See reference
Suppose f is holomorphic on Ω, then for any z0∈Ω and any open disc ¯DR(z0) such that γ:=∂¯DR(z0)⊆Ω, f(z0)=12πi∫γf(ξ)ξ−z0 dξ and f(n)(z0)=n!2πi∫γf(ξ)(ξ−z0)n+1dξ. As a consequence, if f(z)∑k≥0ck(z−z0)k, ck=f(n)(z0)n!=12πi∫γf(ξ)(ξ−z0)n+1dξ.
proof (?):
proof (?):
proof (Alternative):
Exercises
solution:
By Cauchy’s formula, ∫f(z)(z−0)2dz=2πif(1)(0)=2πi.
Without using the residue formula, compute ∫Rf(x)dxf(x):=1x4+16.
solution:
Use a semicircular contour, noting the poles are at ±√2±i√2. Write
- f1(z):=(√2+i√2)f(z)
- f2(z):=(−√2+i√2)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+i√2)=π√2(1−i)32
- For the loop around the left pole: I2=2πif2(√2−i√2)=π√2(1+i)32.
Now show that ∫CR vanishes: parameterize as γ(t)=Reit and use the reverse triangle inequality: |∫CRf|≤∫π01R4−16=πRR4−16→0.
Suppose f is holomorphic on Ω, a simply connected region, and suppose γ⊆Ω. Using the Cauchy integral formula, show that ∫γf′(z)z−adz=∫γf(z)(z−a)2dz. Also prove this when Ω is not simply connected.
solution:
Use the integral formula directly: ∫γf′(z)z−adz=2πif′(a).
On the other hand, use Cauchy’s formula for derivatives: ∫γf(z)(z−a)2dz=2πif(1)(a), and these values are equal.
If Ω is not simply connected, note that by the quotient rule ∂∂zf(z)z−a=f′(z)z−a−f(z)(z−a)2.
Thus ∫γf′(z)z−a−∫γf(z)(z−a)2dz=∫γ(f′(z)z−a−f(z)(z−a)2)dz=∫γ∂∂zf(z)z−adz=G(γ(1))−G(γ(0))=G(p)−G(p)=0, where G(z):=f(z)z−a is a primitive for the integrand by definition.
Evaluate the following integrals using Cauchy’s integral formula:
∫S1cos(z)zdz∫S1sin(z)zdz∫|z|=2z2z−1dz∫S1ezz2dz∫|z|=2z2−1z2+1dz∫|z|=21z2+z+1dz.
solution:
∫S1cos(z)zdz=2πicos(0)=2πi∫S1sin(z)zdz=2πisin(0)=0∫|z|=2z2z−1dz=2πiz2|z=1=2πi∫S1ezz2dz=2πi∂∂zez|z=0=2πi∫|z|=2z2−1z2+1dz=∫D+i(z2−1)/(z+i)z−idz+∫D−i(z2−1)/(z−i)z+idz=2πi(z2−1z+i)|z=i+2πi(z2−1z−i)|z=−i=2πi(−22i)+2πi(−2−2i)=0∫|z|=21z2+z+1dz=∫1(z−ζ3)(z−¯ζ3)dz=∫D+¯ζ31/(z−ζ3)(z−¯ζ3)dz+∫D+ζ31/(z−¯ζ3)(z−ζ3)dz=2πi(1z−ζ3)|z=¯ζ3+2πi(1z−¯ζ3)|z=ζ3=2πi(1¯ζ3−ζ3+1ζ3−¯ζ3)=2πi(1¯ζ3−ζ3−1¯ζ3−ζ3)=0, where for the last we note that ζ3=e2πi3=−1+i√3 and ζ3¯ζ3=2ℜ(ζ3)=2⋅12=1, yielding the factorization of the denominator.