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How do you simplify #(1-2i)/(3-4i)#?

1 Answer

Mar 9, 2018

#(-2i+11)/25#

Explanation:

Multiply by the conjugate of #(3+4i)/(3+4i)# to cancel out the imaginary numbers in the denominator. Also remember that #i^2# is equal to -1.

#=(1-2i)/(3-4i) * (3+4i)/(3+4i)#
#=(3+4i-6i-8i^2)/(9+12i-12i-16i^2)#
#=(-2i+3-8(-1))/(9-16(-1))#
#=(-2i+11)/25#

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