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`Analytic Function  Analytic Function with Zero Derivative Consequences Analytic Functions with Constant Norm A Strange Example`

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# Analytic Function

## Analytic Function with Zero Derivative

By theorem. If 𝑓 is analytic on a domain 𝐷, if 𝑓′(𝑧)=0 for all 𝑧∈𝐷, then 𝑓 is constant in 𝐷.

Recall the 1-dimensional analog:

By Fact, if 𝑓(𝑎,𝑏)→ℝ is differentiable and satisfies 𝑓′(𝑥)=0 for all 𝑥∈(𝑎,𝑏), then 𝑓 is constant on (𝑎,𝑏).

Idea of proof

• First, show that 𝑓 is constant in any disk 𝐵𝑟(𝑎), contained in 𝐷, using 1-dimensional fact.
• Second, use the fact that 𝐷 is connected and the first fact to show that 𝑓 is constant in 𝐷.

Main steps

1. First step. Let 𝐵𝑟(𝑎) be a disk contained in 𝐷, and let 𝑐∈𝐵𝑟(𝑎). Pick the point 𝑏∈𝐵𝑟(𝑎) as in the picture. Since 𝑓′(𝑧)=0 in 𝐷, then 𝑢𝑥=𝑢𝑦=𝑣𝑥=𝑣𝑦=0 in 𝐷. In particular, look at 𝑢 on the horizontal segment from 𝑎 to 𝑏. It depends only on one parameter (namely 𝑥) there, and 𝑢𝑥=0. By the 1-dimensional fact, 𝑢 is constant on the line segment, in particular, 𝑢(𝑎)=𝑢(𝑏). Similarly, 𝑢(𝑏)=𝑢(𝑐), thus 𝑢(𝑎)=𝑢(𝑐). Since 𝑐 was an arbitrary point in 𝐵𝑟(𝑎), 𝑢 is thus constant in 𝐵𝑟(𝑎). Similarly, 𝑣, is constant in 𝐵𝑟(𝑎), thus 𝑓 is constant in 𝐵𝑟(𝑎).
2. Second step. Let 𝑎 and 𝑏 be two arbitrary points in 𝐷. Since 𝐷 is connected, there exist a nice curve in 𝐷, joining 𝑎 and 𝑏. By the previous step, 𝑓 is constant in the disk around the point 𝑎 (see picture). Furthermore, 𝑓 is also constant in the neighboring disk. Since these two disks overlap, the two constants must agree. continue on in this matter until reaching 𝑏. Therefore, 𝑓(𝑎)=𝑓(𝑏). Thus 𝑓 is constant in 𝐷.

## Consequences

The previous theorem, together with the Cauchy-Riemann equations, has strong consequences.

• Suppose that 𝑓=𝑢+𝑖𝑣 is analytic in a domain 𝐷. Suppose furthermore, that 𝑢 is constant in 𝐷. Then 𝑓 must be constant in 𝐷.

Proof: 𝑢 constant in 𝐷 implies that 𝑢𝑥=𝑢𝑦=0 in 𝐷. Since 𝑓 is analytic, the Cauchy-Riemann equations now imply that 𝑣𝑥=𝑣𝑦=0 as well. Thus 𝑓′=𝑢𝑥+𝑖𝑣𝑥=0 in 𝐷. By our theorem, 𝑓 is constant in 𝐷.

• Similarly, if 𝑓=𝑢+𝑖𝑣 is analytic in a domain 𝐷 with 𝑣 being constant, then 𝑓 must be constant in 𝐷.
• Suppose next that 𝑓=𝑢+𝑖𝑣 is analytic in a domain 𝐷 with |𝑓| being constant in 𝐷. This too implies that 𝑓 itself must be constant.

## Analytic Functions with Constant Norm

Let 𝑓=𝑢+𝑖𝑣 be analytic in a domain 𝐷, and suppose that |𝑓| is constant in 𝐷. Then |𝑓|2 is also constant, i.e. there exists 𝑐∈ℂ such that

`|𝑓(𝑧)|2=𝑢2(𝑧)+𝑣2(𝑧)=𝑐 for all 𝑧∈𝐷`
• If 𝑐=0 then 𝑢 and 𝑣 must be equal to zero everywhere, and so 𝑓 is equal to zero in 𝐷.
• If 𝑐≠0 then in fact 𝑐>0. Taking the partial derivative with respect to 𝑥 (and similarly with respect to 𝑦) of the above equation yields:

`2𝑢𝑢𝑥+2𝑣𝑣𝑥=0 and 2𝑢𝑢𝑦+2𝑣𝑣𝑦=0`

Substituting 𝑣𝑥=−𝑢𝑦 in the first and 𝑣𝑦=𝑢𝑥 in the second equation gives

`𝑢𝑢𝑥−𝑣𝑢𝑦=0 and 𝑢𝑢𝑦+𝑣𝑢𝑥=0`

Multiplying the first equation by 𝑢 and the second by 𝑣, find

`𝑢2𝑢𝑥−𝑢𝑣𝑢𝑦=0 and 𝑢𝑣𝑢𝑦+𝑣2𝑢𝑥=0`

`𝑢2𝑢𝑥+𝑣2𝑢𝑥=0`

Since 𝑢2+𝑣2=𝑐, this last equation becomes

`𝑐𝑢𝑥=0`

But 𝑐>0, so it must be the case that 𝑢𝑥=0 in 𝐷. And, 𝑢𝑦=0 in 𝐷 can be found similarly. Using the Cauchy-riemann equations, 𝑢𝑥=𝑣𝑦=0 in 𝐷 can be obtained. Hence 𝑓′(𝑧)=0 in 𝐷, and the theorem yields that 𝑓 is constant in 𝐷. Note: The assumption of 𝐷 being connected is important.

## A Strange Example

Let

`𝑓:ℂ→ℂ, 𝑓(𝑧)={𝑒-1𝑧4, 𝑧≠00, 𝑧=0`
• One can find 𝑢, 𝑣, 𝑢𝑥, 𝑢𝑦, 𝑣𝑥, 𝑣𝑦 and they actually satisfy the Cauchy-Riemann equations in .
• Clearly, 𝑓 is analytic in ℂ\{0}.
• At the origin, one can show that 𝑢𝑥(0)=𝑢𝑦(0)=𝑢𝑥(0)=𝑢𝑦(0)=0
• However, 𝑓 is not differentiable at 0. How is this possible?
• The function 𝑓 isn't even continuous at the origin.
• Consider 𝑧 approaching the origin along the real axis, i.e. 𝑧=𝑥+𝑖⋅0→0. Then

`𝑓(𝑧)=𝑓(𝑥)=𝑒-1𝑥4→0`
• Next, consider 𝑧 approaching the origin along the imaginary zxis, i.e. 𝑧=0+𝑖𝑦→0. Then

`𝑓(𝑧)=𝑓(𝑖𝑦)=𝑒-1𝑖4𝑦4→0`
• However, consider 𝑧=𝑟𝑒𝑖𝜋4→0. Then 𝑧4=𝑟4𝑒𝑖𝜋4⋅4=-𝑟4, so

`𝑓(𝑧)=𝑒-1-𝑟4=𝑒1𝑟4→∞≠𝑓(0)`

Although the functions 𝑢 and 𝑣 satisfy theCauchy-Riemann equations, yet, 𝑓 is not differentable at the origin. This is because the partial derivatives are not continuous at 0, so the assumptions of the theorem of Cauchy-Riemann equations are not satisified.

ID: 190400012 Last Updated: 12/4/2019 Revision: 0 Home 5

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