Suggestions for a Complex Number Proof I am trying to work on a seemingly forward proof but I am not sure if I am taking the right approach. It is $$\mathrm ZW=0 $$ where Z and W are any complex numbers implies that $\mathrm Z=0$ or $\mathrm W=0 $.
What I have tried,
Let Z=(a+bi) and  let W=(c+di)
Then ZW=(a+bi)(c+di)=(ac+adi+bdi-bd)=0
(ac-bd)+(ad+bd)i=0
Now I know for ZW to be zero, then both the real and imaginary part must be zero ie ac-bd=0 and ad+bc=0. Im not sure how to go about proving anything from just this though? Any hints , answers, help, suggestions?
Thanks all
 A: There are many ways. You reached (with a small typo) the fact that $ac-bd=0$ and $ad+bc=0$. Let us go on from there.
Note the following important identity
$$(ac-bd)^2+(ad+bc)^2=(a^2+b^2)(c^2+d^2).$$
This identity, sometimes called the Brahmagupta-Fibonacci Identity, can be verified by multiplying out both sides. More nicely, it can be obtained  as per the comment of Lubin.
So from your equations we conclude that $(a^2+b^2)(c^2+d^2)=0$. Thus either $a^2+b^2=0$, forcing $a=b=0$, or $c^2+d^2=0$, forcing $c=d=0$.
Remark: Possibly you have already met the notion of norm of a complex number $a+bi$. The Brahmagupta-Fibonacci identity says that the norm of a product of two complex numbers is the product of the norms.
A: Suppose $z\ne 0$ and $w\ne 0$, then there are numbers $z',w'$ such that $zz'=1$ and $ww'=1$
therefore $(zw)(z'w')=1$ and therefore $zw\ne 0$ since $a\cdot 0=0$ for any $a\in\Bbb C$.
A: In the end it always boils down to the fact that for $x,y\in\Bbb R$ the equation $x^2+y^2=0$ implies $x=y=0$. One possible approach is to use linear algebra. The complex numbers for a real vector space with basis formed by $1$ and $\def\ii{\mathbf i}\ii$. The operation of multiplying by $Z=a+b\ii$ is linear, and has matrix
$$
  \begin{pmatrix}a&-b\\b&a\end{pmatrix}
$$
with respect to the mentioned basis. It has determinant $a^2+b^2$, which if $Z\neq0$ is nonzero, so that the linear operator is injective, and the only element in its kernel is $W=0$.
If one does not like the asymmetry in $Z,W$ of this argument, one can multiply the matrices of multiplication by $Z$ and $W$, and use $\det(AB)=\det(A)\det(B)$. This gives the Brahmagupta-Fibonacci identity, and the proof in the answer by André Nicolas.
