Question

(2+2i).(1-5i)+(1+i)²

Answer 1

Resposta:  $\large 12 - 6i$

                             $\Large Multiplicac_{\!\!,}\tilde ao ~N\acute{u}meros ~ Complexos$

Para:

Primeira parte,  aplicar a propriedade distributiva.

$\large (2+2i).(1-5i)$

Segunda parte:

Quadrado da soma (a + b)² = a² + 2ab + b²

$\large (1+i)^2 \\\\\large 1^2 + 2 ~. ~1i + i^2 \\\\\large \Big/ \mkern -15mu1 + 2i ~+ ~~ \Big/ \mkern -25mu( -1) \\\\\large \Rightarrow 2i$

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Regra para a multiplicação  ( distributiva ).

$\large \left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i\\ ~$

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$\large (2+2i).(1-5i)+(1+i)^2 \\\\\\\large \left(2\cdot \:1-2\left(-5\right)\right)+\left(2\left(-5\right)+2\cdot \:1\right)i + (2i) \\\\\\\large (2 + 10) + (-10 + 2)i + (2i) \\\\\\\large (12) -8i + (2i) \\\\\\ \boxed{ ~~\large 12 - 6i ~~}$

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