Hello, Joe!

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 + biandc + di, and their product is zero.

Then: . . . . (a + bi)(c + di) .= .0

And: .(ac - bd) + (ad + bc)i .= .0

Hence: .ac - bd .= .0 .[1]

and: . . ad + bc .= .0 .[2]

Square [1]: . a²c² - 2abcd + b²d² .= .0

Square [2]: . 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 tworealnumbers equal to zero.

. . Hence, one of the factors must equal zero.

Sincea² + b²is nonnegative,a² + b² = 0only ifa = 0andb = 0.

Similarly, ifc² + d² = 0, thenc = 0andd = 0.

Therefore, one of the complex numbers,a + biorc + di, must be zero.