Question

6 Sabendo que "a"é um numero real e que a parte ima
ginaria do numero complexo (2+i)/(a+2i)é zero,então
a)- 4 b)-2 c)1 d)2 e)4​

Answer 1

$\dfrac{2+i}{a+2i}~=\\\\\\\\Multiplicando~numerador~e~denominador~pelo~conjugado~do~denominador:\\\\\\\\~=\dfrac{2+i}{a+2i}~.~\dfrac{a-2i}{a-2i}\\\\\\\\=~\dfrac{2\,.\,a~+~2\,.\,(-2i)~+~i\,.\,a~+~i\,.\,(-2i)}{a\,.\,a~+~a\,.\,(-2i)~+~2i\,.\,(a)~+~2i\,.\,(-2i)}\\\\\\\\=~\dfrac{2a~-~4i~+~ai~-~2i^2}{a^2~-~2ai~+~2ai~-~4i^2}$

$=~\dfrac{2a~+~(-4+a)i~-~2\,.\,(-1)}{a^2~-~4\,.\,(-1)}\\\\\\\\=~\boxed{\dfrac{(2a+2)~+~(-4+a)i}{a^2+4}}\\\\\\\\Re\left\{\dfrac{(2a+2)~+~(-4+a)i}{a^2+4}\right\}~=~\dfrac{2a+2}{a^2+4}\\\\\\\\Im\left\{\dfrac{(2a+2)~+~(-4+a)i}{a^2+4}\right\}~=~\dfrac{-4+a}{a^2+4}$

O enunciado nos diz que a parte imaginaria do numero (2+i)/(a+2i) vale 0, logo:

$\dfrac{-4+a}{a^2+4}~=~0\\\\\\-4+a~=~0~.~(a^2+4)\\\\\\-4+a~=~0\\\\\\a~=~0+4\\\\\\\boxed{a~=~4}$