How do you divide #(-1+3i)/(-4-8i)#?

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Answer 1 · 2016-12-10T07:28:47

The answer is #=-1/4-1/4i#

#z_1/z_2#

You multiply the numerator and denominator by the conjugate of the denominator

#(z_1*barz_2)/(z_2*barz_2)#

If #z=a+ib#

Then, #barz=a-ib#

and #i^2=-1#

#(a+b)(a-b)=a^2-b^2#

So,

#(-1+3i) /(-4-8i) = ((-1+3i)(-4+8i)) / ((-4-8i)(-4+8i)) #

#=(4-8i-12i+24i^2)/(16-64i^2)#

#=(-20-20i)/(80)#

#=-1/4-1/4i#