# Complex Numbers

• Sep 14th 2006, 09:04 PM
JoeDardeno23
Complex Numbers
How would you go about proving that if a product of complex numbers is equal to zero then at least one of the numbers is zero? Seems simple, but I don't know where to start.

Greatly appreciate any help. Thanks in advance.
• Sep 14th 2006, 09:59 PM
Soroban
Hello, Joe!

I devised a proof . . . but there must be a shorter way.

Quote:

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 . [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 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.