How do you divide #(5-6i)/(-5+10i)#?

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Answer 1 · 2016-09-04T14:21:33

#- (17) / (25) - (4) / (25) i#

We have: #(5 - 6 i ) / (- 5 + 10 i)#

Let's multiply both the numerator and the denominator by the complex conjugate of the denominator:

#= (5 - 6 i ) / (- 5 + 10 i) cdot (- 5 - 10 i) / (- 5 - 10 i)#

#= ((5) (-5) + (5) (-10 i) + (- 6 i) (- 5) + (- 6 i) (- 10 i) ) / ((- 5)^(2) - (10 i)^(2))#

#= (-25 - 50 i + 30 i + 60 i^(2)) / (25 - 100 i^(2))#

Let's apply the fact that #i^(2) = - 1#:

#= (- 25 + (60 cdot - 1) - 20 i) / (25 - (100 cdot - 1))#

#= (- 85 - 20 i) / (125)#

#= (- 17 - 4 i) / (25)#

#= - (17) / (25) - (4) / (25) i#