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S_David
Mar7-10, 08:31 AM
Hello,

Why when dealing with complex numbers, as with multiplication, we use the complex conjugate operator?

Regards

Studiot
Mar7-10, 09:20 AM
So we can do complex 'division' and fractions.

Compare what happens in the following

\frac{{a + ib}}{{c + id}}*\frac{{c - id}}{{c - id}}

with

\frac{{a + ib}}{{c + id}}*\frac{{c + id}}{{c + id}}

S_David
Mar7-10, 09:33 AM
So we can do complex 'division' and fractions.

Compare what happens in the following

\frac{{a + ib}}{{c + id}}*\frac{{c - id}}{{c - id}}

with

\frac{{a + ib}}{{c + id}}*\frac{{c + id}}{{c + id}}

I didn't get it. I mean, in telecommunication systems, when we deal with baseband signals, we deal with complex numbers, and all the time we use the complex conjugate operator, but I don't understand why and what it is mean physically.

Studiot
Mar7-10, 09:38 AM
I did expect you to work my examples out.

Which one contained the conjugate and which one leads to a single complex number result?

If you apply a formula in real analysis say 27*3 you want the simple answer 81, not something more difficult than you started with such as

{\left( {\sqrt 9 } \right)^2}*{\left( {\sqrt 9 } \right)^2}

The same is true of complex numbers.

What does simple multiplication by a conjugate yield by the way ( a real number)?

HallsofIvy
Mar7-10, 09:42 AM
The product of a complex number and its conjugate has the nice property that is a real number- and for any z other than 0 z*\overline{z} is a positive real number.