervals and the results are added.

The factor $\gamma'(t)$ matters: it records the direction and speed of traversal. In particular, this is not generally the same as integrating $f$ against arc length. Arc length involves $|\gamma'(t)|\,dt$, whereas the complex differential $dz$ contributes $\gamma'(t)\,dt$.

Basic Rules for Curves

Joining curves adds their integrals, and traversing a curve in the opposite direction changes the sign. These rules follow from splitting or reversing the real parameter integral. They allow computations on a path made of several smooth pieces to be handled one piece at a time.

Proposition (Additivity and Reversal): Suppose a curve $\gamma$ is the concatenation of two piecewise $C^1$ curves $\gamma_1$ and $\gamma_2$, with the endpoint of $\gamma_1$ equal to the starting point of $\gamma_2$. Then $$ \int_\gamma f(z)\,dz = \int_{\gamma_1}f(z)\,dz+\int_{\gamma_2}f(z)\,dz. $$ If $\gamma^{-}(t)=\gamma(a+b-t)$ is the reversed curve, then $$ \int_{\gamma^{-}}f(z)\,dz=-\int_\gamma f(z)\,dz. $$

Proof. For concatenation, use a parameter interval divided at the joining point. The definition splits the integral over that interval into its integrals on the two subintervals, which are exactly the integrals along $\gamma_1$ and $\gamma_2$, up to linear changes of parameter. For reversal, substitute $s=a+b-t$ in the defining integral. Then $dt=-ds$, and the reversed curve has derivative $-\gamma'(a+b-t)$ wherever the derivative exists. Thus

\int_a^b f(\gamma(a+b-t))\bigl(-\gamma'(a+b-t)\bigr)\,dt = -\int_a^b f(\gamma(s))\gamma'(s)\,ds.

The finitely many points where a piecewise derivative may fail to exist do not affect these real integrals. This proves both rules. $\square$

The value of the integral does not depend on the speed used to trace a curve, provided the new parametrization preserves its direction. More precisely, let $\phi:[c,d]\to[a,b]$ be a continuously differentiable, increasing bijection with $\phi(c)=a$ and $\phi(d)=b$. The curve $\widetilde{\gamma}(s)=\gamma(\phi(s))$ traces $\gamma$ in the same direction, and the substitution rule gives

\int_c^d f(\widetilde{\gamma}(s))\widetilde{\gamma}'(s)\,ds = \int_c^d f(\gamma(\phi(s)))\gamma'(\phi(s))\phi'(s)\,ds = \int_a^b f(\gamma(t))\gamma'(t)\,dt.

This change-of-parameter rule explains why the geometric path and its orientation are usually more important than the particular speed of traversal. Reversing orientation, in contrast, changes the sign.

Worked Examples

Worked Example: A Line Segment

Compute $\int_\gamma z^2\,dz$, where $\gamma$ is the line segment from $0$ to $1+i$. Use $\gamma(t)=t(1+i)$ for $0\leq t\leq1$. Then $\gamma'(t)=1+i$, and

\gamma(t)^2=t^2(1+i)^2=2it^2, \qquad \gamma(t)^2\gamma'(t)=2it^2(1+i)=(-2+2i)t^2.

Therefore

\int_\gamma z^2\,dz = \int_0^1(-2+2i)t^2\,dt = \frac{-2+2i}{3}.

The computation uses the parametrization directly: the factor $1+i$ is the complex differential contributed by the segment’s direction.

Worked Example: The Same Endpoints Along Two Segments

Let $\gamma$ go from $0$ to $1$ along the real axis and then from $1$ to $1+i$ along a vertical segment. Compute $\int_\gamma z\,dz$ piece by piece. On the first segment, take $z=t$, $0\leq t\leq1$, so $dz=dt$:

\int_0^1 t\,dt=\frac12.

On the second segment, take $z=1+it$, $0\leq t\leq1$, so $dz=i\,dt$. Its contribution is

\int_0^1(1+it)i\,dt = \int_0^1(i-t)\,dt = i-\frac12.

By additivity, the total is

\int_\gamma z\,dz = \frac12+i-\frac12 = i.

This path has the same endpoints as the straight segment in the previous example, but the integrand is different. The result here will also follow from the primitive $F(z)=z^2/2$, since $\big((1+i)^2-0^2\big)/2=i$.

Worked Example: A Closed Circle

Let $\gamma(t)=r e^{it}$ for $0\leq t\leq2\pi$, where $r>0$. This parametrizes the circle of radius $r$ counterclockwise. For $f(z)=\overline{z}$, we have $\overline{\gamma(t)}=r e^{-it}$ and $\gamma'(t)=ir e^{it}$. Hence

\int_\gamma \overline{z}\,dz = \int_0^{2\pi}(r e^{-it})(ir e^{it})\,dt = \int_0^{2\pi}i r^2\,dt = 2\pi i r^2.

The integrand $\overline{z}$ is not a holomorphic function on an open region containing the circle: its values depend on the complex conjugate of $z$. This example illustrates that the definition of a line integral does not require a holomorphic integrand. It also shows that a closed curve’s integral need not be zero.

Estimating the Integral by Curve Length

A useful estimate bounds an integral using the largest size of the integrand along the curve and the curve’s length. For a piecewise $C^1$ curve, define its length by

L(\gamma)=\int_a^b|\gamma'(t)|\,dt.

This definition sums the speed over the parameter interval. The following estimate is often called the length estimate, or the $ML$ inequality, when $|f|\leq M$ along the curve.

Proof. First, for any continuous complex-valued function $h$ on a real interval, the triangle inequality for its integral gives $\left|\int h(t)\,dt\right|\leq\int|h(t)|\,dt$. To see why, write $I=\int h(t)\,dt$. If $I=0$, the inequality is immediate. If $I\neq0$, let $u=I/|I|$, so $|u|=1$ and $\overline{u}I=|I|$. Taking real parts yields

|I| = \int \operatorname{Re}(\overline{u}h(t))\,dt \leq \int|\overline{u}h(t)|\,dt = \int|h(t)|\,dt.

Apply this inequality on each smooth part of the curve to $h(t)=f(\gamma(t))\gamma'(t)$, and add the resulting bounds. Since $|h(t)|=|f(\gamma(t))|\,|\gamma'(t)|$, this gives the first estimate. If $|f|\leq M$ along the curve, then

\int_a^b|f(\gamma(t))|\,|\gamma'(t)|\,dt \leq M\int_a^b|\gamma'(t)|\,dt = M L(\gamma).

This proves both inequalities. $\square$

The estimate is useful when an exact integral is difficult to compute but the integrand and path length can be bounded. It depends on the curve’s length, not the total time spent tracing it: slowing down changes $|\gamma'|$ but does not change the length or the integral’s value.

Primitives and Endpoint Evaluation

A primitive can turn a line integral into an endpoint calculation. The relevant chain rule says that if $F$ has continuous complex derivative and $\gamma$ is differentiable, then the derivative of $F(\gamma(t))$ is $F'(\gamma(t))\gamma'(t)$. Applying the real fundamental theorem of calculus to the real and imaginary parts gives the following result.

Theorem (Evaluation by a Primitive): Let $U\subseteq\mathbb{C}$ be open, let $F:U\to\mathbb{C}$ have continuous complex derivative, and let $f=F'$. For every piecewise $C^1$ curve $\gamma:[a,b]\to U$, $$ \int_\gamma f(z)\,dz=F(\gamma(b))-F(\gamma(a)). $$

Proof. On each subinterval where $\gamma$ is continuously differentiable, the real-variable chain rule, applied to the real and imaginary parts, gives

\frac{d}{dt}F(\gamma(t)) = F'(\gamma(t))\gamma'(t) = f(\gamma(t))\gamma'(t).

The right-hand side is continuous on each such subinterval because $f$ and $\gamma'$ are continuous there. The real fundamental theorem of calculus, applied separately to the real and imaginary parts, shows that the integral over that subinterval equals the difference of $F\circ\gamma$ at its endpoints. Adding these differences over the finite partition cancels the values at the intermediate endpoints. What remains is

\int_a^b f(\gamma(t))\gamma'(t)\,dt = F(\gamma(b))-F(\gamma(a)).

By the definition of the line integral, this is the required formula. $\square$

If an integrand has a primitive on a region, every two piecewise $C^1$ paths in that region with the same starting point and endpoint have the same integral. This is path independence, and it follows immediately because the formula uses only the two endpoint values of the primitive. The hypothesis that a primitive exists is important: the circle example above demonstrates that a closed-curve integral can be nonzero for an integrand that does not have a primitive of the required kind on a surrounding region.

For instance, the function $f(z)=z^2$ has primitive $F(z)=z^3/3$. Along any piecewise $C^1$ path from $0$ to $1+i$, the integral therefore equals

F(1+i)-F(0) = \frac{(1+i)^3}{3} = \frac{-2+2i}{3}.

This agrees with the direct parametrization in the first worked example. In general, parametrization is the definition and a reliable way to compute an integral; the primitive theorem is a shortcut available when its hypotheses hold. The length estimate provides a different tool: it controls the size of an integral without evaluating it exactly.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

How is $\int_\gamma f(z)\,dz$ defined in terms of a parametrization $\gamma$?
What happens to a line integral when a curve is traversed in the opposite direction?
State the length estimate and identify the meaning of $L(\gamma)$.
Why does the existence of a primitive imply path independence for paths with the same endpoints?
For a curve $\gamma(t)=r e^{it}$, $0\leq t\leq2\pi$, what is $\int_\gamma \overline{z}\,dz$?

What You'll Learn