How do you simplify # (1+3i) div (5-3i) #?

1 Answer
Nov 22, 2015

Answer:

#-1/9+1/2i#

Explanation:

Write as #(1+3i)/(5-3i)#. Multiply by the complex conjugate of the denominator.

#(1+3i)/(5-3i)((5+3i)/(5+3i))=(5+3i+15i+9i^2)/(25cancel(-15i+15i)-9i^2)=(5+18i+9i^2)/(25-9i^2)#

Recall that #i=sqrt(-1)#, so #color(red)(i^2=-1#.

#(5+18i+9(-1))/(25-9(-1))=(-4+18i)/36=-1/9+1/2i#

Note that the answer is written in the #a+bi# form of a complex number.