Pre Cal

Simplify the following:
(3+2 i)/(2+3 i)

Multiply numerator and denominator of (3+2 i)/(2+3 i) by 2-3 i:
((3+2 i) (2-3 i))/((2+3 i) (2-3 i))

(2+3 i) (2-3 i) = 2×2+2 (-3 i)+3 i×2+3 i (-3 i) = 4-6 i+6 i+9 = 13:
((3+2 i) (2-3 i))/13

(3+2 i) (2-3 i) = 3×2+3 (-3 i)+2 i×2+2 i (-3 i) = 6-9 i+4 i+6 = 12-5 i:
Answer: |12 -  5i/13

The trick here is to multiply both the numerator and the denominator with what I think is called the complex conjugate of the denominator.

In this case the conjugate is 2 - 3i. Why we do that is because when you have \((a+b)*(a-b)=(a^2-b^2)\)

Then you are left with a real denominator, because \(i^2=-1\)

Here is what I get.

\({{3+2i}\over{2+3i}}*{{2-3i}\over{2-3i}}={{6-9i+4i-6i^2}\over{4-6i+6i-9i^2}}={6-5i+6\over{4+9}}={12-5i\over{13}}\)