Cauchy’s Integral Formula

See reference

theorem (Cauchy Integral Formula):

Suppose f is holomorphic on Ω, then for any z0Ω and any open disc ¯DR(z0) such that γ:=¯DR(z0)Ω, f(z0)=12πiγf(ξ)ξz0 dξ and f(n)(z0)=n!2πiγf(ξ)(ξz0)n+1dξ. As a consequence, if f(z)k0ck(zz0)k, ck=f(n)(z0)n!=12πiγf(ξ)(ξz0)n+1dξ.

proof (?):

figures/image_2021-05-27-16-54-06.png

proof (?):

figures/image_2021-05-27-16-56-39.png

figures/image_2021-05-27-16-56-52.png

proof (Alternative):

figures/2021-12-14_16-49-17.png

figures/2021-12-14_16-49-36.png

Exercises

exercise (Integral computation):

Compute Dezz2dz.

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solution:

By Cauchy’s formula, f(z)(z0)2dz=2πif(1)(0)=2πi.

exercise (Integral computation):

Without using the residue formula, compute Rf(x)dxf(x):=1x4+16.

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solution:

Use a semicircular contour, noting the poles are at ±2±i2. Write

  • f1(z):=(2+i2)f(z)
  • f2(z):=(2+i2)f(z).

Break the curve up into two integrals I1,I2 enclosing the poles, by Cauchy one gets

  • For the loop around the right pole: I1=2πif1(2+i2)=π2(1i)32
  • For the loop around the left pole: I2=2πif2(2i2)=π2(1+i)32.

Now show that CR vanishes: parameterize as γ(t)=Reit and use the reverse triangle inequality: |CRf|π01R416=πRR4160.

exercise (Equality of different integrals):

Suppose f is holomorphic on Ω, a simply connected region, and suppose γΩ. Using the Cauchy integral formula, show that γf(z)zadz=γf(z)(za)2dz. Also prove this when Ω is not simply connected.

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solution:

Use the integral formula directly: γf(z)zadz=2πif(a).

On the other hand, use Cauchy’s formula for derivatives: γf(z)(za)2dz=2πif(1)(a), and these values are equal.

If Ω is not simply connected, note that by the quotient rule zf(z)za=f(z)zaf(z)(za)2.

Thus γf(z)zaγf(z)(za)2dz=γ(f(z)zaf(z)(za)2)dz=γzf(z)zadz=G(γ(1))G(γ(0))=G(p)G(p)=0, where G(z):=f(z)za is a primitive for the integrand by definition.

exercise (Evaluating integrals):

Evaluate the following integrals using Cauchy’s integral formula:

S1cos(z)zdzS1sin(z)zdz|z|=2z2z1dzS1ezz2dz|z|=2z21z2+1dz|z|=21z2+z+1dz.

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solution:

S1cos(z)zdz=2πicos(0)=2πiS1sin(z)zdz=2πisin(0)=0|z|=2z2z1dz=2πiz2|z=1=2πiS1ezz2dz=2πizez|z=0=2πi|z|=2z21z2+1dz=D+i(z21)/(z+i)zidz+Di(z21)/(zi)z+idz=2πi(z21z+i)|z=i+2πi(z21zi)|z=i=2πi(22i)+2πi(22i)=0|z|=21z2+z+1dz=1(zζ3)(z¯ζ3)dz=D+¯ζ31/(zζ3)(z¯ζ3)dz+D+ζ31/(z¯ζ3)(zζ3)dz=2πi(1zζ3)|z=¯ζ3+2πi(1z¯ζ3)|z=ζ3=2πi(1¯ζ3ζ3+1ζ3¯ζ3)=2πi(1¯ζ3ζ31¯ζ3ζ3)=0, where for the last we note that ζ3=e2πi3=1+i3 and ζ3¯ζ3=2(ζ3)=212=1, yielding the factorization of the denominator.

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