Question

URGENTE questão anexada

Answer 1

Explicação passo-a-passo:

$\sf z=\dfrac{a+bi}{1+i}$

$\sf z=\dfrac{a+bi}{1+i}\cdot\dfrac{1-i}{1-i}$

$\sf z=\dfrac{a+bi-ai-bi^2}{1^2-i^2}$

$\sf z=\dfrac{a+bi-ai-b\cdot(-1)}{1-(-1)}$

$\sf z=\dfrac{a+bi-ai+b}{1+1}$

$\sf z=\dfrac{a+b+(b-a)i}{2}$

A parte real é igual ao dobro da parte imaginária

Assim:

$\sf a+b=2\cdot(b-a)$.

$\sf a+b=2b-2a$

$\sf 2b-b=a+2a$

$\sf b=3a$

Substituindo:

$\sf z=\dfrac{a+3a+(3a-a)i}{2}$

$\sf z=\dfrac{4a+2ai}{2}$

$\sf z=2a+ai$

• Módulo

$\sf |z|=\sqrt{(2a)^2+a^2}$

$\sf |z|=\sqrt{4a^2+a^2}$

$\sf |z|=\sqrt{5a^2}$

$\sf |z|=a\sqrt{5}$

$\sf a\sqrt{5}=1$

$\sf a=\dfrac{1}{\sqrt{5}}$

$\sf a=\dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}$

$\sf a=\dfrac{\sqrt{5}}{5}$

Assim:

$\sf b=3a$

$\sf b=3\cdot\dfrac{\sqrt{5}}{5}$

$\sf b=\dfrac{3\sqrt{5}}{5}$

Logo:

$\sf (a\cdot b)^2=\left(\dfrac{\sqrt{5}}{5}\cdot\dfrac{3\sqrt{5}}{5}\right)^2$

$\sf (a\cdot b)^2=\left(\dfrac{3\cdot5}{25}\right)^2$

$\sf (a\cdot b)^2=\left(\dfrac{15}{25}\right)^2$

$\sf (a\cdot b)^2=\left(\dfrac{3}{5}\right)^2$

$\sf (a\cdot b)^2=\dfrac{9}{25}$

Letra D

Answer 2

$\dfrac{a+bi}{1+i}=\dfrac{a+bi}{1+i}\cdot\dfrac{1-i}{1-i}=\dfrac{(a+bi)(1-i)}{1^2-i^2}=\dfrac{a-ai+bi-bi^2}{2}=\dfrac{(a+b)+(-a+b)i}{2}$

Parte Real igual ao dobro da parte imaginaria, portanto:

$\dfrac{a+b}{2}=-a+b\\\\a+b=-2a+2b\\\\3a=b$

Substituindo, temos:

$\dfrac{(a+b)+(-a+b)i}{2}=\dfrac{(a+3a)+(-a+3a)i}{2}=\dfrac{4a+2ai}{2}=2a+ai$

Modulo igual a 1, portanto:

$\sqrt{(2a)^2+a^2}=1\\\\4a^2+a^2=1\\\\5a^2=1\\\\a^2=\dfrac15$

Por ultimo precisamos apenas substituir tudo:

$(ab)^2=(a\cdot3a)^2=(3a^2)^2=\left(\dfrac35\right)^2=\boxed{\dfrac{9}{25}}$