Signals and Systems · 144 h · Topic 6 of 34

Line integrals in the complex plane

Between two points of the plane there are infinitely many paths, and the integral is defined along one of them. Each term of the sum is a vector, so the integral can be seen: it is a chain of little arrows whose end point is the result.

Contour integral Parametrize 2πj ML bound Bromwich

01Integrating along a path

For a real variable, \( \int_a^b \) has only one possible route: from the point \( a \) to the point \( b \) along the line. In the plane there are infinitely many paths between two points, and the integral is defined along one of them. The notation says so:

\[ \int_C f(z)\,dz = \lim_{n \to \infty} \sum_{k=1}^{n} f(z_k)\,\Delta z_k \] The same Riemann sum as always, except that now each term is a product of complex numbers: it has a modulus and a phase.

That detail changes everything. Each term \( f(z_k)\,\Delta z_k \) is a vector in the plane, and the integral is the sum of those vectors placed one after the other. In the lab it looks like this: a chain of little arrows whose final end point is the value of the integral.

Why it matters to an electronics engineer

The inverse Laplace transform is a line integral in the complex plane —the Bromwich integral, along a vertical line to the right of all the poles—. Nobody computes it that way in practice, but the method that is used, residues, is derived from here. Everything that follows in the course goes through this topic.

02How to compute it: parametrize

There is nothing new to invent. You describe the path with a real parameter \( t \), substitute, and end up with an ordinary integral:

\[ z = z(t), \quad t \in [a,b] \;\Longrightarrow\; \int_C f(z)\,dz = \int_a^b f\big(z(t)\big)\,z'(t)\,dt \]

The example you have to know by heart is the circle traversed once counterclockwise. With \( z = r\,e^{jt} \) we get \( dz = j\,r\,e^{jt}\,dt \), and then:

\[ \oint_{|z| = r} \frac{dz}{z} = \int_0^{2\pi} \frac{j\,r\,e^{jt}}{r\,e^{jt}}\,dt = j\int_0^{2\pi} dt = 2\pi j \] It does not depend on the radius. That \( 2\pi j \) will appear in every theorem from here on.

And the rest of the powers, over the same circle, give zero:

\[ \oint_{|z| = r} z^{n}\,dz = 0 \qquad \text{for every integer } n \ne -1 \]

The reason is that all those powers have an antiderivative —\( z^{n+1}/(n+1) \)— and an antiderivative makes the integral over a closed path vanish. The exception is \( n = -1 \), because its antiderivative would be the logarithm, which is multivalued and does not return to the same value after going around.

03The integral, path by path

Lab · the sum that gives the integral

On the left, the path and the points where the function is evaluated. On the right, the terms \( f(z_k)\,\Delta z_k \) chained together: the end of the chain is the value of the integral. Click or tap the left plane to move the end point.

Subdivisions = 24

04When the path does not matter

Let us try \( f(z) = \bar{z} \), from \( 0 \) to \( 1 + j \). Along the straight segment, with \( z = t(1+j) \):

\[ \int_0^1 t(1-j)\,(1+j)\,dt = \int_0^1 2t\,dt = 1 \]

And along the two-leg path, first from \( 0 \) to \( 1 \) and then from \( 1 \) to \( 1+j \):

\[ \underbrace{\int_0^1 t\,dt}_{1/2} \;+\; \underbrace{\int_0^1 (1 - jt)\,j\,dt}_{\tfrac12 + j} \;=\; 1 + j \]

Two paths, two different results. By contrast, with \( f(z) = z \) both give \( j \), and the same happens with \( z^2 \) and with \( e^z \). The difference is the usual one: \( \bar{z} \) is not analytic, and the others are.

If \( f \) is…	The integral between two points…	Over a closed path…
Analytic on the whole region	Does not depend on the path	Is zero
Analytic except at an enclosed pole	Depends on which side you pass	Equals \( 2\pi j \) times the residue
Not analytic	Depends on the path	Can be anything

The first row is Cauchy’s integral theorem, which is the next topic. It is worth checking it beforehand in the lab: with \( z^2 \) the three paths end at the same point, and with \( \bar{z} \) they do not.

05The ML bound

Often there is no need to compute the integral, only to bound it, especially to show that a piece of contour contributes zero when the radius goes to infinity. The tool is elementary:

\[ \left| \int_C f(z)\,dz \right| \le M\,L, \qquad M = \max_{z \in C} |f(z)|, \qquad L = \text{length of } C \]

With it one proves, for example, that the integral of \( 1/z^2 \) over a semicircle of radius \( R \) is bounded by \( (1/R^2)(\pi R) = \pi/R \), which tends to zero. That argument is what allows contours to be closed at infinity, keeping only the part of interest, which is exactly what is done when computing an inverse transform.

06In the lab

Activity 1 · Verifying the 2πj

Compute by hand \( \oint dz/z \) over the circle of radius 2 parametrizing with \( z = 2e^{jt} \). Repeat with radius 0.5 and check that it gives the same result. Compare it with the lab by choosing the closed path.

Activity 2 · The two paths of the conjugate

Redo with pencil the two calculations of section 04 and verify that they give 1 and \( 1+j \). Then look in the lab for an end point where the difference between paths is largest.

Activity 3 · The bound, measured

For \( f(z) = 1/(z^2+1) \) over the upper semicircle of radius \( R = 10 \), estimate \( M \) and \( L \), compute the bound \( ML \) and compare it with the numerical value of the integral. Check how conservative it is.

07Common mistakes

Forgetting the \( z'(t) \). When parametrizing you have to multiply by the derivative of the path, not just change the variable.
Taking for granted that the integral does not depend on the path. It holds only if \( f \) is analytic on the whole region between the two paths.
Getting the direction of traversal wrong. Reversing it changes the sign; the positive direction is counterclockwise.
Believing that \( \oint z^n dz = 0 \) always. It fails precisely at \( n = -1 \), which is the interesting case.
Using the ML bound as if it were the value. It is an upper bound, and it usually lies far above.

08Self-assessment

What is \( \oint dz/z \) over \( |z| = 3 \)?

\( 2\pi j \). It does not depend on the radius: the \( r \) cancels when parametrizing.

And \( \oint z^5\,dz \) over the same circle?

Zero: it has the antiderivative \( z^6/6 \), and over a closed path the antiderivative returns to the same value.

Why is \( n = -1 \) the exception?

Because its antiderivative is the logarithm, which is multivalued: after one trip around, the argument has changed by \( 2\pi \).

What does \( \int \bar{z}\,dz \) give from 0 to \( 1+j \) along the straight segment?

1. Along the two-leg path it gives \( 1 + j \): the conjugate is not analytic and the path matters.

What is the ML bound for?

To prove that an arc contributes zero when the radius grows, so that contours can be closed at infinity.

What line integral is behind the inverse Laplace transform?

The Bromwich integral, along a vertical line to the right of all the poles. It is solved by closing the contour and applying residues.

09Further reading

Ruel V. Churchill and James W. Brown. Complex Variables and Applications. 7th ed. (Spanish edition, “Variable compleja y aplicaciones”), McGraw-Hill, 2004. The chapter on integrals, with the ML bound and path independence carefully proved.
Murray R. Spiegel. Complex Variables. Schaum’s Outline Series (Spanish edition, “Variable compleja”), McGraw-Hill, 2011. Dozens of solved contour integrals, which is the only way to get the hang of them.
A. David Wunsch. Complex Variables with Applications. 2nd ed. (Spanish edition, “Variable compleja con aplicaciones”), Addison-Wesley Iberoamericana, 1997. Connects the contour integral with the inverse Laplace transform earlier than any other.

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