I devised a proof . . . but there must be a shorter way.
Prove: if a product of complex numbers is equal to zero,
. . . . . then at least one of the numbers is zero.
We have two complex numbers, a + bi and c + di, and their product is zero.
Then: . . . . (a + bi)(c + di) .= .0
And: .(ac - bd) + (ad + bc)i .= .0
Hence: .ac - bd .= .0 . 
and: . . ad + bc .= .0 . 
Square : . a²c² - 2abcd + b²d² .= .0
Square : . a²d² + 2abcd + b²c² .= .0
Add: . .a²c² + a²d² + b²c² + b²d² .= .0
Factor: . a²(c² + d²) + b²(c² + d²) .= .0
Factor: . . . . . . (a² + b²)(c² + d²) .= .0
We have the product of two real numbers equal to zero.
. . Hence, one of the factors must equal zero.
Since a² + b² is nonnegative, a² + b² = 0 only if a = 0 and b = 0.
Similarly, if c² + d² = 0, then c = 0 and d = 0.
Therefore, one of the complex numbers, a + bi or c + di, must be zero.