Question

O valor da potência (*raiz de 2 sobre 1+i*) *elevado a 91* é:

a) *raiz de 2* elevado a 91 + i
b) *raiz de 2 sobre 2 (1+i)
c) b1 + i sobre *raiz de 2*
d) (*raiz de dois*) elevado a 91.i
e) - *raiz de 2 sobre 2* (1+i)

Answer 1

Trabalhando na fração:

Vamos multiplicar o numerador e o denominador pelo conjugado do denominador (1 - i):

$\frac{\sqrt{2}}{1+i}=\frac{\sqrt{2}*(1-i)}{(1+i)*(1-i)}$

Sabemos que $(a+b)*(a-b)=a^{2}-b^{2}$

$\frac{\sqrt{2}}{1+i}=\frac{\sqrt{2}*(1-i)}{1^{2}-i^{2}}\\\\\frac{\sqrt{2}}{1+i}=\frac{\sqrt{2}*(1-i)}{1-(-1)}\\\\\frac{\sqrt{2}}{1+i}=\frac{\sqrt{2}*(1-i)}{2}$

$(\frac{\sqrt{2}}{1+i})^{91}=(\frac{\sqrt{2}*(1+i)}{2})^{91}=\frac{\sqrt{2}^{91}*(1-i)^{91}}{2^{91}}\\\\(\frac{\sqrt{2}}{1+i})^{91}=\frac{\sqrt{2}^{90}*\sqrt{2}^{1}*(1-i)^{91}}{2^{91}}\\\\(\frac{\sqrt{2}}{1+i})^{91}=\frac{2^{45}*\sqrt{2}*(1-i)^{91}}{2^{91}}\\\\(\frac{\sqrt{2}}{1+i})^{91}=2^{45-91}*\sqrt{2}*(1-i)^{91}\\(\frac{\sqrt{2}}{1+i})^{91}=2^{-46}*\sqrt{2}*(1-i)^{91}$

Agora, trabalharemos em (1 - i)⁹¹:

$(1-i)^{91}=(1-i)^{90}*(1-i)^{1}\\(1-i)^{91}=([1-i]^{2})^{45}*(1-i)$

$(1 - i)^{2}=1^{2}-2*1*i+i^{2}\\(1-i)^{2}=1-2i-1\\(1-i)^{2}=-2i$

Substituindo (1 - i)² por - 2i:

$(1-i)^{91}=(-2i)^{45}*(1-i)\\(1-i)^{91}=(-2)^{45}*i^{45}*(1-i)$

45 ÷ 4 = 11 + 1 de resto ---> i⁴⁵ = i¹ = i

$(1-i)^{91}=(-2)^{45}*i*(1-i)\\(1-i)^{91}=(-2^{45})*(i-i^{2})\\(1-i)^{91}=(-2^{45})*(i-[-1])\\(1-i)^{91}=(-2^{45})*(i+1)$
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$(\frac{\sqrt{2}}{1+i})^{91}=2^{-46}*\sqrt{2}*(1-i)^{91}\\(\frac{\sqrt{2}}{1+i})^{91}=2^{-46}*\sqrt{2}*(-2)^{45}*(1+i)\\(\frac{\sqrt{2}}{1+i})^{91}=-2^{-46+45}*\sqrt{2}*(1+i)\\(\frac{\sqrt{2}}{1+i})^{91}=-2^{-1}*\sqrt{2}*(1+i)\\(\frac{\sqrt{2}}{1+i})^{91}=-\frac{\sqrt{2}*(1+i)}{2}\\\\\boxed{\boxed{(\frac{\sqrt{2}}{1+i})^{91}=-\frac{\sqrt{2}}{2}*(1+i)}}$

Letra E