Question

o conjugado de 1+(1-i)^16/(1+i)^16 - 2^-8

Answer 1

Usarei os seguintes fatos:

$\bullet\,\,(1-i)^{2}=1^{2}-2\cdot1\cdot i+i^{2}=1-2i+i^{2}=1-2i+(-1)=-2i\\\\\bullet\,\,(1+i)^{2}=1^{2}+2\cdot1\cdot i+i^{2}=1+2i+i^{2}=1+2i+(-1)=2i$
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Estamos interessados em achar a forma algébrica do número complexo dado

$z=\dfrac{1+(1-i)^{16}}{(1+i)^{16}}-2^{-8}\\\\\\z=\dfrac{1}{(1+i)^{16}}+\dfrac{(1-i)^{16}}{(1+i)^{16}}-2^{-8}\\\\\\z=\dfrac{1}{(1+i)^{2\cdot8}}+\dfrac{(1-i)^{2\cdot8}}{(1+i)^{2\cdot8}}-2^{-8}\\\\\\z=\dfrac{1}{[(1+i)^{2}]^{8}}+\bigg[\dfrac{(1-i)^{2}}{(1+i)^{2}}\bigg]^{8}-2^{-8}\\\\\\z=\dfrac{1}{(2i)^{8}}+\bigg[\dfrac{-2i}{\,\,\,2i}\bigg]^{8}-2^{-8}\\\\\\z=\dfrac{1}{2^{8}\cdot i^{8}}+\big[-1\big]^{8}-2^{-8}\\\\\\z=\dfrac{1}{2^{8}\cdot(i^{2})^{4}}+1-2^{-8}\\\\\\z=\dfrac{1}{2^{8}\cdot(-1)^{4}}+1-2^{-8}$

$z=\dfrac{1}{2^{8}\cdot1}+1-2^{-8}\\\\\\z=\dfrac{1}{2^{8}}+1-2^{-8}\\\\\\z=2^{-8}+1-2^{-8}\\\\\\\boxed{\boxed{z=1=\mathbf{1}+\mathbf{0}i}}$

Encontramos a forma algébrica de $z$ ($z$ é um número real, portanto tem parte imaginária nula)

O conjugado de um número complexo com forma algébrica dada por

$z=a+bi$

é definido por $\bar{z}=a-bi$

Portanto, para $z=\mathbf{1}+\mathbf{0}i$, temos o conjugado dado por

$\bar{z}=1-0i\\\\\\\boxed{\boxed{\bar{z}=1}}$