Resolução de uma pergunta de Números Complexos
19 jun 2015, 15:17
Estou com dificuldades em resolver o problema que envio.
- Anexos
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Re: Resolução de uma pergunta de Números Complexos
19 jun 2015, 16:37
|z|= r
\(\large z=r\, cis\left ( \frac{\Pi }{6} \right )=r\, \left ( \cos \frac{\Pi }{6}+i\, \sin \frac{\Pi }{6} \right )=r\, \left ( \frac{\sqrt{3}}{2}+\frac{1}{2}\, i \right )=\frac{\sqrt{3}}{2}\, r+\frac{1}{2}\, r\, i\)
\(\large 2\times \frac{\sqrt{3}}{2}\, r+2\times \frac{1}{2}\, r=16\, \Leftrightarrow \, r\left ( \sqrt{3}+1 \right )=16\, \Leftrightarrow\, r=\frac{16}{\sqrt{3}+1}\, \Leftrightarrow r=8\sqrt{3}-8\)
\(\large z=\left ( 12-4\sqrt{3} \right )+\left ( -4+4\sqrt{3} \right )i\)
\(\large Re\left ( z \right )+Im\left ( z \right )+|z|=\left ( 12-4\sqrt{3} \right )+\left ( -4+4\sqrt{3} \right )+\left ( 8\sqrt{3}-8 \right )=8\sqrt{3}\)
\(\large z=r\, cis\left ( \frac{\Pi }{6} \right )=r\, \left ( \cos \frac{\Pi }{6}+i\, \sin \frac{\Pi }{6} \right )=r\, \left ( \frac{\sqrt{3}}{2}+\frac{1}{2}\, i \right )=\frac{\sqrt{3}}{2}\, r+\frac{1}{2}\, r\, i\)
- complexos.jpg (11.14 KiB) Visualizado 1962 vezes
\(\large 2\times \frac{\sqrt{3}}{2}\, r+2\times \frac{1}{2}\, r=16\, \Leftrightarrow \, r\left ( \sqrt{3}+1 \right )=16\, \Leftrightarrow\, r=\frac{16}{\sqrt{3}+1}\, \Leftrightarrow r=8\sqrt{3}-8\)
\(\large z=\left ( 12-4\sqrt{3} \right )+\left ( -4+4\sqrt{3} \right )i\)
\(\large Re\left ( z \right )+Im\left ( z \right )+|z|=\left ( 12-4\sqrt{3} \right )+\left ( -4+4\sqrt{3} \right )+\left ( 8\sqrt{3}-8 \right )=8\sqrt{3}\)