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:

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

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