Question

Se I é um número complexo e Ī o seu conjugado, então, o número de soluções da equação Ī = I2 é? me ajudeeem por favor!

Answer 1

Olá Adriana, chamarei o número complexo de Z ao invés de I, ok?

$\mathrm{Z=a+bi\ \ \|\ \ \overline{Z}=a-bi}\\\\ \mathrm{\overline{Z}=Z^2\ \to\ a-bi=(a+bi)^2\ \to\ a-bi=a^2+2abi+b^2i^2}\\ \mathrm{a-bi=a^2+2abi+b^2(-1)\ \to\ a-bi=a^2-b^2+2abi}\\\\ \textbf{Comparando as partes reais e imagin\'arias:}\\\\ \mathrm{\Re{}=\Re{}\ \to\ a=a^2-b^2\ \to\ a=(a-b)(a+b)}\\\\ \mathrm{\mathbf{i)}\ a=a-b\ \to\ b=0\ \ \|\ \ \mathbf{ii)}\ a=a+b\ \to\ b=0}\\\\ \mathrm{a=a^2-0^2\ \to\ a^2-a=0\ \to\ a(a-1)=0}\\ \mathrm{a=0\ \ \|\ \ a-1=0\ \to\ a=1}\\\\ \boxed{\mathrm{(a,b)=\{(0,0);(1,0)\}}}$

$\mathrm{\Im{}=\Im{}\ \to\ -b=2ab\ \to\ 2a=-1\ \to\ a=\dfrac{-1}{2}}\\\\ \mathrm{b^2=a^2-a=\bigg(\dfrac{-1}{2}\bigg)^2-\bigg(\dfrac{-1}{2}\bigg)=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1}{4}+\dfrac{2}{4}}\\\\ \mathrm{b^2=\dfrac{3}{4}\ \to\ b=\pm\sqrt{\dfrac{3}{4}}\ \to\ b=\pm\dfrac{\sqrt{3}}{2}}\\\\ \boxed{\mathrm{(a,b)=\bigg\{\bigg(\dfrac{-1}{2},\dfrac{\sqrt{3}}{2}\bigg);\bigg(\dfrac{-1}{2},\dfrac{-\sqrt{3}}{2}\bigg)\bigg\}}}$

$\textbf{Conjunto de pares ordenados}\ \mathbf{(a,b)}\ \textbf{que}\\ \textbf{satisfazem a igualdade:}\\\\ \boxed{\boxed{\mathrm{(a,b)=\bigg\{(0,0);(1,0);\bigg(\dfrac{-1}{2},\dfrac{\sqrt{3}}{2}\bigg);\bigg(\dfrac{-1}{2},\dfrac{-\sqrt{3}}{2}\bigg)\bigg\}}}}$

Resposta: 4 soluções.