• Matéria: Matemática
  • Autor: sgtamin1
  • Perguntado 3 anos atrás

O determinante a seguir define um número complexo. Qual é o seu módulo?​

Anexos:

Respostas

respondido por: CyberKirito
2

\large\boxed{\begin{array}{l}\rm O~determinante\begin{vmatrix}\sf1&\sf i&\sf1\\\sf i&\sf1&\sf i\\\sf 1+i&\sf 1-i&\sf0\end{vmatrix}~d~\!\!efine~um~n\acute umero~complexo.\\\\\rm Seu~m\acute odulo~\acute e:\\\tt I~~~~~~\rm\sqrt{6}\\\tt II~~~~~\rm2\sqrt{2}\\\tt III~~~~\rm\sqrt{2}\\\tt IV~~~~~\rm2\sqrt{5}\end{array}}\large\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao:}\\\\\begin{vmatrix}\sf1&\sf i&\sf1\\\sf i&\sf1&\sf i\\\sf 1+i+&\sf1-i&\sf0\end{vmatrix}\\\sf=1\cdot(0-i+i^2)-i\cdot(0-i-i^2)+1\cdot(\diagup\!\!\!i-i^2-1-\diagup\!\!\!i)\\\sf=-i+i^2+i^2+i^3-i^2-1\\\sf=-i-1-1-i+\backslash\!\!\!1-\backslash\!\!\!1\\\sf=-2i-2\end{array}}

\Large\boxed{\begin{array}{l}\sf z=-2i-2\\\sf \rho=\sqrt{(-2)^2+(-2)^2}\\\sf\rho=\sqrt{4+4}\\\sf \rho=\sqrt{2\cdot4}\\\sf\rho=2\sqrt{2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~opc_{\!\!,}\tilde ao~II}}}}\end{array}}


sgtamin1: Muito obrigado moço
CyberKirito: De nada :)
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