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

1 Answer
mason m
Dec 31, 2015

#-2/13+3/13i#

Explanation:

Divide each term by #3#.

#=>i/(3-2i)#

Now, to remove the imaginary part from the denominator and write the answer as a complex number in the form #a+bi#, multiply the fraction by the complex conjugate of the denominator.

#=>i/(3-2i)((3+2i)/(3+2i))#

Distribute. Notice that the bottom will form a difference of squares.

#=>(3i+2i^2)/(9-4i^2)#

To continue simplifying, rewrite #i^2# as #-1#. #i^2=-1# since #i=sqrt(-1)#.

#=>(3i+2(-1))/(9-4(-1))=(-2+3i)/13#

Split apart the numerator to write in #a+bi# form.

#=>-2/13+3/13i#

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