How do you divide # (6-4i) / (9-4i) #?

1 Answer
Apr 10, 2018

Answer:

See answer below.

Explanation:

Always get rid of the complex numbers on the bottom. This can be done by multiplying the denominator and the numerator by the complex conjugate of the denominator.

Allow me to demonstrate:

#{6-4i}/{9-4i}={6-4i}/{9-4i} times {9+4i}/{9+4i}#

Doing so gets rid of the imaginary part of the denominator.

#{6-4i}/{9-4i} times {9+4i}/{9+4i}= {(6-4i)(9+4i)}/{(9)^2-(4i)^2} #

#{(6-4i)(9+4i)}/{(9)^2-(4i)^2} = {(6-4i)(9+4i)}/{81-16( -1)}= {(6-4i)(9+4i)}/{97}#

#{(6-4i)(9+4i)}/{97} = {54 -36i +24i + 16}/{97}={70 -12i}/{97}#

#{70 -12i}/{97}=70/97 -12/97i#