How do you divide #(-3-2i)/(-10-3i)#?

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Answer 1 · 2016-12-02T15:19:46

The answer is #=36/109+11/109i#

We multiply numerator and denominator by the conjugate of the denominator.

If #z=a+ib#

The conjugate is #barz=a-ib#

and #i^2=-1#

So,

#(-3-2i)/(-10-3i)=((-3-2i)(-10+3i))/((-10-3i)(-10+3i))#

#=(30-9i+20i-6i^2)/(100-9i^2)#

#=(36+11i)/(109)#

#=36/109+11/109i#