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

1 Answer
Jan 8, 2017

Answer:

I multiply by 1 in form of the complex conjugate of the denominator divided by itself, use the F.O.I.L. method to multiply the numerator, and then simplify.

Explanation:

Given: #(-6 - i)/(1 + 5i)#

Multiply by 1 in the form of #(1 - 5i)/(1 - 5i)#:

#(-6 - i)/(1 + 5i)(1 - 5i)/(1 - 5i)#

The denominator becomes the difference of two squares:

#((-6 - i)(1 - 5i))/(1^2 - (5i)^2)#

#((-6 - i)(1 - 5i))/(1 - 25i^2)#

Use the F.O.I.L method to multiply the numerator:

#(-6 + 30i - i + 5i^2)/(1 - 25i^2)#

Substitute -1 for #i^2#

#(-6 + 30i - i - 5)/(1 + 25)#

Combine like terms:

#(-11 + 29i)/26#

Divide each term:

#-11/26 + 29/26i#