How do you simplify #(8+10i)/(8+6i)# and write in a+bi form?

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Answer 1 · 2018-03-12T15:17:42

#(31+8i)/(21)#

Multiply the numerator and denominator by the conjugate of the denominator :

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

#=(64-48i+80i-60i^2)/(64-36i^2)#

#=(64+60(-i^2)−48i+80i)/(64+36(-i^2))#

#(124+32i)/(100)#

simplifiying:

#(31+8i)/(25)#