# A complex equation

I want to solve the following equation:

$g(s)f(s)=0$

where $f$ and $g$ are defined in the complex plane with real values and they are not analytic.

My question is:

If I assume that $f(s)≠0$, can I deduce that $g(s)=0$ without any further complications?

I am a little confused about this case: if $f=x+iy$ and $g=u+iv$, then $fg=ux-vy+i(uy+vx)$ and $fg$ can be zero if $ux-vy=0,uy+vx=0$ without the implications: $x=y=0,u=v=0$

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Are you confusing $f, g$ as functions $f, g: \mathbb{C} \rightarrow \mathbb{R}$ with being constants later $f = x + iy, g = u+iy$? – Calvin Lin Jan 1 '13 at 6:28
You may be interested in the notion of zero divisors: en.wikipedia.org/wiki/Zero_divisors – blah Jan 1 '13 at 10:15
@ blash: Yes, thank you very much. – ZE1 Jan 1 '13 at 10:17

## 3 Answers

The point is that the complex numbers form a field. Doing it with real and imaginary parts is the hard way. Simpler: note that $z^{-1} = \overline{z}/|z|^2$ is the multiplicative inverse for $z \ne 0$, since $z \overline{z} = |z|^2$ (here if $z = x + i y$, $\overline{z} = x - i y$ and $|z| = \sqrt{x^2 + y^2}$). So if $wz = 0$ and $z \ne 0$, multiplying both sides by $z^{-1}$ we get $w = 0$.

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If $g(s) f(s) = 0$, then either $g(s) = 0$ or $f(s) = 0$. If we assume that $f(s) \neq 0$, then we must have $g(s) = 0$.

For the case where $f= x+iy, g= u+iv$ are constants, then $fg = 0$ if and only if $f=0$ and/or $g=0$. Introducing complex coordinates doesn't change this fact. This is because the values must satisfy $ux-vy = 0$ AND $uy + vx = 0$, and so $0 = (ux-vy)^2 + (uy+vx)^2 = (u^2 + v^2)(x^2 + y^2)$, which implies that $u^2 +v^2 = 0$ or $x^2 + y^2 = 0$, which implies that $(u,v)=(0,0)$ or $(x,y) = (0,0)$.

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First, note that your question is really just about individual complex numbers, not about complex-valued functions.

Now, as you note, if $(x + i y)(u + i v) = 0$ then this implies that $x u - v y = x v + u y = 0$. However, the only way these equations can hold is if either $x + i y = 0$ or $u + i v = 0$.

Multiplying the first equation through by $v$ and the second by $u$, we have $x u v - v^2 y = 0$ and $x u v + u^2 y = 0$. Subtracting, $(u^2 + v^2)y = 0$. Since $u, v, y \in \mathbb{R}$, this means that either $y = 0$ or $u = v = 0$. In the latter case, we have $u + i v = 0$. In the former, since $y = 0$ we have $x (u + i v) = 0$, so $x u = x v = 0$. Either $x = 0$, in which case $x + i y = 0$, or $u = v = 0$, in which case $u + i v = 0$ again.

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