How do you evaluate #-2i ( - 2+ 3i )#?

1 Answer
Fleur
Aug 2, 2017

#6+4i#

Explanation:

To evaluate this, distribute #-2i# through the expression #-2+3i#.

#-2i(-2+3i)#

#(-2i)(-2)+(-2i)(3i)#

#4i+(-6i^2)#

#4i-6i^2#

Since we know that #i=sqrt(-1)#, #i^2=(sqrt(-1))^2 = -1#.

#4i-6(-1)#

#4i+6#

Rewriting this in #a+bi# form, we get #6+4i# as the answer.

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