source/reference:
https://www.youtube.com/channel/UCaTLkDn9_1Wy5TRYfVULYUw/playlists

Complex Integration

Integration in ℝ

Let 𝑓:[𝑎,𝑏]→ℝ be continuous. Then 𝑏∫𝑎𝑓(𝑡)𝑑𝑡= Lim𝑛→∞ 𝑛−1∑𝑗=0𝑓(𝑡𝑗)(𝑡𝑗+1−𝑡𝑗) where 𝑎=𝑡₀<𝑡₁<⋯<𝑡_𝑛=𝑏

• if 𝑓≥0 on [𝑎,𝑏] then 𝑏∫𝑎𝑓(𝑡)𝑑𝑡 is the "area under the curve".
• Otherwise: sum of the areas above the x-axis minus sum of the areas below the x-axis.

The Fundamental Theorem of Calculus

TheoremLet 𝑓:[𝑎,𝑏]→ℝ be continuous, and define 𝐹(𝑥)=𝑥∫𝑎𝑓(𝑡)𝑑𝑡. Then 𝐹 is differentiable and 𝐹′(𝑥)=𝑓(𝑥) for 𝑥∈[𝑎,𝑏].

Antiderivatives

Let 𝑓:[𝑎,𝑏]→ℝ as above. A function 𝑓:[𝑎,𝑏]→ℝ that satisfies that 𝐹′(𝑥)=𝑓(𝑥) for all 𝑥∈[𝑎,𝑏] is called an antiderivative of 𝑓.

Note: If 𝐹 and 𝐺 are both antiderivatives of the same function 𝑓, then (𝐺−𝐹)′(𝑥)=𝐺′(𝑥)−𝐹′(𝑥)=𝑓(𝑥)−𝑓(𝑥)=0 for all 𝑥∈[𝑎,𝑏], and so 𝐺−𝐹 is constant.

Conclusion: Let 𝐺 be any antiderivative of 𝑓. Then

𝑏∫𝑎𝑓(𝑡)𝑑𝑡=𝐺(𝑏)−𝐺(𝑎)

Generalization to ℂ

Instead of integrating over an interval [𝑎,𝑏]⊂ℝ, integrating in ℂ will be integrating ovver curves. Recall: A curve is a smooth or piecewise smooth function

𝛾:[𝑎,𝑏]→ℂ, 𝛾(𝑡)=𝑥(𝑡)+𝑖𝑦(𝑡)

If 𝑓 is complex-valued on 𝛾, define  ∫𝛾𝑓(𝑧)𝑑𝑧= Lim𝑛→∞ 𝑛−1∑𝑗=0𝑓(𝑧𝑗)(𝑧𝑗+1−𝑧𝑗) where 𝑧_𝑗=𝛾(𝑡_𝑗) and 𝑎=𝑡₀<𝑡₁<⋯<𝑡_𝑛=𝑏

The Path Integral

 ∫𝛾𝑓(𝑧)𝑑𝑧= Lim𝑛→∞ 𝑛−1∑𝑗=0𝑓(𝑧𝑗)(𝑧𝑗+1−𝑧𝑗)

where 𝑧_𝑗=𝛾(𝑡_𝑗) and 𝑎=𝑡₀<𝑡₁<⋯<𝑡_𝑛=𝑏

One can show: If 𝛾:[𝑎,𝑏]→ℂ is a smooth curve and 𝑓 is continuous on 𝛾, then

 ∫𝛾𝑓(𝑧)𝑑𝑧= 𝑏∫𝑎𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡 Proof Idea 𝑛−1∑𝑗=0𝑓(𝑧𝑗)(𝑧𝑗+1−𝑧𝑗) =𝑛−1∑𝑗=0𝑓(𝛾(𝑡𝑗)) 𝛾(𝑡𝑗+1)−𝛾(𝑡𝑗)𝑡𝑗+1−𝑡𝑗(𝑡𝑗+1−𝑡𝑗)  → 𝑏∫𝑎𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡 as 𝑛→∞

Integrals over Complex-valued Functions

Note: If 𝑔:[𝑎,𝑏]→ℂ, 𝑔(𝑡)=𝑢(𝑡)+𝑖𝑣(𝑡), then

𝑏∫𝑎𝑔(𝑡)𝑑𝑡= 𝑏∫𝑎𝑢(𝑡)𝑑𝑡+ 𝑖𝑏∫𝑎𝑣(𝑡)𝑑𝑡

Examples

• 𝜋∫0ℯ^𝑖𝑡𝑑𝑡= 𝜋∫0cos 𝑡+ 𝑖𝜋∫0sin 𝑡𝑑𝑡= sin 𝑡=sin 𝑡|^𝜋₀−𝑖 cos 𝑡=sin 𝑡|^𝜋₀=0−𝑖(−1−1)=2𝑖
• Alternatively: 𝜋∫0ℯ^𝑖𝑡𝑑𝑡 =−𝑖ℯ^𝑖𝑡𝜋｜0=−𝑖ℯ^𝑖𝜋+𝑖ℯ⁰=2𝑖
• 1∫0(𝑡+𝑖)𝑑𝑡 =12𝑡²+𝑖𝑡 1｜0=12𝑡²+𝑖

• 𝛾(𝑡)=𝑡+𝑖𝑡, 0≤𝑡≤1, 𝛾′(𝑡)=1+𝑖, 𝑓(𝑧)=𝑧². Then

   ∫𝛾𝑓(𝑧)𝑑𝑧= 1∫0𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡= 1∫0(𝑡+𝑖𝑡)2(1+𝑖)𝑑𝑡  = 1∫0(𝑡2+2𝑖𝑡2−𝑡2)(1+𝑖)𝑑𝑡= 1∫0(2𝑖𝑡2−2𝑡2)𝑑𝑡  = −21∫0𝑡2𝑑𝑡+ 2𝑖1∫0𝑡2𝑑𝑡  =−23𝑡31｜0 +2𝑖3𝑡31｜0  =−23+𝑖23=23(−1+𝑖)
•  ∫|𝑧|=11𝑧𝑑𝑧=?

  Let 𝛾(𝑡)=ℯ^𝑖𝑡, 0≤𝑡≤2𝜋. Then 𝛾′(𝑡)=𝑖ℯ^𝑖𝑡, so:

   ∫|𝑧|=11𝑧𝑑𝑧 =2𝜋∫01𝛾(𝑡)𝛾′(𝑡)𝑑𝑡  =2𝜋∫01ℯ𝑖𝑡𝑖ℯ𝑖𝑡𝑑𝑡  =𝑖2𝜋∫0𝑑𝑡  =𝑖𝑡|2𝜋0=2𝜋𝑖
•  ∫|𝑧|=1𝑧𝑑𝑧=?

  Let 𝛾(𝑡)=ℯ^𝑖𝑡, 0≤𝑡≤2𝜋. Then 𝛾′(𝑡)=𝑖ℯ^𝑖𝑡, so:

   ∫|𝑧|=1𝑧𝑑𝑧 =2𝜋∫0𝛾(𝑡)𝛾′(𝑡)𝑑𝑡=2𝜋∫0 ℯ𝑖𝑡𝑖ℯ𝑖𝑡𝑑𝑡  =𝑖2𝜋∫0ℯ2𝑖𝑡𝑑𝑡= 12ℯ2𝑖𝑡2𝜋｜0  =12(ℯ4𝜋𝑖−ℯ0)=0
•  ∫|𝑧|=11𝑧²𝑑𝑧=?

  Let 𝛾(𝑡)=ℯ^𝑖𝑡, 0≤𝑡≤2𝜋. Then 𝛾′(𝑡)=𝑖ℯ^𝑖𝑡, so:

   ∫|𝑧|=11𝑧2𝑑𝑧 =2𝜋∫01𝛾2(𝑡)𝛾′(𝑡)𝑑𝑡= 2𝜋∫0𝑖ℯ𝑖𝑡ℯ2𝑖𝑡𝑑𝑡  =2𝜋∫0𝑖ℯ−𝑖𝑡𝑑𝑡=−ℯ−𝑖𝑡2𝜋｜0  =−ℯ−2𝜋𝑖+ℯ0

• In general,

   ∫|𝑧|=1𝑧𝑚𝑑𝑧=2𝜋𝑖, if 𝑚=-1 0, otherwise

• Let 𝛾:[𝑎,𝑏]→ℂ be a smooth curve, and let 𝑓 be complex-valued and continuous on 𝛾. Then

   ∫𝛾𝑓(𝑧)𝑑𝑧= 𝑏∫𝑎𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡

  Let 𝛾(𝑡)=1−𝑡(1−𝑖), 0≤𝑡≤1, and let 𝑓(𝑧)=Re 𝑧. Then

   ∫𝛾𝑓(𝑧)𝑑𝑧= 1∫0Re(1−𝑡(1−𝑖))(−1)(1−𝑖)𝑑𝑡  =(𝑖−1) 1∫0Re(1−𝑡)𝑑𝑡  =(𝑖−1)𝑡−12𝑡21｜0  =(𝑖−1)1−12=𝑖−12

• Let 𝛾(𝑡)=𝑟ℯ^𝑖𝑡, 0≤𝑡≤2𝜋. Then 𝛾′(𝑡)=𝑟𝑖ℯ^𝑖𝑡. Let 𝑓(𝑧)=𝑧

   ∫𝛾𝑓(𝑧)𝑑𝑧= ∫𝛾𝑧𝑑𝑧= 2𝜋∫0𝛾(𝑡)𝛾′(𝑡)𝑑𝑡  = 2𝜋∫0𝑟ℯ−𝑖𝑡𝑟𝑖ℯ𝑖𝑡𝑑𝑡  =𝑟2𝑖 2𝜋∫0𝑑𝑡  =2𝜋𝑖𝑟2=(2𝑖)∙area(𝐵𝑟(0))

Integraton by substitution

Let [𝑎,𝑏] and [𝑐,𝑑] be intervals in ℝ and let ℎ:[𝑐,𝑑]→[𝑎,𝑏] be smooth. Suppose that 𝑓:[𝑎,𝑏]→ℝ is a continuous function. Then

ℎ(𝑑)∫ℎ(𝑐)𝑓(𝑡)𝑑𝑡=𝑑∫𝑐𝑓(ℎ(𝑠))ℎ′(𝑠)𝑑𝑠

Examples

𝑡=ℎ(𝑠)=𝑠³+1, ℎ′(𝑠)=3𝑠² 4∫2𝑠2(𝑠3+1)4𝑑𝑠= 13ℎ(4)∫ℎ(2)𝑡4𝑑𝑡  = 1365∫9𝑡4𝑑𝑡  =13𝑡551｜0  =115(655−95)

Fact: Independence of Parametrization

Let 𝛾:[𝑎,𝑏]→ℂ be a smooth curve, and let 𝛽:[𝑐,𝑑]→ℂ be another smooth parametrization of the same curve, given by 𝛽(𝑠)=𝛾(ℎ(𝑠)), where ℎ:[𝑐,𝑑]→[𝑎,𝑏] is a smooth bijection.

Let 𝑓 be a complex-valued function, defined on 𝛾. Then  ∫𝛽𝑓(𝑧)𝑑𝑧=𝑑∫𝑐𝑓(𝛽(𝑠))𝛽′(𝑠)𝑑𝑠  =𝑑∫𝑐𝑓(𝛾(ℎ(𝑠)))𝛾′(ℎ(𝑠))ℎ′(𝑠)𝑑𝑠  =𝑑∫𝑐𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡= ∫𝛾𝑓(𝑧)𝑑𝑧 Therefore, the commplex path integral is independent of the parametrization

Fact: Piecewise Smooth Curves

Let 𝛾=𝛾₁+𝛾₂+⋯+𝛾_𝑛 be a piecewise smooth curve (i.e. 𝛾_𝑗+1 starts where 𝛾_𝑗 ends). Then

 ∫𝛾𝑓(𝑧)𝑑𝑧= ∫𝛾1𝑓(𝑧)𝑑𝑧+ ∫𝛾2𝑓(𝑧)𝑑𝑧+⋯+ ∫𝛾𝑛𝑓(𝑧)𝑑𝑧

Reverse Paths

If 𝛾:[𝑎,𝑏]→ℂ be a curve, then a curve (−𝛾):[𝑎,𝑏]→ℂ is defined by (−𝛾)(𝑡)=𝛾(𝑎+𝑏−𝑡)

Note that (−𝛾)′(𝑡)=𝛾′(𝑎+𝑏−𝑡)(−1). If 𝑓 is continuous and complex-valued on 𝛾, then

 ∫(−𝛾)𝑓(𝑧)𝑑𝑧=𝑏∫𝑎𝑓((−𝛾)(𝑠))(−𝛾)′(𝑠)𝑑𝑠=−𝑏∫𝑎𝑓(𝛾(𝑎+𝑏−𝑠))𝛾′(𝑎+𝑏−𝑠)𝑑𝑠  =𝑎∫𝑏𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡=−𝑏∫𝑎𝑓(𝛾(𝑡))𝛾′(𝑡)𝑑𝑡  =− ∫𝛾𝑓(𝑧)𝑑𝑧

Fact

If 𝛾 is a curve, 𝑐 is a complex constant and 𝑓, 𝑔 are continuous and complex-valued on 𝛾, then

•  ∫𝛾(𝑓(𝑧)+𝑔(𝑧))𝑑𝑧= ∫𝛾𝑓(𝑧)𝑑𝑧+ ∫𝛾𝑔(𝑧)𝑑𝑧
•  ∫𝛾𝑐𝑓(𝑧)𝑑𝑧=𝑐 ∫𝛾𝑓(𝑧)𝑑𝑧
•  ∫−𝛾𝑐𝑓(𝑧)𝑑𝑧=− ∫𝛾𝑓(𝑧)𝑑𝑧

Arc Length

Given a curve 𝛾:[𝑎,𝑏]→ℂ. Let 𝑎=𝑡₀<𝑡₁<⋯<𝑡_𝑛=𝑏. Then length(𝛾)≈𝑛∑𝑗=0|𝛾(𝑡𝑗+1)−𝛾(𝑡𝑗)| If the limit exists as 𝑛→∞, then this is the length of 𝛾:[𝑎,𝑏]→ℂ 𝑛∑𝑗=0|𝛾(𝑡𝑗+1)−𝛾(𝑡𝑗)|= 𝑛∑𝑗=0 |𝛾(𝑡𝑗+1)−𝛾(𝑡𝑗)|𝑡𝑗+1−𝑡𝑗(𝑡𝑗+1−𝑡𝑗)→ 𝑏∫𝑎|𝛾′(𝑡)|𝑑𝑡 Thus:

length(𝛾)=𝑏∫𝑎|𝛾′(𝑡)|𝑑𝑡

Examples

• Let 𝛾(𝑡)=𝑅ℯ^𝑖𝑡, 0≤𝑡≤2𝜋, for some 𝑅>0. Then 𝛾′(𝑡)=𝑅𝑖ℯ^𝑖𝑡, and so

  length(𝛾)= 2𝜋∫0|𝑅𝑖ℯ𝑖𝑡|𝑑𝑡= 2𝜋∫0𝑅𝑑𝑡=2𝜋𝑅

• Let 𝛾(𝑡)=𝑡+𝑖𝑡, 0≤𝑡≤1. Then 𝛾′(𝑡)=1+𝑖, and so

  length(𝛾)= 1∫0|1+𝑖|𝑑𝑡= 1∫02𝑑𝑡=2

Integration with respect to Arc Length

DefinitionLet 𝛾 be a smooth curve, and let 𝑓 be a complex-valued and continuous function on 𝛾. Then  ∫𝛾𝑓(𝑧)|𝑑𝑧|= 𝑏∫𝑎𝑓(𝛾(𝑡))|𝛾′(𝑡)|𝑑𝑡 is the integral of 𝑓 over 𝛾 with respect to arc length

Examples

• length(𝛾)=  ∫𝛾|𝑑𝑧|.
•  ∫|𝑧|=1𝑧|𝑑𝑧|= 2𝜋∫0ℯ^𝑖𝑡⋅1𝑑𝑡=−𝑖ℯ^𝑖𝑡 2𝜋｜0=0

Note: Piecewise smooth curves are allowed as well (break up the integral into a sum over smooth pieces).

The 𝑀𝐿-Estimate

TheoremIf 𝛾 is a curve and 𝑓 is continuous on 𝛾 then  ∫𝛾𝑓(𝑧)𝑑𝑧≤  ∫𝛾|𝑓(𝑧)||𝑑𝑧|. In particular, if |𝑓(𝑧)|≤𝑀 on 𝛾, then  ∫𝛾𝑓(𝑧)𝑑𝑧≤𝑀⋅length(𝛾)

Examples

• Let 𝛾(𝑡)=ℯ^𝑖𝑡, 0≤𝑡≤2𝜋.  ∫|𝑧|=11𝑧|𝑑𝑧|= 2𝜋∫0ℯ−𝑖𝑡𝑑𝑡=𝑖ℯ−𝑖𝑡 2𝜋｜0=0
• Let 𝛾(𝑡)=𝑡+𝑖𝑡, 0≤𝑡≤1. The upper bound for  ∫𝛾𝑧²𝑑𝑧 can be found as following.

  First use the second part of the theorem:  ∫𝛾𝑓(𝑧)𝑑𝑧≤𝑀⋅length(𝛾). For 𝑓(𝑧)=𝑧², and having that |𝑓(𝑧)|=|𝑧|²≤(2)²=2 on 𝛾, so 𝑀=2. Also, recall that length(𝛾)=2. Thus

   ∫𝛾𝑧2𝑑𝑧≤22
• Let 𝛾(𝑡)=𝑡+𝑖𝑡, 0≤𝑡≤1, 𝛾′(𝑡)=1+𝑡, |𝛾′(𝑡)|=2, 𝑓(𝑧)=𝑧². A better estimate for  ∫𝛾𝑧²𝑑𝑧 can be found as following.

  Using the first part of the theorem:  ∫𝛾𝑓(𝑧)𝑑𝑧≤  ∫𝛾|𝑓(𝑧)||𝑑𝑧| Thus  ∫𝛾𝑧2𝑑𝑧≤  ∫𝛾|𝑧|2|𝑑𝑧|= 1∫0|𝛾(𝑡)|2|𝛾′(𝑡)|𝑑𝑡= 1∫0|𝑡+𝑖𝑡|22𝑑𝑡  = 1∫02𝑡22𝑑𝑡  = 223𝑡31｜0 =232 Note:  ∫𝛾𝑧²𝑑𝑧=23(−1+𝑖)

Antiderivatives and Primitives

FactFrom the fundamental theorem of calculus, if 𝑓:[𝑎,𝑏]→ℝ is continuous and has an antiderivative 𝐹:[𝑎,𝑏]→ℝ, then 𝑏∫𝑎𝑓(𝑥)𝑑𝑥=𝐹(𝑏)−𝐹(𝑎) For a complex equivalent. DefinitionLet 𝐷⊂ℂ be a domain, and let 𝑓:𝐷⊂ℂ be a continuous function. A primitive of 𝑓 on 𝐷 is an analytic function 𝐹:𝐷→ℂ such that 𝐹′=𝑓 on 𝐷.

Functions with Primitives

An analytic function 𝐹:𝐷→ℂ such that 𝐹′=𝑓 is a primitive of 𝑓 in 𝐷

TheoremIf 𝑓 is continuous on a domain 𝐷 and if 𝑓 has a primitive 𝐹 in 𝐷, then for any curve 𝛾:[𝑎,𝑏]→𝐷. Thus have that  ∫𝛾𝑓(𝑧)𝑑𝑧=𝐹(𝛾(𝑏))−𝐹(𝛾(𝑎))

Note:

• The integral only depends on the initial point and the terminal point of 𝛾.
• Big 'hidden' assumption: 𝑓 needs to have a primitive in 𝐷.
• Under what assumptions does 𝑓 have a primitive?

Examples

• Let 𝛾:[𝑎,𝑏]→ℂ be the line segment from 0 to 1+𝑖. What is  ∫𝛾𝑧²𝑑𝑧

  The function 𝑓(𝑧)=𝑧² has a primitive in ℂ, namely 𝐹(𝑧)=13𝑧³. Therefore, 1+𝑖∫0𝑧2𝑑𝑧=  ∫𝛾𝑓(𝑧)𝑑𝑧=𝐹(𝛾(𝑏))−𝐹(𝛾(𝑎))  =𝐹(1+𝑖)−𝐹(0)  =13(1+𝑖)3−0  =13(1+3𝑖−3−𝑖)=23(−1+𝑖)

• Can  ∫|𝑧|=11𝑧𝑑𝑧 using a primitive?

  The function 𝐹(𝑧)=Log 𝑧 satisfies that 𝐹′(𝑧)=1𝑧, but not in all of ℂ.

  • 𝐹 is analytic in ℂ\(−∞,0]
  • Let 𝛾:[𝑎,𝑏]→ℂ be the part of the unit circle, started just below the negative x-axis, to just above the negative x-axis.
  • Then  ∫𝛾1𝑧𝑑𝑧≈ ∫𝛾1𝑧𝑑𝑧=Log(𝛾(𝑏))−Log(𝛾(𝑎))  ≈𝜋𝑖−(−𝜋𝑖)=2𝜋𝑖
• Let 𝛾 be any curve in ℂ from 𝑖 to 𝑖2. Then  ∫𝛾ℯ𝜋𝑧𝑑𝑧=1𝜋ℯ𝜋𝑧𝑖/2｜𝑖  =1𝜋ℯ𝜋𝑖/2−1𝜋ℯ𝜋𝑖  =1𝜋(𝑖+1)
• Let 𝛾 be any path in ℂ from −𝜋𝑖 to 𝜋𝑖. Then  ∫𝛾cos 𝑧𝑑𝑧=sin𝜋𝑖｜−𝜋𝑖=sin(𝜋𝑖)−sin(−𝜋𝑖) But sin(𝜋𝑖)−sin(−𝜋𝑖)=sin(𝜋𝑖)+sin(𝜋𝑖)  =2sin(𝜋𝑖)  =2ℯ𝑖𝜋𝑖−ℯ−𝑖𝜋𝑖2𝑖 =−𝑖(ℯ−𝜋−ℯ𝜋)=𝑖(ℯ𝜋−ℯ−𝜋)

Primitive

When does 𝑓 have a primitive?

Theorem (Goursat) Let 𝐷 be a simply connected domain in ℂ, and let 𝑓 be analytic in 𝐷. Then 𝑓 has a primitive in 𝐷. Moreover, a primitve is given explicitly by picking 𝑧0∈𝐷 and letting 𝐹(𝑧)=𝑧∫𝑧0𝑓(𝑤)𝑑𝑤where the integral is taken over an arbitrary curve in 𝐷 from 𝑧0 to 𝑧 One way to prove this theorem is as follows First, show Morera's Theorem: If 𝑓 is continuous on a simply connected domain 𝐷, and if  ∫𝛾𝑓(𝑧)𝑑𝑧=0 for any triangular curve 𝛾 in 𝐷, then 𝑓 has a primitive in 𝐷. Next, show the Cauchy Theorem for Triangles: For any triangle 𝑇 that fits into 𝐷 (including its boundary),  ∫∂𝑇𝑓(𝑧)𝑑𝑧=0.

The Cauchy Theorem for Triangles

Theorem (Cauchy for Triangles) Let 𝐷 be an open set in ℂ, and let 𝑓 be analytic in 𝐷. Let 𝑇 be a triangle that fits into 𝐷 (including its boundary), and let ∂𝑇 be its boundary, oriented positively. Then  ∫∂𝑇𝑓(𝑧)𝑑𝑧=0 Proof idea Subdivide the triangle into four equal-sized triangles.The integral of 𝑓 over ∂𝑇 is the same as the sum of the four integrals over the boundaries of the smaller triangles. Use the 𝑀𝐿-estimate and delicate balancing of boundary length of triangles and the fact that𝑓(𝑧)=𝑓(𝑧0)+(𝑧−𝑧0)𝑓′(𝑧0)+𝜀(𝑧−𝑧0)for 𝑧 near a point 𝑧0) inside 𝑇.

Morera's Theorem

Theorem (Morera)If 𝑓 is continuous on a simply connected domain 𝐷, and if  ∫𝛾𝑓(𝑧)𝑑𝑧=0 for any triangular curve in 𝐷, then 𝑓 has a primitive in 𝐷. Proof idea First, show Morera's theorem in a disk (the proof is not hard and resembles the proof of the real-valued fundamental theorem of calculus).Extending the result to arbitrary simply connected domains is not that easy. This part of the proof requires the use of Cauchy's Theorem for simply connected domains.