How do you simplify #(-5+6i)/(5+4i)#?

1 Answer
Tony B
Jan 13, 2016

#(-1+50i)/41#

Explanation:

The automatic response is to try change the denominator such that it is no longer of type Complex. We do this by using what is called its 'Conjugate'. That is: #5-4i#.

To do this we multiply the whole by the value of 1 but in the form of

#(5-4i)/(5-4i)# so that we do not effectively change the value of the whole.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#(-5+6i)/(5+4i) xx (5-4i)/(5-4i)#

#(-25+20i+30i-24i^2)/(25-20i+20i-16i^2)#

But #i^2=-1# giving:

#(-1+50i)/41#

Related questions
See all questions in Division of Complex Numbers
Impact of this question
4559 views around the world
You can reuse this answer
Creative Commons License