Question #8380d

1 Answer
Feb 27, 2018

Did you mean to put the negative sign after the divide sign?
If so, #(4-3i+2+6i)/-((3-i)(3-i))=-3/10-3/5i#.

Explanation:

First, simplify the numerator by combining the integer terms and combining the #i# terms: #4-3i+2+6i=3i+6#.

Secondly, the denominator: #-((3-i)(3-i))=-(9-3i-3i+i^2)#. Since #i=sqrt(-1)#, #i^2=(sqrt(-1))^2=-1#.
Plugging that into the denominator and combining like terms, we get #-(9-6i+(-1))=-(8-6i)=-8+6i=6i-8#

Our fraction is now #(3i+6)/(6i-8)#

This may be okay for your teacher, but the standard form for complex numbers is #a+-bi#, so let's get our answer in that format by multiplying the numerator and denominator by the conjugate of the denominator: #((3i+6)(6i+8))/((6i-8)(6i+8))#
We get #(18i^2+24i+36i+48)/(-36-64)=(30+60i)/-100# and put it in the form of #a+-bi# then reduce it:
#30/-100-(60i)/-100=-3/10-3/5i#

Hope that helps! Have fun; imaginary numbers are cool!