Estude! Com Prof. Caju

Calcular a soma:

• [tex3]\Large \frac{\cos\frac{\pi}{4}}{2}\large +\Large \frac{\cos\frac{2\pi}{4}}{2^2}\large + \cdots +\Large \frac{\cos\frac{n\pi}{4}}{2^n}\large[/tex3]

Seja [tex3]z[/tex3] um número complexo, [tex3]z \neq 1[/tex3]. Então,

• [tex3]z + z^{2} + \cdots + z^{n} = \frac{z - z^{n+1}}{1 - z}[/tex3].

Seja [tex3]z = \frac{1}{2}\cos \frac{\pi}{4} + \frac{i}{2}\text{sen}\frac{\pi}{4}[/tex3]. Então,

• [tex3]\frac{\cos \frac{\pi}{4} }{2} + \frac{\cos \frac{2\pi}{4} }{2^{2}} + \frac{\cos \frac{3\pi}{4} }{2^{3}} + \cdots + \frac{\cos \frac{n\pi}{4} }{2^{n}} =\text{Re}(z + z^{2} + \cdots + z^{n}).[/tex3]

Logo, basta calcular [tex3]\text{Re} \left[ \frac{z - z^{n+1}}{1 - z} \right].[/tex3]

Note que,

• [tex3](z - z^{n+1})(1 - \overline{z}) =\left[\frac{1}{2}\cos \frac{\pi}{4} + \frac{i}{2}\text{sen}\frac{\pi}{4} - \frac{1}{2^{n+1}}\cos \frac{\pi}{4}(n+1) - \frac{i}{2^{n+1}} \text{sen}\frac{\pi}{4}(n+1) \right]\left[1 - \frac{1}{2}\cos \frac{\pi}{4} + \frac{i}{2}\text{sen}\frac{\pi}{4}\right].[/tex3]

Assim

• [tex3]\text{Re} [(z - z^{n+1})(1 - \overline{z})] =[/tex3]

• • [tex3]\left[\frac{1}{2}\cos \frac{\pi}{4} - \frac{1}{2^{n+1}}\cos \frac{\pi}{4}(n+1) \right]\left( 1 - \frac{1}{2}\cos \frac{\pi}{4}\right) - \left[\frac{1}{2}\text{sen} \frac{\pi}{4} - \frac{1}{2^{n+1}}\text{sen} \frac{\pi}{4}(n+1) \right]\left(\frac{1}{2}\text{sen} \frac{\pi}{4} \right).[/tex3]

Logo,

• [tex3]\text{Re} [(z - z^{n+1})(1 - \overline{z})] = \left[\frac{\sqrt{2}}{4} - \frac{1}{2^{n+1}}\cos \frac{\pi}{4}(n+1) \right]\left(1 - \frac{\sqrt{2}}{4}\right) - \frac{\sqrt{2}}{4} \left[\frac{\sqrt{2}}{4} - \frac{1}{2^{n+1}}\text{sen} \frac{\pi}{4}(n+1)\right].[/tex3]

Note também que

• [tex3](1 - z)\overline{(1 - z)} = |1 - z|^{2} = \left(1 - \frac{\sqrt{2}}{4} \right)^{2} + \left(\frac{\sqrt{2}}{4} \right)^{2} = \frac{5 - 2\sqrt{2}}{4}.[/tex3]

Como

• [tex3]\text{Re} \left[\frac{z - z^{n+1}}{1 - z} \right] = \frac{\text{Re}[(z - z^{n+1})(1 - \overline{z})] }{|1 - z|^{2}},[/tex3]

temos que

• [tex3]\Large\frac{\cos \frac{\pi}{4} }{2}\large + \Large\frac{\cos \frac{2\pi}{4} }{2^{2}}\large + \Large\frac{\cos \frac{3\pi}{4} }{2^{3}}\large + \cdots + \Large\frac{\cos \frac{n\pi}{4} }{2^{n}} \large=\Large\frac{
  \left(4 - \sqrt{2}\right)\left[\frac{\sqrt{2}}{4} - \frac{1}{2^{n+1}}\cos \frac{\pi}{4}(n+1) \right] - \sqrt{2}\left[\frac{\sqrt{2}}{4} - \frac{1}{2^{n+1}}\text{sen} \frac{\pi}{4}(n+1) \right]}{5 - 2\sqrt{2}}\large.[/tex3]

Até mais!

Caramba, John! Não tinha nem pensado nisso: calcular a parte real do complexo resultante da soma das potências dos outros complexos. Muito bem pensado! Um abraço, e até outros tópicos.