Showcasing f(z)=u(x,y)+i*v(x,y) Properties

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Discussion Overview

The discussion focuses on the properties of the complex function f(z) = u(x,y) + i*v(x,y) and aims to demonstrate that under certain conditions, this function can be expressed as f(z) = (e^x) * (cos y + i*sin y). The scope includes theoretical exploration and mathematical reasoning related to complex analysis.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to show that f(z) must take a specific form based on three properties: f(x+0*i)=e^x, f(z) is entire, and f'(z)=f(z) for all z.
  • The same participant mentions a hint from a textbook suggesting to express u(x,y) and v(x,y) in terms of functions g(y) and h(y) that depend on y.
  • Another participant provides a comparison of real and imaginary parts, concluding that g(0)=1 and h(0)=0, which may help in resolving the earlier participant's confusion.
  • The first participant expresses difficulty in reaching the conclusion that g(0) + i*h(0) = 1, indicating a potential misunderstanding or calculation error.

Areas of Agreement / Disagreement

There is no clear consensus in the discussion, as one participant is struggling with a specific calculation while another provides a potential resolution. The discussion remains partially unresolved regarding the final steps to demonstrate the desired properties of f(z).

Contextual Notes

The discussion includes assumptions about the harmonic nature of u(x,y) and the dependence of g(y) and h(y) on real constants A and B, which are not fully explored or resolved.

neginf
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Trying to show any f(z)=u(x,y) + i*v(x,y) with the 3 properties:
1. f(x+0*i)=e^x,
2. f(z) is entire,
3. f ' (z)=f(z) for all z,
has to be f(z)=(e^x) * (cos y + i*siny).

The hint in the book (Complex Variables and Applications, 6 ed., Churchill+Brown) says:
(a) Get u=u sub x and v=v sub y, show there are g(y), h(y) such that u(x,y)=(e^x) * g(y)
and v(x,y)=(e^x) * h(y). did this
(b) Use u(x,y) being harmonic to get g''(y)+g(y)=0 and so g(y)=A*cos y+B*sin y for real
constants A and B. did this
(c) Point out why h(y) = A*sin y-B*cos y and note g(0)+i*h(0)=1, find A and B and conclude
that f(z)=(e^x) * (cos y + i*siny). did the h(y)=-g'(y)=A*sin y-B*cos y
part but cannot get the g(0)+ i*h(0)=1 part. I get g(0)+i*h(0)=A-i*B and get stuck.
 
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[tex]f(z)=u(x,y)+iv(x,y)[/tex]
[tex]=e^x(g(y)+ih(y))[/tex]
[tex]f(x+0i)=e^x(g(0)+ih(0))=e^x[/tex]
Compare real and imaginary parts, g(0)=1, h(0)=0.
 
Thank you very much.
 
That really was an "e^z" question!:biggrin:
 

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