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Products of Complex Conjugates


Date: 05/21/98 at 04:13:45
From: michael garrett
Subject: complex numbers

How do I prove that the complex conjugate of a product is equal to the 
product of the conjugates?


Date: 05/21/98 at 13:11:09
From: Doctor Bruce
Subject: Re: complex numbers

Hello Michael,

Here's how you can prove this. Let two complex numbers be given, call 
them:

   a + b*i
   c + d*i

where a, b, c, d are real numbers, and i has the property that 
i^2 = -1.

When we multiply these two complex numbers together (for example, 
using the FOIL method from quadratics), we get:

   (ac - bd) + (ad + bc)*i

To find the conjugate of a complex number, we negate the sign of the 
part that has "i" attached -- what we call the imaginary part of the 
complex number. So this would give:

   (ac - bd) - (ad + bc)*i

That's the conjugate of the product.  

Now, let's do the product of the conjugates. Go back to the original
complex numbers and take the conjugate of each one. This gives:

   a - b*i
   c - d*i

Multiplying these together, we get:

   (ac - bd) - (ad + bc)*i

which is the same quantity as before. You should multiply this out for
yourself and be very careful with all the minus signs, to see that you 
get this answer.  

So we have proved that the conjugate of the product is equal to the 
product of the conjugates.

-Doctor Bruce, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Imaginary/Complex Numbers

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