Math 2.0

Functions of a complex variable Example

Integral of 1/z around the unit circle

Builds on Circle integralComplex exponential

Example

The circle integral of f(z)=1/zf(z) = 1/z over the unit circle ∣z∣=1|z| = 1 is ∮∣z∣=1dzz=2πi.\oint_{|z| = 1} \frac{dz}{z} = 2 \pi i. Computation: the path is γ(θ)=eiθ\gamma(\theta) = e^{i\theta} with θ∈[0,2π]\theta \in [0, 2\pi], so γ′(θ)=ieiθ\gamma'(\theta) = i e^{i\theta}. Then ∮∣z∣=1dzz=∫02πieiθeiθ dθ=∫02πi dθ=2πi.\oint_{|z| = 1} \frac{dz}{z} = \int_0^{2\pi} \frac{i e^{i\theta}}{e^{i\theta}} \, d\theta = \int_0^{2\pi} i \, d\theta = 2 \pi i. More generally, let c,w∈ℂc, w \in \mathbb{C} and R>0R > 0. If ∣w−c∣<R|w - c| < R, then ∮∣z−c∣=Rdzz−w=2πi\oint_{|z - c| = R} \frac{dz}{z - w} = 2 \pi i. If ∣w−c∣>R|w - c| > R, the integral is 00.

Remarks

For each integer n≠−1n \neq -1, ∮∣z∣=1zn dz=0\oint_{|z| = 1} z^n \, dz = 0, because znz^n has a primitive on ℂ∖{0}\mathbb{C} \setminus \{0\}. The value is not 00, so 1/z1/z has no primitive on ℂ∖{0}\mathbb{C} \setminus \{0\}, and Cauchy's integral theorem does not apply on ℂ∖{0}\mathbb{C} \setminus \{0\}. This integral is the base of Cauchy's integral formula, of the residue and of the winding number.

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  1. Complex numbers object
  2. Complex exponential definition
  3. Contour integral definition
  4. Circle integral definition
  5. Integral of 1/z around the unit circle example

Further reading

  1. Web https://en.wikipedia.org/wiki/Contour_integration, Example
  2. Web https://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem, Discussion
About this page
Tag0SXZ68HV6F
URIm2://0SXZ68HV6F
Aliasescontour-integral-of-1-over-z
Kindexample
WritingW0 stub
MSC30E20
Instance ofCircle integral
UsesCircle integral, Complex exponential

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