How do you divide #(3+9i)/(-6-6i)#?

Question

2845 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-25T18:41:03

#-1-i/2#.

Let #z=(3+9i)/(-6-6i)#

#:. z=(3(1+3i))/(-6(1+i))=-1/2((1+3i)/(1+i))#

#:. z=-1/2((1+3i)/(1+i))((1-i)/(1-i))#

#:.z=-1/2((1-i+3i-3i^2)/(1^2-i^2))#

#:.z=-1/2((1+2i-3(-1)))/(1-(-1))#

#:.z=-1/2((4+2i)/(2))#

#:.z=-1-i/2#.

Answer 2 · 2016-07-25T18:58:56

#-1-1/2i#

You can multiply numerator and denominator by the expression

#-6+6i#:

#((3+9i)(-6+6i))/((-6-6i)(-6+6i))=#

#(-18+18i-54i+54i^2)/(36-36i^2)=#

Since #i^2=-1#, you will have:

#(-18-36i-54)/(36+36)=#

#(-72-36i)/72=#

#-1-1/2i#