Good evening all of our class is having a online debate with the following question, any feedback would be appreciated:

In the Real Number realm,

ab = 0 è a = 0 or b = 0

Is the same theorem true in theComplex Numberrealm? (Why or why not?)

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- January 6th 2009, 09:15 PMAlgebraicallyChallengedReal Number
Good evening all of our class is having a online debate with the following question, any feedback would be appreciated:

In the Real Number realm,

**ab = 0 è a = 0 or b = 0**

Is the same theorem true in the**Complex Number**realm? (Why or why not?) - January 6th 2009, 09:24 PMMush
In the complex number 'realm', for a complex number to be considered equal to , then both it's real and imaginary parts must be equal to zero!

Hence and for

So let's look at the case where . For this and , which means that . So the first equation is satisfied. . So the 2nd is satisfied.

Now look at the case where . For this c=0 and d = 0, which means that . So the first equation is satisfied. . So the 2nd is satisfied. - January 6th 2009, 10:11 PMMush
Ahhh I see what you were asking. But the question remains, are there any combinations of and for which their product is zero, but for which neither of them are 0?!

Well. Let's try this then. For a complex number to be zero, then it's modulus must be zero, yes? So let's find the modulus of our product.

Clearly, if this is zero, then the expression inside the square root is zero!

Hence, by the logic of REAL numbers (a, b, c and d must all be real, remember!) either:

OR

For these to be true:

or

and are always positive numbers, which means that the solutions to these 2 equations for all values of b and d are not real solutions, but purely imaginary solutions. And by the definition of the complex numbers and , a, b, c and d must be REAL numbers. Hence there are no two non-zero complex numbers whose product is zero, and hence if then or