Prove $zw=0\iff z=0 \lor w=0$ for all complex numbers

Question

Question :

If $z$ and $w$ are complex numbers, then prove :

$$zw=0\iff z=0 \lor w=0$$

My proof :

$p\Rightarrow q$ Suppose $z\ne 0$

$\Rightarrow w=\left(z^{-1}z\right)\cdot w=z^{-1}\cdot (zw)=z^{-1}\cdot 0=0$

$p\Leftarrow q$ Suppose $z\ne 0\land w=0$

$\Rightarrow zw=z\cdot 0=0$

I'm not sure with my proof, cz this doesn't involve the facts that $z$ and $w$ have form ($a+bi$).

Answer 1

Per se, if you already know that the multiplication is associative, that $1\cdot w=w\cdot 1=w$ for all $w$ and that for all $z\ne 0$ there is some $z^{-1}$ such that $z^{-1}z=1$, then you can prove $[\Longrightarrow]$ using $[\Longleftarrow]$ and the aforementioned three formal properties.

I'd rather point out the following facts:

1. You are using the $[\Longleftarrow]$ part of the theorem in the proof of the $[\Longrightarrow]$ part, therefore the former should precede the latter.

2. I don't quite see why in your proof of $[\Longleftarrow]$ you start by assuming that exactly one of the two factors is $0$, since you need the proof to encompass the case when both are $0$ as well. Also, technically you are not proving the assertion, you are just taking it for granted.

Answer 2

Let $zw=0$ and $z\neq0$. Every nonzero complex number has an inverse. Namely, if $z=re^{i\theta}$, then $z^{-1}=1/re^{-i\theta}$. Now $w=z^{-1}zw=z^{-1}0=0$.