Question

Dados os complexos z e w, tais que 2z + w = 2 e
z + w = 1+2i / i, i^2 = – 1, o módulo de w é igual a?

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

$\begin{cases}\mathsf{2z + w = 2}\\\mathsf{z + w = \dfrac{1 + 2i}{i}}\end{cases}$

$\mathsf{w = 2 - 2z}$

$\mathsf{z + 2 - 2z = \dfrac{1 + 2i}{i}}$

$\mathsf{z = 2 - \dfrac{1 + 2i}{i}}$

$\mathsf{z = \dfrac{2i - 1 - 2i}{i}}$

$\boxed{\boxed{\mathsf{z = i}}}$

$\boxed{\boxed{\mathsf{w = 2 - 2i}}}$

$\mathsf{|\:w\:| = \sqrt{a^2 + b^2}}$

$\mathsf{|\:w\:| = \sqrt{2^2 + (-2)^2}}$

$\mathsf{|\:w\:| = \sqrt{4 + 4}}$

$\boxed{\boxed{\mathsf{|\:w\:| = \sqrt{8}}}}$

Answer 2

Resposta:

2z + w = 2

z+w=(1+2i)/i =1/i+2 = -i +2

w=a+bi

2*(2-i) +a+bi =2

4+a=2  ==>a=-2

-2i+2bi=0   ==> -2+2b=0    ==>b=1

w = -2 +i

|w|=√[(-2)²+1²] =√5

|w|= √5