What is the Condition for a Unique Solution in a Complex Functional Equation?

[utkarshakash]

Homework Statement

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

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]

 

[pasmith]

Homework Statement

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

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]

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?

 

[utkarshakash]

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?

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]

 

[pasmith]

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]

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?