When -3-2i is multiplied by its conjugate, what is the result?

1 Answer
Dec 17, 2016

Answer:

The result is #13#.

Explanation:

A binomial's conjugate is the same expression but with the opposite sign on the 2nd term. For example: #(a+b)# and #(a-b)# are conjugates. This is useful when we have a fraction with a radical (or imaginary) binomial for a denominator, and we wish to "rationalize" it to remove that.

For any complex number #a+bi#, the complex conjugate is #a-bi#. Again, just the same binomial with the imaginary part negated. When we multiply a complex number #a+bi# by its conjugate #a-bi#, the result (by FOIL-ing) is

#(a+bi)(a-bi)=a^2-(ab)i+(ab)i-b^2i^2#
#color(white)((a+bi)(a-bi))=a^2-cancel((ab)i)+cancel((ab)i)-b^2("-"1)#
#color(white)((a+bi)(a-bi))=a^2+b^2#

See what happened? The imaginary part is no longer there. Multiplying a complex number by its conjugate turns it into a completely real number.

In this example, we're asked to multiply #"-"3-2i# by its conjugate (which is #"-"3+2i#). The result is

#("-"3-2i)("-"3+2i)=9-6i+6i-4i^2#
#color(white)(("-"3-2i)("-"3+2i))=9-cancel(6i)+cancel(6i)-4("-"1)#
#color(white)(("-"3-2i)("-"3+2i))=9+4#
#color(white)(("-"3-2i)("-"3+2i))=13#

Bonus:

The quick trick to find the product of a complex number and its conjugate: square each coefficient, and then add the squares. That'll be your result. (That's actually right from the general formula above.)