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.

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- September 14th 2006, 09:04 PMJoeDardeno23Complex 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. - September 14th 2006, 09:59 PMSoroban
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.