Ensino Médio ⇒ Complexos

[poti]

Prove que [tex3]i^i = e^{\frac{-\pi}{2}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]i^i = e^{\frac{-\pi}{2}}[/tex3]

[F u r u y á]

Solução

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Partindo da fórmula de Euler, temos:

• [tex3]\begin{matrix}{rcl}e^{i\theta} & = & \cos{\theta}\ +\ i\,\sin{\theta} \\
  e^{i\frac{\pi}{2}} & = & \cos{\frac{\pi}{2}}\ +\ i \,\sin{\frac{\pi}{2}} \\
  e^{i\frac{\pi}{2}} & = & 0 \ + \ i \\
  i & = & e^{i\frac{\pi}{2}} \\
  \big( \,i \,\big)^{i} & = & \big( e^{i\frac{\pi}{2}} \big) ^ {i} \\
  i^{i} & = & e^{\left( i^{2} \,\frac{\pi}{2} \right)} \\
  i^{i} & = & e^{-\frac{\pi}{2}}
  \end{matrix}[/tex3]

  Código: Selecionar tudoSHIFT+Click na eq = ZOOM

  [tex3]\begin{matrix}{rcl}e^{i\theta} & = & \cos{\theta}\ +\ i\,\sin{\theta} \\
e^{i\frac{\pi}{2}} & = & \cos{\frac{\pi}{2}}\ +\ i \,\sin{\frac{\pi}{2}} \\
e^{i\frac{\pi}{2}} & = & 0 \ + \ i \\
i & = & e^{i\frac{\pi}{2}} \\
\big( \,i \,\big)^{i} & = & \big( e^{i\frac{\pi}{2}} \big) ^ {i} \\
i^{i} & = & e^{\left( i^{2} \,\frac{\pi}{2} \right)} \\
i^{i} & = & e^{-\frac{\pi}{2}}
\end{matrix}[/tex3]

[poti]

Muito obrigado.