How do you simplify #(-1-6i)/(5+9i)#?

1 Answer
Feb 9, 2017

Answer:

# (-1-6i)/(5+9i) = -59/106-21/106i#

Explanation:

In order to rationalise the denominator (i.e. make the denominator real) we multiply by the complex conjugate of the denominator, we then just multiply out and tidy up terms.

The denominator is #5+9i#, so the complex conjugate is #5-9i#

So we have:

# (-1-6i)/(5+9i) = ((-1-6i))/((5+9i)) * ((5-9i))/((5-9i)) #
# " " = (-5+9i-30i+54i^2)/(25-45i+45i-81i^2) #
# " " = (-5-21i-54)/(25+81) \ \ \ \ (because i^2=-1)#
# " " = (-59-21i)/(106) #
# " " = -59/106-21/106i#