Question

dados os números complexos Z1 = 1 + 2i e Z2 = - 2 - i calcule​:

a)z1+z2

b)z1 z2

2
c) Z
1

2
d)z1+ z
2

Answer 1

a)

Para a soma de números complexos, somamos parte real com parte real e parte imaginaria com parte imaginaria.

$Z_1+Z_2~=~(1+2i)+(-2-i)\\\\\\Z_1+Z_2~=~(1+(-2))~+~(2i+(-i))\\\\\\Z_1+Z_2~=~(1-2)~+~(2i-i)\\\\\\\boxed{Z_1+Z_2~=~-1~+~i}$

b)

Para a multiplicação, basta aplicarmos a propriedade distributiva da multiplicação.

$Z_1.Z_2~=~(1+2i)~.~(-2-i)\\\\\\Z_1.Z_2~=~1~.~(-2)~+~1~.~(-i)~+~2i~.~(-2)~+~2i~.~(-i)\\\\\\Z_1.Z_2~=~-2~-~i~-~4i~-~2i^2\\\\\\Lembrando~que~i=\sqrt{-1}~e,~consequentemente~i^2=-1, ~temos:\\\\\\Z_1.Z_2~=~-2~-~5i~-~2~.~(-1)\\\\\\Z_1.Z_2~=~-2-5i+2\\\\\\Z_1.Z_2~=~0-5i\\\\\\\boxed{Z_1.Z_2~=~-5i}$

c)

Aplicando o mesmo raciocínio da questão anterior:

$Z_1^2~=~Z_1~.~Z_1\\\\\\Z_1^2~=~(1+2i)~.~(1+2i)\\\\\\Z_1^2~=~1~.~1~+~1~.~2i~+~2i~.~1~+~2i~.~2i\\\\\\Z_1^2~=~1~+~2i~+~2i~+~2^2i^2\\\\\\Z_1^2~=~1+4i+4~.~(-1)\\\\\\Z_1^2~=~1+4i-4\\\\\\\boxed{Z_1^2~=~-3+4i}$

d)

$Z_1+Z_2^2~=~(1+2i)~+~(-2-i).(-2-i)\\\\\\Z_1+Z_2^2~=~(1+2i)~+~(-2)~.~(-2)~+~-2~.~(-i)~+~(-i)~.~(-2)~+~(-i)~.~(-i)\\\\\\Z_1+Z_2^2~=~(1+2i)~+~4~+~2i~+~2i~+~i^2\\\\\\Z_1+Z_2^2~=~1~+~2i~+4~+~2i~+~2i~-~1\\\\\\\boxed{Z_1+Z_2^2~=~4+6i}$