# Homework Help: Complex functional equation

1. Nov 10, 2013

### utkarshakash

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 $af(z)+bf(\omega ^2 z)=g(z)$ 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
$af(\omega ^2 z) + bf(\omega z) = g(\omega ^2 z)$

Now if I add both eqns $f(\omega ^2 z)(a+b)+af(z)+bf(\omega z)=g(z)+g(\omega ^2 z)$

2. Nov 10, 2013

### pasmith

Good start.

You have three unknowns $f(z), f(\omega z), f(\omega^2 z)$, so to determine them uniquely you need three equations. You have the original functional equation and the result of substituting $\omega^2 z$ in place of z; can you think of a way to obtain a third equation?

3. Nov 11, 2013

### utkarshakash

I substituted ωz in place of z.

$af(\omega z) + bf(z) = g(\omega z)$

Now If I add all the three eqns I get

$(a+b)(f(z)+f(\omega z)+f(\omega ^2 z))=g(z)+g(\omega z)+g(\omega ^2 z)$

4. Nov 11, 2013

### pasmith

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