How do you divide #(-2-10i)/(9-9i)#?

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Answer 1 · 2017-01-12T07:09:26

The answer is #=4/9-2/3i#

If #z=a+ib#

Then, #barz=a-ib#

#(a+b)(a-b)=a^2-b^2#

#i^2=-1#

To simplify the denominator of the quotient of 2 complex numbers, multiply numerator and denominator by the conjugate of the denominator

#(-2-10i)/(9-9i)#

#=((-2-10i)(9+9i))/((9-9i)(9+9i))#

#=(-18-18i-90i-90i^2)/(81-81i^2)#

#=(-18-108i+90)/(81+81)#

#=(72-108i)/(162)#

#=(36-54i)/(81)#

#=(4-6i)/(9)#

#=4/9-2/3i#