How do you simplify #(-6-8i)/(4+12i) #?

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Answer 1 · 2016-02-12T18:58:38

#(-3+i)/4#

We will multiply the numerator and the denominator by something called a "complex conjugate." It is the opposite of the current denominator.

So, #4-12i# is our conjugate.

This leads to #(-6-8i)/(4+12i)#* #(4-12i)/(4-12i)#

Foil the numerators to get #-24+72i-32i+96i^2#

Foil the denominators to get #16-48i+48i-144i^2#

Remember #i^2=-1#

Now, the numerator becomes #-24+40i+96(-1)#
---> #-24-96+40i#
---> #-120+40i#

The denominator becomes #16-144(-1)#
--->#16+144#
---> #160#

No we have #(-120+40i)/160#

Reducing we get #(-3+i)/4#