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Complex functional equation

  1. Nov 10, 2013 #1

    utkarshakash

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    Gold Member

    1. The problem statement, all variables and given/known data
    Suppose f(z) is a possibly complex valued function of a complex valued function of a complex number z, which satisfies a functional equation of the form [itex]af(z)+bf(\omega ^2 z)=g(z)[/itex] for all z in C, where a and b are some fixed complex numbers and g(z) is some function of z and ω is cube root of unity (ω≠1), then f(z) can be determined uniquely if
    a)a+b=0
    b)a^2+b^2≠0
    c)a^3+b^3≠0
    d)a^3+b^3=0

    3. The attempt at a solution
    If I substitute ω^2z in place of z the equation reduces to
    [itex]af(\omega ^2 z) + bf(\omega z) = g(\omega ^2 z)[/itex]

    Now if I add both eqns [itex]f(\omega ^2 z)(a+b)+af(z)+bf(\omega z)=g(z)+g(\omega ^2 z) [/itex]
     
  2. jcsd
  3. Nov 10, 2013 #2

    pasmith

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    Homework Helper

    Good start.

    You have three unknowns [itex]f(z), f(\omega z), f(\omega^2 z)[/itex], so to determine them uniquely you need three equations. You have the original functional equation and the result of substituting [itex]\omega^2 z[/itex] in place of z; can you think of a way to obtain a third equation?
     
  4. Nov 11, 2013 #3

    utkarshakash

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    I substituted ωz in place of z.

    [itex]af(\omega z) + bf(z) = g(\omega z)[/itex]

    Now If I add all the three eqns I get

    [itex](a+b)(f(z)+f(\omega z)+f(\omega ^2 z))=g(z)+g(\omega z)+g(\omega ^2 z)[/itex]
     
  5. Nov 11, 2013 #4

    pasmith

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    You have
    [tex]
    af(z) + bf(\omega^2 z) = g(z) \\
    af(\omega z) + bf(z) = g(\omega z) \\
    af(\omega^2 z) + bf(\omega z) = g(\omega^2 z)
    [/tex]
    which is a system of linear simultaneous equations for the unknowns [itex]f(z), f(\omega z), f(\omega^2 z)[/itex]. What condition do [itex]a[/itex] and [itex]b[/itex] have to satisfy for this linear system to have a unique solution?
     
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