How do you divide #(-3+10i)/(-6i)#?

1 Answer
Oct 15, 2016

Answer:

#(-10-3i)/6=-5/3-1/2i#

Explanation:

#frac{-3+10i}{-6i}#

Multiply by the conjugate of the denominator over itself.
A conjugate of a complex number #a+bi# is #a-bi#.
The denominator can be written as #0-6i#, so the conjugate is #0+6i# or #6i#

#frac{(-3+10i)}{-6i}*(6i)/(6i)#

Distribute in the numerator and multiply in the denominator.

#frac{-3*6i +10i*6i}{-36i^2}#

#frac{-18i+60i^2}{-36i^2}#

Recall that #i^2=-1#

#frac{-18i+60(-1)}{-36(-1)}#

#(-18i-60)/36#

Factor out a 6 in both numerator and denominator.

#frac{6(-3i-10)}{6(6)}#

#frac{cancel(6)(-3i-10)}{cancel(6)(6)}#

#(-3i-10)/6# or

#(-10-3i)/6= -10/6-(3i)/6=-5/3-i/2=-5/3-1/2i#