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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum