Cauchy Integral Theorem

Cauchy’s Theorems I

Ang M.S. Augustin-Louis Cauchy October 26, 2012 1789 – 1857

References Murray R. Spiegel Complex V ariables with introduction to conformal mapping and its applications Dennis G. Zill , P. D. Shanahan A F irst Course in Complex Analysis with Applications J. H. Mathews , R. W. Howell Complex Analysis F or Mathematics and Engineerng

1 Summary

• Cauchy Integral Theorem

Let f be analytic in a simply connected domain D. If C is a simple closed contour that lies in D , and there is no singular point inside the contour, then ˆ f (z) dz = 0 C • Cauchy Integral Formula (For simple pole)

If there is a singular point z0 inside the contour, then ˛ f(z) dz = 2πj f(z0) z − z0

• Generalized Cauchy Integral Formula (For pole with any order) ˛ f(z) 2πj (n−1) n dz = f (z0) (z − z0) (n − 1)!

• Cauchy Inequality

M · n! f (n)(z ) ≤ 0 rn • Gauss Mean Value Theorem ˆ 2π ( ) 1 jθ f(z0) = f z0 + re dθ 2π 0

1 2 Related Mathematics Review

2.1 Stoke’s Theorem ¨ ˛ ∇ × F¯ · dS = F¯ · dr Σ ∂Σ (The proof is skipped)

¯ Consider F = (FX ,FY ,FZ )   xˆ yˆ zˆ  ∂ ∂ ∂  ∇ × F¯ = det    ∂x ∂y ∂z  FX FY FZ

Let FZ = 0   xˆ yˆ zˆ  ∂ ∂ ∂  ∇ × F¯ = det    ∂x ∂y ∂z  FX FY 0 dS =ndS ˆ , for dS = dxdy , nˆ =z ˆ . By zˆ · zˆ = 1 , consider zˆ component only for ∇ × F¯   xˆ yˆ zˆ ( )  ∂ ∂ ∂  ∂F ∂F ∇ × F¯ = det   = Y − X zˆ  ∂x ∂y ∂z  ∂x ∂y FX FY 0 i.e. ¨ ¨ ( ) ∂F ∂F ∇ × F¯ · dS = Y − X dxdy Σ Σ ∂x ∂y

Consider the RHS (let FZ = 0) ˛ ˛ ˛ ¯ F · dr = (FX ,FY ,FZ ) · (dx, dy, dz) = FX dx + FY dy ∂Σ ∂Σ ∂Σ Thus ¨ ( ) ˛ ∂FY ∂FX − dxdy = FX dx + FY dy Σ ∂x ∂y ∂Σ

2.2 Cauchy-Reimann Condition

For f(z) = u(z) + jv(z) = u(x, y) + jv(x, y) The complex derivative is

f (z0 + h) − f(z0) ′ lim = f (z0) h → 0 h h ∈ C If the limit exists.

2 If the limit exists, limits along real axis should be the same the limit along imaginary axis

f (z + h) − f (z ) ∂f (z ) 1 ∂f (z ) f (z + jh) − f (z ) lim 0 0 = 0 = 0 = lim 0 0 h → 0 h ∂x j ∂y h → 0 jh h ∈ R h ∈ R i.e. The Cauchy-Riemann Equation

∂f (z ) 1 ∂f (z ) 0 = 0 ∂x j ∂y Expand ∂ 1 ∂ [u (x , y ) + jv (x , y )] = [u (x , y ) + jv (x , y )] ∂x 0 0 0 0 j ∂y 0 0 0 0 Equalize real part and imaginary part   ∂ ∂  u (x , y ) = v (x , y ) ∂x 0 0 ∂y 0 0  ∂ ∂  v (x , y ) = − u (x , y ) ∂x 0 0 ∂y 0 0 Or simply as   ∂u ∂v  = ∂x ∂y  ∂v ∂u  = − ∂x ∂y i.e. If f(z) is analytic, then it fulﬁll this condition

2.3 ML Inequality

First, ˆ ˆ b b if g(x) ≤ f(x) , then g(x)dx ≤ f(x)dx a a This is true if

sup g(x) ≤ sup f(x) inf g(x) ≤ inf f(x) Then for any function f , it is true that

−|f| ≤ f ≤ |f| Then apply the inequality above , ˆ ˆ ˆ b b b − |f(x)|dx ≤ f(x)dx ≤ |f(x)|dx a a a Now is the time to show , for bounded f(z) , i.e. |f(z)| ≤ M , ˆ

f(z)dz ≤ ML c

3 Pf. ´ ´ b ≤ b | | By using the a f(x)dx a f(x) dx ˆ ˆ

f(z)dz ≤ |f(z)|dz c c Since f(z) si bounded, ˆ ˆ ˆ ˆ

f(z)dz ≤ |f(z)|dz ≤ Mdz = M dz = ML c c c | C{z } L Therefore ˆ | f(z)dz| ≤ ML c

3 The Cauchy Theorem

f(z) = u(x, y) + jv(x, y) ˛ ˛ ˛ ˛ f (z) dz = [u (x, y) + jv (x, y)] [dx + jdv] = udx − vdy + j vdx + udy C C C C

3.1 The Real Part

Let FX = u (x, y) FY = −u (x, y) and consider the real part of the integral ˛ ˛

udx − vdy = FX dx + FY dy C By the Stoke’ Theorem ˛ ¨ ¨ − − − − − udx + ( v)dy = [( v)x uy] dxdy = [ vx uy] dxdy C Since the function is analytic in the region D , ⇐⇒ the function fulﬁll Cauchy-Reimann Condition ∂v ∂u = − ∂x ∂y Then ¨ ¨

[−vx − uy] dxdy = [uy − uy] dxdy = 0 ´ The real part of C f(z)dz equal zero

4 3.2 The imaginary part

´ Consider the imaginary part of C f(z)dz , ˛ vdx + udy C

Let FX be v(x, y) and FY be u(x, y) , and apply Stoke’s Theorem ˛ ¨

vdx + udy = (ux − vy) dxdy C ∂u ∂v As the function is analytic, so it fulﬁll Cauchy-Reimann Condition : = , thus the integral ∂x ∂y become zero

Finally, Let f be analytic in a simply connected domain D , for the simple closed contour C that lies in D , the close contour integral equal zero ˆ f (z) dz = 0 C

4 The Cauchy Integral Formulas

4.1 Simple Pole ˛ g (z) − dz = 2πjg (z0) C z z0 Condition

• z0 inside C , if z0is outside C , then the integral is just zero • g(z) is analytic in simply connected domain ⇐⇒ domain of g(z) is simply connect closed region, and g(z) fulﬁll CR-Condition there

Proof. Consider the diagram g(z) By concept of path deformation , the integral of z − z0

along path C is equivalent to the integral along path Cϵ that ϵ → 0 ˛ ˛ g(z) g(z) dz = lim dz − ϵ→0 − C z z0 Cϵ z z0 The small circle inside can be expressed as

jθ |z − z0| = ϵe 0 ≤ θ < 2π As the radius ϵ → 0+ , the circle is thus

jθ z − z0 = ϵe 0 ≤ θ < 2π

5 Put this back into the integral ˛ ˛ ˆ ( ) 2π jθ ( ) g(z) g(z) g z0 + ϵe jθ dz = lim dz = lim d z0 + ϵe − ϵ→0 − ϵ→0 jθ − C z z0 Cϵ z z0 0 (z0 + ϵe ) z0 ˆ ( ) ˆ 2π g z + ϵejθ 2π ( ) = lim 0 jϵejθdθ = jlim g z + ϵejθ dθ → jθ → 0 ϵ 0 0 ϵe ϵ 0 0 ˆ ˆ ˆ 2π [ ( )] 2π 2π = j limg z + ϵejθ dθ = j g (z ) dθ = jg (z ) dθ = 2πjg (z ) → 0 0 0 0 0 ϵ 0 0 0 ∴ ˛ g(z) − dz = 2πjg (z0) C z z0

4.2 Generalized Cauchy Integral Formula ˛ f (z0) 2πj (n−1) n dz = 2πj f (z0) (z − z0) (n − 1)! First, the Cauchy Integral Formula for simple pole is ˛ g (z) dz = 2πjg (z ) z − z 0 ( ) C 0 d d Take on both side Not ! dz0 ˛ dz ˛ d g (z) d g (z) 2πj dz = 2πj g (z ) ⇒ dz = g′ (z ) − 0 2 0 dz0 C z z0 dz0 C (z − z0) 1! Repeat, ˛ ˛ d g (z) 2πj d ′ ⇒ g (z) 2πj 2 dz = g (z0) 3 dz = g”(z0) dz0 C (z − z0) 1! dz0 C (z − z0) 2! ˛ ˛ d g (z) 2πj d g (z) 2πj ⇒ (3) 3 dz = g”(z0) 4 dz = g (z0) dz0 C (z − z0) 1! dz0 C (z − z0) 3! Thus, the general form of Cauchy Integral Formula is ˛ g (z) 2πj (n−1) − n dz = − g (z0) C (z z0) (n 1)! This can be proved using Mathematical Induction.

6 5 Consequences of Cauchy’s Integral Formula 5.1 Cauchy Inequality

The generalized Cauchy Integral Formula is ˛ f (z) 2πj (n) n+1 dz = f (z0) C (z − z0) n! Take absolute value ˛ ˛ f (z) 2πj f (z) |2π| · |j| dz = f (n) (z ) ⇒ dz = f (n) (z ) n+1 0 n+1 | | 0 C (z − z0) n! C (z − z0) n! Rearrange ˛ ˛ n! f (z) n! f (z) (n) ≤ f (z0) = n+1 dz n+1 dz 2π C (z − z0) 2π C (z − z0) Since

jθ ⇒ − jθ ⇒ − n+1 | n+1| · | j(n+1)θ| n+1 z = z0 + re (z z0) = re (z z0) = r |e {z } = r 1 And apply ML inequality ˛ ˛ ˛ n! f (z) n! M · n! M · n! f (n) (z ) ≤ dz ≤ Mdz = dz = · 2πr 0 n+1 · n+1 · n+1 · n+1 2π C r 2π r C 2π r | C{z } 2π r 2πr Finally M · n! f (n) (z ) ≤ 0 rn 5.2 Gauss Mean Value Theorem

Recall, the Cauchy Integral Formula for simple pole ˛ f (z) − dz = 2πj f (z0) C z z0 jθ jθ Rearrange, and take z = z0 + re , dz = rje dθ ˛ ˆ ( ) 1 f (z) 1 2π f z + rejθ ( ) f (z ) = dz = 0 rjejθdθ 0 − jθ 2πj C z z0 2πj 0 re Cancel out common factor ˆ 2π ( ) 1 jθ f (z0) = f z0 + re dθ 2π 0

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