lacunary - Mathnotes

Cauchy's Integral Theorem

Theorem: Cauchy's Integral Theorem \@{cauchys-integral-theorem}

If f(z) is analytic in a @simply-connected domain D, then for every simple closed @path C in D,

∮Cf(z)dz=0.

Proof \@{proof-of-cauchys-integral-theorem}

Let f(z)=u(x,y)+i(v,y). By the definition of the contour integral, we have

∫Cf(z)dz=∫C(u+vi)(dx+dyi)=∫C(udx−vdy)+i∫C(vdx+udy).

Because f is analytic, the Cauchy-Riemann Equations tell us that u(x,y) and v(x,y) have continuous first partial derivatives, and therefore, we can apply Green's Theorem. Now, we can take the first integral and transform it

∫C(udx+(−vdy))=−∬R(vx+uy)dxdy.

But, by Cauchy-Riemann Equations, vx=−uy, so this integral evaluates to 0.

Now, for the second integral, using Green's Theorem again, we get

i∫C(vdx+udy)=i∬R(ux−vy)dxdy,

and by Cauchy-Riemann Equations, ux=vy, so this integral also evaluates to 0, which makes the overall integral 0 as well.

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Cauchy-Goursat Theorem

First, a less general theorem that can be shown using Green's theorem:

Theorem: Cauchy Integral Theorem \@{cauchy-integral-theorem}

If the derivative f′ of a complex function f is continuous in a domain containing a simple, closed, piecewise smooth curve C and its interior, then

∮cf(z)dz=0.

The Cauchy-Goursat Theorem is more general and doesn't require continuity of f′:

Theorem: Cauchy-Goursat Theorem \@{cauchy-goursat-theorem}

If f is analytic inside and on a closed, piecewise smooth curve C, then

∮cf(z)dz=0.

Note \@{note-5}

Note that f has to be analytic both on and inside the piecewise smooth curve C. This means that if C encloses a singularity, then Cauchy-Goursat can't be applied directly.

In these cases we can apply an extension of Cauchy-Goursat:

Theorem \@{theorem-6}

Let C be a simple, closed piecewise smooth curve, and C1, C2, …, Cn be disjoint, simple, closed piecewise smooth curves in the interior of C. If f is analytic at all points that are both inside or on C, and outside or on each Cj, then

∮cf(z)dz=∑j=1n∮Cjf(z)dz.

This arms us with a method to find the following contour integral:

∮C1z2−1,C:|z|=4.

Since f(z)=1z2−1 has singularities at −1 and 1 that are completely encircled by C, it's not obvious whether or not f has an antiderivative along C. However, we can apply the theorem above by encircling the singularities:

Curve Around Singularities

Now we can find our contour integral by evaluating the contour integrals on paths around our singularities, as these paths are analytic everywhere outside them but inside C.

A useful theorem to proceed from here is the following:

Theorem \@{theorem-7}

If C is a simple, closed, piecewise smooth curve and z0 is interior to C, then:

∮C1z−z0dz=2πi.

Intuition \@{theorem-7-intuition}

See If C is the unit circle, then... and consider path independence and the principle of path deformation. Consider toy contours, particularly a collapsed keyhole contour around the singularity.

Now, via partial fraction decomposition we have

∮C1z2−1=12∮C1z−1−1z+1dz=12(∮C1z−1dz−∮C1z+1dz)=12(2πi−2πi)=0.

The following theorems illustrate the nice properties of simply connected domains, where we don't have to deal with singularities or holes:

When f is analytic in a simply-connected domain D, the contour integral of f around every closed, piecewise smooth curve in D vanishes.

When f is analytic in a simply-connected domain D, the contour integral of f is independent of path in D.

When f is analytic in a simply-connected domain D, it has an antiderivative therein.