How do you divide #(4-2i)/(3+7i)#?

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Answer 1 · 2016-06-18T06:08:56

#-1/29-17/29*i#

you can multiply numerator and denominator by the expression

#3-7i#

#((4-2i)(3-7i))/((3+7i)(3-7i))#

and, by symplifying:

#(12-28i-6i+14i^2)/(9-49i^2)#

let's substitute #i^2=-1# and sum similar terms:

#(12-34i-14)/(9+49)#

#(-2-34i)/58#

or

#-2/58-(34/58)i#

that's

#-1/29-17/29*i#