Cauchy Integral Formula

Theorem. Let f(z) be analytic in the closed region

DR = {|z| ≤ R }.

Then for |z| < R,

f(z) = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) ζ− z  dζ.

Proof: Fix z. For ϵ small, let C_z,ϵ = {ζ| |ζ− z| = ϵ}. Define g_z(ζ) = [(f(ζ))/(ζ− z)]. Then g_z(ζ) is an analytic function of ζ in the region between the two circles C_z,ϵ and C_R.

By the Two Circles Theorem

⌠ (⎜) ⌡ CR  gz(ζ)   dζ = ⌠ (⎜) ⌡ Cz,ϵ  gz(ζ)  dζ, ⌠ (⎜) ⌡ CR  f(ζ) ζ− z  dζ = ⌠ (⎜) ⌡ Cz,ϵ  f(ζ) ζ− z  dζ.

Since f is analytic at z, for ζ near z,

f(ζ) = f(z) + f′(z) (ζ− z) + o(ζ− z).

Thus

⌠ (⎜) ⌡ Cz,ϵ  f(ζ) ζ− z  dζ = ⌠ (⎜) ⌡ Cz,ϵ  f(z) + f′(z) (ζ− z) + o(ζ− z) ζ− z dζ. = f(z) ⌠ (⎜) ⌡ Cz,ϵ  1 ζ− z  dζ + f′(z) ⌠ (⎜) ⌡ Cz,ϵ  1  dζ + ⌠ (⎜) ⌡ Cz,ϵ  o(1)  dζ = 2 πi f(z) + 0 + o(ϵ).

Now let ϵ→ 0 to obtain

⌠ (⎜) ⌡ CR  f(ζ) ζ− z  dζ = 2 πi f(z).

Power Series Representation of Analytic Functions

We will show that the analytic function has a power series representation with a radius of convergence at least R.

Fix ζ, |ζ| = R, and fix z, |z| < R. Let θ = |z/(ζ)|. Then

1 ζ− z = 1 ζ ⎛ ⎝ 1 − z ζ ⎞ ⎠ = 1 ζ ∞ ∑ n = 0  ⎛ ⎝ z ζ ⎞ ⎠ n   = 1 ζ N ∑ n = 0  ⎛ ⎝ z ζ ⎞ ⎠ n   + 1 ζ ⎛ ⎝ z ζ ⎞ ⎠ N+1   1 − z ζ .

It follows that

f(z) = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) ζ− z  dζ. = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) 1 ζ ∞ ∑ n = 0  ⎛ ⎝ z ζ ⎞ ⎠ n    dζ = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) 1 ζ N ∑ n = 0  ⎛ ⎝ z ζ ⎞ ⎠ n     dζ + 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) 1 ζ ⎛ ⎝ z ζ ⎞ ⎠ N+1   1 − z ζ  dζ = N ∑ n = 0  an zn + RN(z),

where

an = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) 1 ζn+1   dζ, RN(z) = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) 1 ζ ⎛ ⎝ z ζ ⎞ ⎠ N+1   1 − z ζ  dζ, |RN(z)| ≤ max |ζ| = R  |f(ζ)| |θ|N+1 1 − |θ| .

Theorem. Let f(z) be analytic in the closed region

DR = {|z| ≤ R }.

Then for |z| < R:

•

f(z) = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) ζ− z  dζ.

•

f(z) = ∞ ∑ n = 0  an zn, an = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) 1 ζn+1   dζ,

•

For n = 0,1,2, …, f⁽ⁿ⁾(z) exists and

f(n)(z) = n! 2 πi ⌠ (⎜) ⌡ CR  f(ζ) 1 (ζ− z)n+1   dζ, = 1 2πi ⌠ (⎜) ⌡ CR  f(ζ) dn d zn ⎧ ⎨ ⎩ 1 ζ− z ⎫ ⎬ ⎭  dζ.

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File translated from T_EX by T_TH, version 4.03.
On 16 Nov 2013, 21:03.