Complex Analytic Function
Draft for Information Only ContentAnalytic Function β Analytic Function with Zero Derivative βConsequences βAnalytic Functions with Constant Norm βA Strange Example
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Analytic Function
Analytic Function with Zero Derivative
By theorem. If π is analytic on a domain π·, if πβ²(π§)=0 for all π§βπ·, then
π is constant in π·.
Recall the 1dimensional analog:
By Fact, if π(π,π)ββ is differentiable and satisfies πβ²(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 1dimensional fact.
 Second, use the fact that π· is connected and the first fact to show that π is
constant in π·.
Main steps
 First step. Let π΅_{π}(π) be a disk contained in π·, and let πβπ΅_{π}(π).
Pick the point πβπ΅_{π}(π) as in the picture. Since πβ²(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 1dimensional 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).
 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 CauchyRiemann equations, has strong consequences.
 Suppose that π=π’+ππ£ 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 CauchyRiemann equations now imply that π£_{π₯}=π£_{π¦}=0
as well. Thus πβ²=π’_{π₯}+ππ£_{π₯}=0 in π·. By our theorem, π is
constant in π·.

Similarly, if π=π’+ππ£ is analytic in a domain π· with π£ being constant, then π
must be constant in π·.

Suppose next that π=π’+ππ£ 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’+ππ£ be analytic in a domain π·, and suppose that π is constant in π·.
Then π^{2} is also constant, i.e. there exists πββ such that
π(π§)^{2}=π’^{2}(π§)+π£^{2}(π§)=π for all π§βπ·
 If π=0 then π’ and π£ must be equal to zero everywhere, and so π is equal
to zero in π·.

If πβ 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₯}=βπ’_{π¦} in the first and π£_{π¦}=π’_{π₯}
in the second equation gives
π’π’_{π₯}βπ£π’_{π¦}=0 and π’π’_{π¦}+π£π’_{π₯}=0
Multiplying the first equation by π’ and the second by π£, find
π’^{2}π’_{π₯}βπ’π£π’_{π¦}=0 and π’π£π’_{π¦}+π£^{2}π’_{π₯}=0
Add the twoe equations
π’^{2}π’_{π₯}+π£^{2}π’_{π₯}=0
Since π’^{2}+π£^{2}=π, this last equation becomes
ππ’_{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 Cauchyriemann equations, π’_{π₯}=π£_{π¦}=0
in π· can be obtained. Hence πβ²(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:βββ, 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 CauchyRiemann 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₯+πβ
0β0. Then
π(π§)=π(π₯)=π^{1π₯4}β0

Next, consider π§ approaching the origin along the imaginary zxis, i.e. π§=0+ππ¦β0. Then
π(π§)=π(ππ¦)=π^{1π4π¦4}β0

However, consider π§=ππ^{ππ4}β0. Then π§^{4}=π^{4}π^{ππ4β
4}=π^{4}, so
π(π§)=π^{1π4}=π^{1π4}βββ π(0)
Although the functions π’ and π£ satisfy theCauchyRiemann 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 CauchyRiemann equations are not satisified.
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ID: 190400012 Last Updated: 2019/4/12 Revision:

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