How do you divide #(-4+3i)/(-10+7i)#?

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Answer 1 · 2016-08-01T09:31:13

#61/149-2/149i#

Since #(a+b)(a-b)=a^2-b^2 and i^2=-1#

you can multiply numerator and denominator of the fraction by #+10+7i# and have:

#((-4+3i)(10+7i))/((-10+7i)(10+7i))=#

#(-40-28i+30i+21i^2)/(-100+49i^2)=#

#(-40+2i-21)/(-100-49)=#

#(-61+2i)/-149=#

#61/149-2/149i# in standard form