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Complex Analysis

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


Analytic Function

Analytic Function with Zero Derivative

By theorem. If 𝑓 is analytic on a domain 𝐷, if 𝑓′(𝑧)=0 for all π‘§βˆˆπ·, then 𝑓 is constant in 𝐷.

Recall the 1-dimensional analog:

By Fact, if 𝑓(π‘Ž,𝑏)→ℝ is differentiable and satisfies 𝑓′(π‘₯)=0 for all π‘₯∈(π‘Ž,𝑏), then 𝑓 is constant on (π‘Ž,𝑏).

Idea of proof

  • First, show that 𝑓 is constant in any disk π΅π‘Ÿ(π‘Ž), 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 π΅π‘Ÿ(π‘Ž) be a disk contained in 𝐷, and let π‘βˆˆπ΅π‘Ÿ(π‘Ž). Pick the point π‘βˆˆπ΅π‘Ÿ(π‘Ž) as in the picture. Since 𝑓′(𝑧)=0 in 𝐷, then 𝑒π‘₯=𝑒𝑦=𝑣π‘₯=𝑣𝑦=0 in 𝐷. In particular, look at 𝑒 on the horizontal segment from π‘Ž to 𝑏. It depends only on one parameter (namely π‘₯) there, and 𝑒π‘₯=0. By the 1-dimensional fact, 𝑒 is constant on the line segment, in particular, 𝑒(π‘Ž)=𝑒(𝑏). Similarly, 𝑒(𝑏)=𝑒(𝑐), thus 𝑒(π‘Ž)=𝑒(𝑐). Since 𝑐 was an arbitrary point in π΅π‘Ÿ(π‘Ž), 𝑒 is thus constant in π΅π‘Ÿ(π‘Ž). Similarly, 𝑣, is constant in π΅π‘Ÿ(π‘Ž), thus 𝑓 is constant in π΅π‘Ÿ(π‘Ž).
  2. Second step. Let π‘Ž and 𝑏 be two arbitrary points in 𝐷. Since 𝐷 is connected, there exist a nice curve in 𝐷, joining π‘Ž and 𝑏. By the previous step, 𝑓 is constant in the disk around the point π‘Ž (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(π‘Ž)=𝑓(𝑏). Thus 𝑓 is constant in 𝐷.


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

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

    Proof: 𝑒 constant in 𝐷 implies that 𝑒π‘₯=𝑒𝑦=0 in 𝐷. Since 𝑓 is analytic, the Cauchy-Riemann equations now imply that 𝑣π‘₯=𝑣𝑦=0 as well. Thus 𝑓′=𝑒π‘₯+𝑖𝑣π‘₯=0 in 𝐷. By our theorem, 𝑓 is constant in 𝐷.

  • Similarly, if 𝑓=𝑒+𝑖𝑣 is analytic in a domain 𝐷 with 𝑣 being constant, then 𝑓 must be constant in 𝐷.
  • Suppose next that 𝑓=𝑒+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=𝑒+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 π‘βˆˆβ„‚ such that

|𝑓(𝑧)|2=𝑒2(𝑧)+𝑣2(𝑧)=𝑐 for all π‘§βˆˆπ·
  • If 𝑐=0 then 𝑒 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 π‘₯ (and similarly with respect to 𝑦) of the above equation yields:

    2𝑒𝑒π‘₯+2𝑣𝑣π‘₯=0 and 2𝑒𝑒𝑦+2𝑣𝑣𝑦=0

    Substituting 𝑣π‘₯=βˆ’π‘’π‘¦ in the first and 𝑣𝑦=𝑒π‘₯ in the second equation gives

    𝑒𝑒π‘₯βˆ’π‘£π‘’π‘¦=0 and 𝑒𝑒𝑦+𝑣𝑒π‘₯=0

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

    𝑒2𝑒π‘₯βˆ’π‘’π‘£π‘’π‘¦=0 and 𝑒𝑣𝑒𝑦+𝑣2𝑒π‘₯=0

    Add the twoe equations


    Since 𝑒2+𝑣2=𝑐, this last equation becomes


    But 𝑐>0, so it must be the case that 𝑒π‘₯=0 in 𝐷. And, 𝑒𝑦=0 in 𝐷 can be found similarly. Using the Cauchy-riemann equations, 𝑒π‘₯=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


𝑓:β„‚β†’β„‚, 𝑓(𝑧)={𝑒-1𝑧4, 𝑧≠00, 𝑧=0
  • One can find 𝑒, 𝑣, 𝑒π‘₯, 𝑒𝑦, 𝑣π‘₯, 𝑣𝑦 and they actually satisfy the Cauchy-Riemann equations in β„‚.
  • Clearly, 𝑓 is analytic in β„‚\{0}.
  • At the origin, one can show that 𝑒π‘₯(0)=𝑒𝑦(0)=𝑒π‘₯(0)=𝑒𝑦(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=π‘₯+𝑖⋅0β†’0. Then

  • Next, consider 𝑧 approaching the origin along the imaginary zxis, i.e. 𝑧=0+𝑖𝑦→0. Then

  • However, consider 𝑧=π‘Ÿπ‘’π‘–πœ‹4β†’0. Then 𝑧4=π‘Ÿ4π‘’π‘–πœ‹4β‹…4=-π‘Ÿ4, so

    𝑓(𝑧)=𝑒-1-π‘Ÿ4=𝑒1π‘Ÿ4β†’βˆžβ‰ π‘“(0)

Although the functions 𝑒 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: 2019/4/12 Revision:


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