Necessary condition of complex numbers zero product $w,z$ are complex numbers. Prove that if $w\cdot z=0$ then $w=0$ or $z=0$
My proof is:
Suppose that there exist $w\neq0$ and $z\neq0$ such that $wz=0$.
Thus $w=\frac0z=0$. Contradiction.
Is this proof correct?
Can you suggest anything better?
 A: In this entire post, I'm assuming you can use the fact that you can divide by non-zero real numbers.
The fact that you can divide by non-zero things actually depends on the zero-product principle. I would show that $(a + bi)(c + di) = 0, \ a + bi \ne 0 \implies c + di = 0$.
If $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$ is zero, then $ac = bd$ and $ad + bc = 0$. We know that either $a$, $b$, or both is non-zero.
Case $a \ne 0$: Multiply the former by $c$ and the latter by $d$ to get $ac^2 = bcd$ and $ad^2 + bcd = 0$. Substitute to get $a(c^2 + d^2) = 0$, then divide by $a$.
Case $b \ne 0$: Multiply the former by $d$ and the latter by $c$ to get $acd = bd^2$ and $acd + bc^2 = 0$. Substitute to get $b(c^2 + d^2) = 0$, then divide by $b$.
Either way, $c^2 + d^2 = 0$. Since both are real, $c^2, d^2 \ge 0$. So $c = d = 0$.
EDIT: If you need to prove that $a + bi = 0$ implies $a = b = 0$: Assume $a + bi = 0$ and $b \ne 0$. This gives $-a = bi$, and since $b$ is real and non-zero, $-\frac{a}{b} = i$. But because the left side is real and the right side is not, this is absurd, so $b = 0$. Then we have $a + 0i = 0 \implies a = 0$.
A: $wz=0$ iff $|wz|=0$ iff $|w||z| = 0$ iff $|w|=0$ or $|z|=0$.
A: Discovering this 6 years too late, but, nonetheless:
The exercise appears in https://people.math.gatech.edu/~cain/winter99/ch1.pdf , in the context of constructing the complex numbers as ordered pairs of reals, and defining the complex product as $(x, y)(u, v) = (xu - yv, xv + yu)$.  The proof suggested in the question is incorrect, because we don't yet know those assumptions for complex product and division; indeed, that it was we're proving.  Remember, at this point, although the complex product is written similarly to real product, we have no reason to believe they have similar properties.
Rather, the proof is:
$(x, y) * (u, v) = (xu - yv, xv + yu) = (0, 0)$ means $xu = yv$ and $xv = -yu$. If $y$ is zero, either $x$ is zero, which is QED, or $u$ is zero.  If $y$ is zero and $x$ is nonzero, $v$ must also be zero, which again is QED.  If $y$ is nonzero, $u$ must be zero, since otherwise $v = xu/y$, so $x^2u/y = -yu$, and $x^2 = -y^2$, which is impossible since $x$ and $y$ are real.  If $u$ is zero and $y$ nonzero, $v$ must be zero, QED.
Note that the problem could be restated in high school real algebra, "If $xu - yv = 0$ and $xv + yu = 0$, show that $x$ and $y$ are zero or $u$ and $z$ are zero," which is precisely the point: to show that statements about complex numbers are just a convenient shorthand for statements about real numbers.  Also note that the property depends on the fact that $x$, $y$, $u$, and $v$ are real.  If we would try to define a "super-complex" number $(x, y)$ where $x$ and $y$ are complex, we'd have nonzero $w$ and $z$ but $wz = 0$, e.g. $(i,1)$ and $(1,i)$.
