Cauchy’s integral theorem
If the function is analytic inside, the integral over any closed curve is zero. From that comes what is used most: a contour can be deformed at will as long as it does not cross a singularity.
01The statement
It is the result that organizes the whole second half of complex variable, and it can be said in one line.
Two words in the statement do all the work. Analytic, whose meaning is already known. And simply connected: a domain with no holes, where any closed curve can be shrunk to a point without leaving it. A disk is simply connected; an annulus is not, because the hole in the middle prevents shrinking the curves that surround it.
It is what allows you to move an integration contour without changing the result. In the Nyquist criterion, the contour is deformed with semicircles to avoid poles on the imaginary axis, and the reasoning holds because the function is analytic between the old contour and the new one. The same happens when computing an inverse transform: the Bromwich line can be shifted at will as long as it does not cross a pole.
02Where it comes from
Splitting the integral into real and imaginary parts, with \( f = u + jv \) and \( dz = dx + j\,dy \):
Each of the two is a real line integral, and for those there is Green’s theorem, which converts them into area integrals:
Applying it to both, the integrands that appear are \( -v_x - u_y \) and \( u_x - v_y \). And those are exactly the two residuals of Cauchy-Riemann: both vanish if \( f \) is analytic. The two area integrals are zero, and with them the contour integral.
The proof above, which is Cauchy’s, needs \( f' \) to be continuous in order to use Green. Édouard Goursat later proved that this hypothesis is superfluous: it is enough for \( f \) to be differentiable. That is why the theorem appears in books with both surnames.
03Deforming the contour
The practical consequence is more useful than the theorem itself. If two closed contours enclose the same singularities and the function is analytic between them, their integrals are equal. The contour can be stretched, wrinkled or turned into a square: as long as it does not cross a singularity, the number does not change.
With that, any contour that goes once around the origin gives the same value as the unit circle. And since we have already computed \( \oint dz/z = 2\pi j \), that number holds for all of them.
The contour is traversed counterclockwise and the integral is computed point by point. Change the shape, the size and the position: as long as it does not cross an X, the result stays put. Click or tap the canvas to move the center.
04When there are holes
If the domain has singularities, it is no longer simply connected and the theorem does not apply directly. What does hold is a splitting: a contour enclosing several singularities is equivalent to the sum of small contours, one around each.
The trick of the proof is to join the large contour to each of the small ones by means of cuts traversed back and forth. The cuts are traversed twice in opposite directions and cancel, and what is left is a contour that encloses no singularity: by Cauchy, zero.
From there on it is all a matter of knowing how much each singularity contributes, and that is the residue theorem. The lab already anticipates it with \( 1/(z^2-1) \): enclosing one pole gives \( \pi j \), enclosing both gives zero, because the contributions cancel.
With \( f(z) = 1/z^2 \) the origin is a genuine singularity, and yet any contour that surrounds it gives zero. The reason is that \( 1/z^2 \) has an antiderivative —\( -1/z \)— which does return to the same value after going around. The only exponent that escapes is \( -1 \).
05In the lab
Compute by hand \( \oint dz/z \) over the square with vertices \( \pm 1 \pm j \), parametrizing the four sides separately. Verify that it gives \( 2\pi j \), the same as the circle, and compare with the lab.
With \( f(z) = 1/(z^2-1) \), do the partial fraction decomposition and check that the residues at \( z = 1 \) and \( z = -1 \) are \( 1/2 \) and \( -1/2 \). Explain why a contour that encloses both gives zero.
Draw the Nyquist contour for a system with a pole at the origin, with its semicircle of radius \( \varepsilon \) going around it. Justify with Cauchy’s theorem why that detour does not change the result, and compute how much the semicircle contributes as \( \varepsilon \to 0 \).
06Common mistakes
- Applying the theorem without checking for singularities inside. It is the number one cause of zero results that are not zero.
- Forgetting the simply connected condition. On an annulus the theorem does not hold directly.
- Assuming that every enclosed singularity contributes \( 2\pi j \). It depends on the residue: \( 1/z^2 \) contributes zero.
- Losing track of the direction of traversal. Clockwise changes the sign of everything.
- Believing that the theorem computes integrals. It says when they are zero; for the rest you need Cauchy’s formula and residues.
07Self-assessment
What does the theorem say, in one line?
That the integral of an analytic function over any closed curve in a simply connected domain is zero.
What role does Cauchy-Riemann play in the proof?
The two integrands that Green’s theorem leaves are precisely the two residuals of Cauchy-Riemann, and they vanish.
What is \( \oint e^z\,dz \) over any closed curve?
Zero: \( e^z \) is entire, analytic on the whole plane.
And \( \oint dz/z^2 \) over \( |z| = 1 \)?
Also zero, even though there is a singularity inside: \( 1/z^2 \) has the antiderivative \( -1/z \).
Why does \( \oint dz/(z^2-1) \) give zero over \( |z| = 2 \)?
Because it encloses both poles and their residues, \( 1/2 \) and \( -1/2 \), cancel. Enclosing only one would give \( \pi j \).
What is deforming a contour good for?
For replacing a complicated one with a convenient one —a small circle around each pole— without changing the value. It is the basis of the residue method and of the Nyquist contour.
08Further reading
- Ruel V. Churchill and James W. Brown. Complex Variables and Applications. 7th ed. (Spanish edition, “Variable compleja y aplicaciones”), McGraw-Hill, 2004. The Cauchy-Goursat chapter, with the version for multiply connected domains and the cuts.
- Murray R. Spiegel. Complex Variables. Schaum’s Outline Series (Spanish edition, “Variable compleja”), McGraw-Hill, 2011. Contour deformation problems, which is where the theorem stops being abstract.
- Katsuhiko Ogata. Modern Control Engineering. 5th ed. (Spanish edition, “Ingeniería de control moderna”), Pearson, 2010. The Nyquist contour and its detours around poles, which is this theorem put to use.