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`Cauchy's Theorem and Integral Formula Cauchy's Theorem  Example A First conclusion Examples The Cauchy Integral Formula The Proof of the Cauchy Integral Formula Examples Analyticity of the Derivative The Cauchy Integral Formula for Derivatives  Examples Summary Cauchy's Estimate Liouville's Theorem  Example Use Liouville to Prove Fundamentatl Theorem of Algebra Factoring of Polynomials The Maximum Principle  Example`

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# Cauchy's Theorem and Integral Formula

## Cauchy's Theorem

```Theorem (Cauchy's Theorem for Simply Connected Domains)Let 𝐷 be a simply connected domain in ℂ, and let 𝑓 be analytic in 𝐷. Let 𝛾:[𝑎,𝑏]→𝐷 be a piecewise smooth, closed curve in 𝐷 (i.e. 𝛾(𝑏)=𝛾(𝑎)). Then  ∫𝛾𝑓(𝑧)𝑑𝑧=0```

### Example

𝑓(𝑧)=ℯ(𝑧3) is analytic in ℂ, and ℂ is simply connected. Therefore,  𝛾𝑓(𝑧)𝑑𝑧=0 for any closed, piecewise smooth curve in ℂ.

`Proof ideaSince 𝐷 has no holes, 𝛾 can be deformed continuously to a point in 𝐷. Show that the integral does not change along the way by using the Cauchy Theorem in a disk.`

## A First conclusion

```CorollaryLet 𝛾1 and 𝛾2 be two simple closed curves (i.e. neither of the curves intersects itself), oriented counterclockwise, where 𝛾2 is inside 𝛾1. If 𝑓 is analytic in a domain 𝐷 that contains both curves as well as the region between them, then ∫𝛾1𝑓(𝑧)𝑑𝑧= ∫𝛾2𝑓(𝑧)𝑑𝑧``` `Proof ideaA neat trick: Form a "joint curve" 𝛾 as in the picture below. As 𝑓 is analytic in a simply connected region, containing 𝛾, thus have  ∫𝛾𝑓(𝑧)𝑑𝑧=0`

## Examples

• Let 𝑅 be a rectangle, centered at 𝑧0. Claim:  ∂𝑅1𝑧−𝑧0𝑑𝑧=2𝜋𝑖.`ProofCalculate that  ∫|𝑧−𝑧0|=𝑟1𝑧−𝑧0𝑑𝑧=2𝜋𝑖. Since 𝑓 is analytic "between" the two curves, the integrals must agree.`
•  |𝑧|=11𝑧2+2𝑧𝑑𝑧=? `1𝑧2+2𝑧=1𝑧(𝑧+2)=12(𝑧+2)−𝑧𝑧(𝑧+2)=1212−1𝑧+2` Thus``` ∫|𝑧|=11𝑧2+2𝑧𝑑𝑧=12 ∫|𝑧|=11𝑧𝑑𝑧− ∫|𝑧|=11𝑧+2𝑑𝑧  =12(2𝜋𝑖−0)=𝜋𝑖``` Since the function 𝑧↦1𝑧+2 is analytic in the simply connected domain 𝐵1.5(0), which in turn contains the curve we're integrating over.

## The Cauchy Integral Formula

```Theorem (Cauchy Integral Formula)Let 𝐷 be a simply connected domain, bounded by a piecewise smooth curve 𝛾, and let 𝑓 be analytic in a set 𝑈 that contains the closure of 𝐷 (i.e. 𝐷 and 𝛾). Then 𝑓(𝑤)=12𝜋𝑖 ∫𝛾𝑓(𝑧)𝑧−𝑤𝑑𝑧 for all 𝑤∈𝐷 ```

## The Proof of the Cauchy Integral Formula

The proof of the Cauchy Integral Formula goes as follows:

• Let 𝑤∈𝐷, pick 𝜀>0 such that 𝐵𝜀(𝑤)⊂𝐷
• Using Cauchy's theorem, then`12𝜋𝑖 ∫𝛾𝑓(𝑧)𝑧−𝑤𝑑𝑧=12𝜋𝑖 ∫∂𝐵𝜀(𝑤)𝑓(𝑧)𝑧−𝑤𝑑𝑧`Since the integrand is analytic in a region contraining these two curves and the area in between them.
• It is easily seen that`12𝜋𝑖 ∫∂𝐵𝜀(𝑤)𝑓(𝑧)𝑧−𝑤𝑑𝑧=12𝜋2𝜋∫0𝑓(𝑤+𝜀ℯ𝑖𝑡)𝑑𝑡`
• This is true for any (small) 𝜀>0, and as 𝜀→0, the right-hand side approaches 𝑓(𝑤)

## Examples

Let 𝑓(𝑤)=12𝜋𝑖 𝛾𝑓(𝑧)𝑧−𝑤𝑑𝑧 for all 𝑤∈𝐷

•  |𝑧|=2𝑧2𝑧−1𝑑𝑧=?

Here, 𝑓(𝑧)=𝑧2 and 𝑤=1, and so`12𝜋𝑖 ∫|𝑧|=2𝑓(𝑧)𝑧−1𝑑𝑧=𝑓(1)=1`Hence ` ∫|𝑧|=2𝑧2𝑧−1𝑑𝑧=2𝜋𝑖`

•  |𝑧|=1𝑧2𝑧−2𝑑𝑧=?

First thought:𝑓(𝑧)=𝑧2 and 𝑤=2⋯. However: 𝑤=2 is not inside the curve! But: The function 𝑧↦𝑧2𝑧−2 is analytic in 𝐵1.5(0), and this implies (using Cauchy's theorem) that` ∫|𝑧|=1𝑧2𝑧−2𝑑𝑧=0`

•  |𝑧|=1Log(𝑧+ℯ)𝑧𝑑𝑧=?

Here, 𝑓(𝑧)=Log(𝑧+ℯ), and 𝑤=0. The function 𝑓 is analytic in {Re 𝑧>−ℯ}, which contains the curve integrating over along with its inside. Also, 𝑤 is inside the curve. Thus ` ∫|𝑧|=1Log(𝑧+ℯ)𝑧𝑑𝑧=2𝜋𝑖Log(0+ℯ)=2𝜋𝑖`

## Analyticity of the Derivative

Here is an amazing consequence of the Cauchy Integral Formula:

`TheoremIf 𝑓 is analytic in an open set 𝑈, then 𝑓′ is also analytic in 𝑈.` ```Idea of ProofFirst use the Cauchy Integral Formula to show that for any 𝑤∈𝑈, the derivative 𝑓′(𝑤) can be found via 𝑓′(𝑤)=12𝜋𝑖 ∫𝛾𝑓(𝑧)(𝑧−𝑤)2𝑑𝑧 where 𝛾 is the boundary of a small disk, centered at 𝑤; small enough so that if fits into 𝑈.Next show that the right-hand side defines an analytic function in 𝑤, and therefore 𝑓′ must be analytic.```

## The Cauchy Integral Formula for Derivatives

Repeated application of the previous theorem shows that an analytic function has infinitely many derivatives!

Continuing along the same lines as the previous proof yields the following extension of the Cauchy Integral Formula:

```Theorem (Cauchy Integral Formula for Derivatives)Let 𝐷 be a simply connected domain, bounded by a piecewise smooth curve 𝛾, and let 𝑓 be analytic in a set 𝑈 that contains the closure of 𝐷 (i.e. 𝐷 and 𝛾). Then 𝑓(𝑘)(𝑤)=𝑘!2𝜋𝑖 ∫𝛾𝑓(𝑧)(𝑧−𝑤)𝑘+1𝑑𝑧 for all 𝑤∈𝐷, 𝑘≥0```

where, 𝑓(𝑘) denotes the 𝑘th derivative of 𝑓.

### Examples

Let `𝑓(𝑘)(𝑤)=𝑘!2𝜋𝑖 ∫𝛾𝑓(𝑧)(𝑧−𝑤)𝑘+1𝑑𝑧 for all 𝑤∈𝐷, 𝑘≥0`

•  |𝑧|=2𝜋𝑧2sin 𝑧(𝑧−𝜋)3𝑑𝑧=?

Here, having 𝑓(𝑧)=𝑧2sin 𝑧, 𝑤=𝜋, and 𝑘=2. Thus need to find 𝑓″(𝜋)!```𝑓′(𝑧)=2𝑧sin 𝑧+𝑧2cos 𝑧 𝑓″(𝑧)=2sin 𝑧+2𝑧cos 𝑧+2𝑧cos 𝑧−𝑧2sin 𝑧  =2sin 𝑧+4𝑧cos 𝑧−𝑧2sin 𝑧 so 𝑓″(𝜋)=−4𝜋``` Thus ` ∫|𝑧|=2𝜋𝑧2sin 𝑧(𝑧−𝜋)3𝑑𝑧=2𝜋𝑖2!𝑓(2)(𝜋)=2𝜋𝑖(−4𝜋)2=−4𝜋2𝑖`

•  |𝑧|=2𝑧(𝑧+1)2𝑑𝑧=?

Here, having 𝑓(𝑧)=ℯ𝑧, 𝑤=−1, and 𝑘=1. Thus need to find 𝑓′(-1)! `𝑓′(𝑧)=ℯ𝑧, so 𝑓′(-1)=ℯ−1=1ℯ` Thus ` ∫|𝑧|=2ℯ𝑧(𝑧+1)2𝑑𝑧=2𝜋𝑖1!𝑓′(-1)=2𝜋𝑖ℯ`

## Summary

```Theorem (Cauchy's Theorem for Simply Connected Domains)Let 𝐷 be a simply connected domain in ℂ, and let 𝑓 be analytic in 𝐷. Let 𝛾:[𝑎,𝑏]→𝐷 be a piecewise smooth, closed curve in 𝐷 (i.e. 𝛾(𝑏)=𝛾(𝑎)). Then  ∫𝛾𝑓(𝑧)𝑑𝑧=0``` and ```Theorem (Cauchy Integral Formula for Derivatives)Let 𝐷 be a simply connected domain, bounded by a piecewise smooth curve 𝛾, and let 𝑓 be analytic in a set 𝑈 that contains the closure of 𝐷 (i.e. 𝐷 and 𝛾). Then 𝑓(𝑘)(𝑤)=𝑘!2𝜋𝑖 ∫𝛾𝑓(𝑧)(𝑧−𝑤)𝑘+1𝑑𝑧 for all 𝑤∈𝐷, 𝑘≥0```

## Cauchy's Estimate

`Theorem (Cauchy's Extimate)Suppose that 𝑓 is analytic in an open set that contains 𝐵𝑟(𝑧0), and that |𝑓(𝑧)|≤𝑚 holds on ∂𝐵𝑟(𝑧0) for some constant 𝑚. Then for all 𝑘≥0, |𝑓(𝑘)(𝑧0)|≤𝑘!𝑚𝑟𝑘` ```ProofBy the Cauchy Integral Formula, having that |𝑓(𝑘)(𝑧0)|=𝑘!2𝜋 ∫|𝑧-𝑧0|=𝑟𝑓(𝑧)(𝑧-𝑧0)𝑘+1𝑑𝑧  ≤𝑘!2𝜋 ∫|𝑧-𝑧0|=𝑟|𝑓(𝑧)|(𝑧-𝑧0)𝑘+1𝑑𝑧  ≤𝑘!𝑚2𝜋𝑟𝑘+1⋅2𝜋𝑟=𝑘!𝑚𝑟𝑘+1```

## Liouville's Theorem

`Theorem (Liouville)Let 𝑓 be analytic in the complex plane (thus 𝑓 is an entire function). If 𝑓 is bounded then 𝑓 must be constant.` `ProofSuppose that |𝑓(𝑧)|≤𝑚 for all 𝑧∈ℂ. Pick 𝑧0∈ℂ. Since ℂ contains 𝐵𝑟(𝑧0) for any 𝑟>0, obtaining from Cauchy's estimate:|𝑓′(𝑧0)|≤𝑚𝑟for any 𝑟>0. Letting 𝑟→∞, finding that 𝑓′(𝑧0)=0. Since 𝑧0 is arbitary, 𝑓′(𝑧0)=0 for all 𝑧, hence 𝑓 is constant.`

### Example

`ExampleSuppose that 𝑓 is an entire function, 𝑓=𝑢+𝑖𝑣, and suppose that 𝑢(𝑧)≤0 for all 𝑧∈ℂ. Then 𝑓 must be constant.` ```ProofConsider the function 𝑔(𝑧)=ℯRe 𝑓(𝑧). Then 𝑔 is an entire function as well. Furthermore, |𝑔(𝑧)|=ℯRe 𝑓(𝑧)=ℯ𝑢(𝑧)≤ℯ0=1 Thus 𝑔 is an entire and bounded function, and so Liouville's theorem implies that 𝑔 is constant. This now implies that 𝑓 is constant (look at 𝑔′). ```

## Use Liouville to Prove Fundamentatl Theorem of Algebra

`Theorem (Fundamental Theorem of Algebra)Any polynomial 𝑝(𝑧)=𝑎0+𝑎1𝑧+⋯+𝑎𝑛𝑧𝑛 (with 𝑎0, ⋯,𝑎𝑛∈ℂ, 𝑛≥1 and 𝑎𝑛≠0) has a zero in ℂ, i.e. there exists 𝑧0 such that 𝑝(𝑧0)=0.` ```ProofSuppose to the contrary that there exists a polynomial 𝑝 as in the theorem that has no zeros. Then 𝑓(𝑧)=1𝑝(𝑧) is an entire function! Goal: Apply Liouville's theorem to 𝑓! 𝑝(𝑧)=𝑧𝑛𝑎𝑛+𝑎𝑛-1𝑧+⋯+𝑎0𝑧𝑛 so |𝑝(𝑧)|≥|𝑧|𝑛|𝑎𝑛|−|𝑎𝑛-1||𝑧|−⋯−|𝑎0||𝑧|𝑛|𝑧|→∞→=∞ Thus |𝑓(𝑧)|→0 as |𝑧|→∞, and so 𝑓 is bounded in ℂ. By Liouville, 𝑓 is constant, and so 𝑝 is constant. This is a contradiction.```

## Factoring of Polynomials

Consequence of the Fundamental Theorem of Algebra: Polynomials can be factored in ℂ `𝑝(𝑧)=𝑎𝑛(𝑧−𝑧1)(𝑧−𝑧2)⋯(𝑧−𝑧𝑛)` where 𝑧1, 𝑧2, ⋯, 𝑧𝑛∈ℂ are the zeros of 𝑝 (and not necessarity distinct).

`Expample𝑝(𝑥)=𝑥2+1 has no zeros in ℝ, thus cannot be factored in ℝ. However, in ℂ, 𝑝(𝑧)=𝑧2+1 has two zeros 𝑖 and −𝑖 and thus factors as 𝑝(𝑧)=(𝑧−𝑖)(𝑧+𝑖)`

## The Maximum Principle

Another consequence of the Cauchy Integral Formula is the following powerful result.

`Theorem (Maximum Principle)Let 𝑓 be analytic in a domain 𝐷 and suppose there exists a point 𝑧0∈𝐷 such that |𝑓(𝑧)|≤|𝑓(𝑧0)| for all 𝑧∈𝐷. Then 𝑓 is constant in 𝐷.` `ConsequenceIf 𝐷⊂ℂ is a bounded domain, and if 𝑓:𝐷→ℂ is continuous in 𝐷 and analytic in 𝐷, then |𝑓| reaches its maximum on ∂𝐷.`

### Example

Let 𝑓(𝑧)=𝑧2-2𝑧. What is max|𝑓(𝑧)| on the square 𝑄={𝑧=𝑥+𝑖𝑦:0≤𝑥,𝑦≤1}?

since 𝑓 is analytic inside 𝑄 and continuous on 𝑄, the maximum of |𝑓| occurs on ∂𝑄.

• On 𝛾1:0≤𝑥≤1, 𝑦=0, so `|𝑓(𝑧)|=|𝑓(𝑥)|=|𝑥2−2𝑥|=|𝑥(𝑥−2)|` The maximum on 𝛾1 occurs at 𝑥=1, so |𝑓(𝑧)|≤|𝑓(𝑥)|=1 on 𝛾1
• On 𝛾2:0≤𝑦≤1, 𝑥=1, so `|𝑓(𝑧)|=|𝑓(1+𝑖𝑦)|=|1−𝑦2+2𝑖𝑦−2−2𝑖𝑦|=|−1−𝑦2|=𝑦2+1` The maximum on 𝛾2 occurs at 𝑦=1, so |𝑓(𝑧)|≤|𝑓(1+𝑖)|=2 on 𝛾2
• On 𝛾3, 𝛾4: Similarly, one sees that |𝑓(𝑧)|≤|𝑓(𝑖)|=|−1−2𝑖|=5 on 𝛾3 and 𝛾4
Thus |𝑓(𝑧)|≤|𝑓(𝑖)|=5 on 𝑄.

ID: 190500002 Last Updated: 2/5/2019 Revision: 0 Home 5

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