Question

2z+z(conjugado)+6 = -5i...

Answer 1

1)

Temos que descobrir o número complexo z. Para achá-lo, faremos o seguinte:

$z=a+bi$

Sabemos também que o conjugado de z é dado por:

$\overline{z}=a-bi$
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$2z+\overline{z}-6=-5i\\2(a+bi)+(a-bi)=-5i+6\\2a+2bi+a-bi=6-5i\\3a+bi=6-5i$

2 números complexos só são iguais se suas partes reais forem iguais, assim como as imaginárias:

$3a=6\\a=6/3\\a=2\\\\b=-5$

Escrevendo 'z':

$z=a+bi~~~\therefore~~~\boxed{\boxed{z=2-5i}}$
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3)

Para tirarmos i do denominador de uma fração, multiplicamos o numerador e o denominador pelo conjugado do denominador

Trabalhando na primeira fração:

$\dfrac{1+2i}{3i}=\dfrac{(1+2i)(-3i)}{(3i)(-3i)}\\\\\\\dfrac{1+2i}{3i}=\dfrac{-3i-6i^{2}}{-9i^{2}}\\\\\\\dfrac{1+2i}{3i}=\dfrac{-3i+6}{9}\\\\\\\dfrac{1+2i}{3i}=\dfrac{3(-i+2)}{3\cdot3}\\\\\\\dfrac{1+2i}{3i}=\dfrac{2-i}{3}$

Trabalhando na segunda fração:

$\dfrac{1}{1-2i}=\dfrac{1(1+2i)}{(1-2i)(1+2i)}\\\\\\\dfrac{1}{1-2i}=\dfrac{1+2i}{1^{2}-(2i)^{2}}\\\\\\\dfrac{1}{1-2i}=\dfrac{1+2i}{1-4i^{2}}\\\\\\\dfrac{1}{1-2i}=\dfrac{1+2i}{1+4}\\\\\\\dfrac{1}{1-2i}=\dfrac{1+2i}{5}$

Logo:

$\dfrac{1+2i}{3i}+\dfrac{1}{1-2i}=\dfrac{2-i}{3}+\dfrac{1+2i}{5}\\\\\\\dfrac{1+2i}{3i}+\dfrac{1}{1-2i}=\dfrac{5(2-i)+3(1+2i)}{3\cdot5}\\\\\\\dfrac{1+2i}{3i}+\dfrac{1}{1-2i}=\dfrac{10-5i+3+6i}{15}\\\\\\\dfrac{1+2i}{3i}+\dfrac{1}{1-2i}=\dfrac{13+i}{15}\\\\\\\dfrac{1+2i}{3i}+\dfrac{1}{1-2i}=\dfrac{13}{15}+\dfrac{i}{15}\\\\\\\boxed{\boxed{\dfrac{1+2i}{3i}+\dfrac{1}{1-2i}=\dfrac{13}{15}+\dfrac{1}{15}i}}$