Question

Por favor ajuda na questão 2!!!

Answer 1

Resposta:

n=2

Explicação passo-a-passo:

$seja: z_{1} =a+ib \\a=1 , b=1\\\lambda=\sqrt{2}$

$cos(\eta)= \frac{a}{\lambda} =\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2} \\sin(\eta)=\frac{b}{\lambda} =\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2}$

$onde: \\\eta=\frac{\pi }{4} \\assim:$

$z_{1} =cos(\frac{\pi }{4} )+i.sin(\frac{\pi }{4} )\\z_{2} =cos(\frac{n.\pi }{8} )+i.sin(\frac{n.\pi }{8})\\assim:\\z_{1} .z_{2} =[cos(\frac{\pi }{4} )+i.sin(\frac{\pi }{4} )].[cos(\frac{n.\pi }{8} )+i.sin(\frac{n.\pi }{8})]=[cos(\frac{\pi }{4} )cos(\frac{n\pi }{8} )-sin(\frac{\pi }{4} )sin(\frac{n\pi }{8} )]+i.[sin(\frac{n\pi }{8} ).cos(\frac{\pi }{4} )+sin(\frac{\pi }{4} ).cos(\frac{n\pi }{8})]$para que tal número seja imaginário puro essa primeira parte deve ser zero.

daí:

$cos(\frac{n\pi}{8} )=sin(\frac{n\pi}{8} )\\cos^{2} x+sin^{2} x=1\\2cos^{2}[\frac{n\pi}{8}]=1\\cos^{2} [\frac{n\pi}{8} ]=\frac{1}{2} \\cos(\frac{n\pi}{8} )=\frac{\sqrt{2} }{2} =sin(\frac{n\pi}{8} )\\assim:\\\frac{n\pi}{8} =\frac{\pi }{4} \\n=2.$