 output.to from Sideway
Complex Analysis

Draft for Information Only

# Content

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

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

# 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`

Add the twoe equations

`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.

©sideway

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

Business

Management

HBR (3)

Information

Recreation

Hobbies (7)

Culture

Chinese (1097)

English (336)

Reference (66)

Computer

Hardware (149)

Software

Application (187)

Digitization (24)

Numeric (19)

Programming

Web (644) CSS (SC)

ASP.NET (SC)

HTML

Knowledge Base

Common Color (SC)

Html 401 Special (SC)

OS (389) MS Windows

Windows10 (SC)

.NET Framework (SC)

DeskTop (7)

Knowledge

Mathematics

Formulas (8)

Number Theory (206)

Algebra (20)

Trigonometry (18)

Geometry (18)

Calculus (67)

Complex Analysis (21)

Engineering

Tables (8)

Mechanical

Mechanics (1)

Rigid Bodies

Statics (92)

Dynamics (37)

Fluid (5)

Control

Acoustics (19)

Biology (1)

Geography (1)

Latest Updated Links

Copyright © 2000-2019 Sideway . All rights reserved Disclaimers last modified on 10 Feb 2019