Question

Como resolver essa expressão de números complexos?

Answer 1

$x=1-i$


Dessa forma,

$\bullet\;\;x^{-1}=(1-i)^{-1}\\\\ =\dfrac{1}{1-i}\\\\\\ =\dfrac{1\cdot (1+i)}{(1-i)\cdot (1+i)}\\\\\\ =\dfrac{1+i}{1+i-i-i^2}\\\\\\ =\dfrac{1+i}{1-(-1)}\\\\\\ =\dfrac{1+i}{2}\\\\\\ \therefore~~\boxed{\begin{array}{c}x^{-1}=\dfrac{1}{2}+\dfrac{1}{2}\,i \end{array}}$


$\bullet\;\;x^2=(1-i)^2\\\\ =1^2-2i+i^2\\\\ =1-2i+(-1)\\\\\\ \therefore~~\boxed{\begin{array}{c}x^2=-2i \end{array}}$

________________

Portanto,

$E=x^{-1}+x^2\\\\ =\left(\dfrac{1}{2}+\dfrac{1}{2}\,i \right )+(-2i)\\\\\\ =\dfrac{1}{2}+\dfrac{1}{2}\,i-2i\\\\\\ =\dfrac{1}{2}+\left(\dfrac{1}{2}-2 \right )i\\\\\\ =\dfrac{1}{2}+\left(\dfrac{1}{2}-\dfrac{4}{2} \right )i\\\\\\ =\dfrac{1}{2}-\dfrac{3}{2}\,i$


Resposta: alternativa $\text{e) }\dfrac{1}{2}-\dfrac{3}{2}\,i.$


Bons estudos! :-)

Answer 2

Letra  E
A solução está no anexo.