How do you simplify #(3+i)/(3-i)# and write the complex number in standard form?

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Answer 1 · 2017-01-05T18:54:26

#4/5 +i 3/5#

Multiply the expression in the numerator and the denominator by the conjugate of the number in the denominator, to make the denominator a real number

#(3+i)/(3-i) * (3+i)/(3+i)#

=#(9+6i+i^2)/(9-i^2)#

=#(9+6i-1)/(9+1)#

= #(8+6i)/10#

=#8/10 +(6i)/10#

=#4/5 +i 3/5#