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`Analytic Functionβ Analytic Function with Zero DerivativeβConsequencesβAnalytic Functions with Constant NormβA Strange Example`

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# Analytic Function

## Analytic Function with Zero Derivative

By theorem. If π is analytic on a domain π·, if πβ²(π§)=0 for all π§βπ·, then π is constant in π·.

Recall the 1-dimensional analog:

By Fact, if π(π,π)ββ is differentiable and satisfies πβ²(π₯)=0 for all π₯β(π,π), then π is constant on (π,π).

Idea of proof

• First, show that π is constant in any disk π΅π(π), contained in π·, using 1-dimensional fact.
• Second, use the fact that π· is connected and the first fact to show that π is constant in π·.

Main steps

1. First step. Let π΅π(π) be a disk contained in π·, and let πβπ΅π(π). Pick the point πβπ΅π(π) as in the picture. Since πβ²(π§)=0 in π·, then π’π₯=π’π¦=π£π₯=π£π¦=0 in π·. In particular, look at π’ on the horizontal segment from π to π. It depends only on one parameter (namely π₯) there, and π’π₯=0. By the 1-dimensional fact, π’ is constant on the line segment, in particular, π’(π)=π’(π). Similarly, π’(π)=π’(π), thus π’(π)=π’(π). Since π was an arbitrary point in π΅π(π), π’ is thus constant in π΅π(π). Similarly, π£, is constant in π΅π(π), thus π is constant in π΅π(π).
2. Second step. Let π and π be two arbitrary points in π·. Since π· is connected, there exist a nice curve in π·, joining π and π. By the previous step, π is constant in the disk around the point π (see picture). Furthermore, π is also constant in the neighboring disk. Since these two disks overlap, the two constants must agree. continue on in this matter until reaching π. Therefore, π(π)=π(π). Thus π is constant in π·.

## Consequences

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

• Suppose that π=π’+ππ£ is analytic in a domain π·. Suppose furthermore, that π’ is constant in π·. Then π must be constant in π·.

Proof: π’ constant in π· implies that π’π₯=π’π¦=0 in π·. Since π is analytic, the Cauchy-Riemann equations now imply that π£π₯=π£π¦=0 as well. Thus πβ²=π’π₯+ππ£π₯=0 in π·. By our theorem, π is constant in π·.

• Similarly, if π=π’+ππ£ is analytic in a domain π· with π£ being constant, then π must be constant in π·.
• Suppose next that π=π’+ππ£ is analytic in a domain π· with |π| being constant in π·. This too implies that π itself must be constant.

## Analytic Functions with Constant Norm

Let π=π’+ππ£ be analytic in a domain π·, and suppose that |π| is constant in π·. Then |π|2 is also constant, i.e. there exists πββ such that

`|π(π§)|2=π’2(π§)+π£2(π§)=π for all π§βπ·`
• If π=0 then π’ and π£ must be equal to zero everywhere, and so π is equal to zero in π·.
• If πβ 0 then in fact π>0. Taking the partial derivative with respect to π₯ (and similarly with respect to π¦) of the above equation yields:

`2π’π’π₯+2π£π£π₯=0 and 2π’π’π¦+2π£π£π¦=0`

Substituting π£π₯=βπ’π¦ in the first and π£π¦=π’π₯ in the second equation gives

`π’π’π₯βπ£π’π¦=0 and π’π’π¦+π£π’π₯=0`

Multiplying the first equation by π’ and the second by π£, find

`π’2π’π₯βπ’π£π’π¦=0 and π’π£π’π¦+π£2π’π₯=0`

`π’2π’π₯+π£2π’π₯=0`

Since π’2+π£2=π, this last equation becomes

`ππ’π₯=0`

But π>0, so it must be the case that π’π₯=0 in π·. And, π’π¦=0 in π· can be found similarly. Using the Cauchy-riemann equations, π’π₯=π£π¦=0 in π· can be obtained. Hence πβ²(π§)=0 in π·, and the theorem yields that π is constant in π·. Note: The assumption of π· being connected is important.

## A Strange Example

Let

`π:βββ, π(π§)={π-1π§4, π§β 00, π§=0`
• One can find π’, π£, π’π₯, π’π¦, π£π₯, π£π¦ and they actually satisfy the Cauchy-Riemann equations in β.
• Clearly, π is analytic in β\{0}.
• At the origin, one can show that π’π₯(0)=π’π¦(0)=π’π₯(0)=π’π¦(0)=0
• However, π is not differentiable at 0. How is this possible?
• The function π isn't even continuous at the origin.
• Consider π§ approaching the origin along the real axis, i.e. π§=π₯+πβ0β0. Then

`π(π§)=π(π₯)=π-1π₯4β0`
• Next, consider π§ approaching the origin along the imaginary zxis, i.e. π§=0+ππ¦β0. Then

`π(π§)=π(ππ¦)=π-1π4π¦4β0`
• However, consider π§=ππππ4β0. Then π§4=π4πππ4β4=-π4, so

`π(π§)=π-1-π4=π1π4βββ π(0)`

Although the functions π’ and π£ satisfy theCauchy-Riemann equations, yet, π 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: 2019/4/12 Revision:

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