lacunary - Mathnotes

Contour Integrals

A contour integral is essentially a line integral in the complex plane.

Definition: Contour Integral \@{contour-integral}

Given a smooth curve C in a domain D and f(z)=u(x,y)+v(x,y)i a continuous function defined along C, we say that

Cf(z)dz=C(u+vi)(dx+dyi)=C(udxvdy)+iC(vdx+udy).

We can parameterize C as

C:z=z(t)=x(t)+y(t)i,αtβ

Then, z(t)=x(t)+iy(t),

and by the chain rule,

x(t)=dxdtdt,y(t)=dydtdt.

Therefore,

z(t)=dzdt=dxdt+idydt

or equivalently

dz=z(t)dt.

Then we have

Cf(z)dz=αβf[z(t)]z(t)dt.

One trick we can use is that when the modulus of a function is constant, we can use the argument of z as a parameter by writing C:z=eθi,argαθargβ

When the path of integration in a contour integral is a closed curve, we place a circle on the integral sign

Cf(z)dz.

You can add an arrow to the circle to indicate clockwise vs counterclockwise direction, but that's not supported in MathJax so I don't have it here!

Some additional properties:

ReCf(z)dz=CRe[f(z)dz];

ImCf(z)dz=CIm[f(z)dz];

|Cf(z)dz|C|f(z)||dz|ML

where M is the maximum value of f(z) for z on C, and L is the length of C.

Here's an example problem: evaluate Cezdz where C is the closed (triangular) path from 0 to 1 (C1) to 1+i (C2) back to 0 (C3):

Curve Path

Cezdz=C1ezdz+C2ezdz+C3ezdz,

where

C1:y=0,0x1;

C2:x=1,0y1;

C3:x=t,y=t,1t0;

Thus,

Cezdz=01exdx+01e1+yiidy+10et+ti(dt+dt i)

={ex}01+{e1+yi}01+{e(1+i)t}10

=(e1)+(e1+ie)+(1e1+i)=0.

Some Important Contour Integrals

First, we'll compute C1zdz where C is the unit circle. We'll call this out as a theorem, because it's a super important result that gets used all the time in complex analysis.

Theorem \@{integral-of-one-over-z-around-unit-circle}

If C is the unit circle, then C1zdz=2πi.

Proof \@{proof-of-integral-of-one-over-z-around-unit-circle}

First, note that we can parameterize C, the unit circle, as

z(θ)=eiθ,0θ<2π.

Now, making the substitution,

f(z(θ))=1e(iθ)=eiθ.

Now, using the definition of the contour integral,

Cf(z)dz=αβf[z(t)]z(t)dt

we have, noting that z(θ)=ieiθ,

C1zdz=02πeiθieiθdθ=i02πdθ=2πi.

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