Question

prove that (1+8/√2)^⁸+(1-i/√2)^8=2

Answer 1

Answer:

Option 1

Convert to polar form  r(cos θ + i sin θ).

Since cos(π/4) = sin(π/4) = 1/√2, this gives

• (1 + i)/√2  =  cos(π/4) + i sin(π/4)
• (1 - i)/√2  =  cos(π/4) - i sin(π/4)

De Moivre's formula says that (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ).  So...

• [(1 + i)/√2]⁸  =  cos(2π) + i sin(2π)  = 1
• [(1 - i)/√2]⁸  =  cos(2π) - i sin(2π)  = 1

Therefore...

• [(1 + i)/√2]⁸ + [(1 - i)/√2]⁸  =  1 + 1  =  2

Option 2

By the Binomial Theorem,

• (1 + x)⁸ = 1 + 8x + 28x² + 56x³ + 70x⁴ + 56x⁵ + 28x⁶ + 8x⁷ + x⁸
• (1 - x)⁸ = 1 - 8x + 28x² - 56x³ + 70x⁴ - 56x⁵ + 28x⁶ - 8x⁷ + x⁸

so...

• (1 + x)⁸ + (1 - x)⁸  =  2(1 + 28x² + 70x⁴ + 28x⁶ + x⁸).

Then...

   [(1 + i)/√2]⁸ + [(1 - i)/√2]⁸

=  (1/√2)⁸ × [ (1 + i)⁸ + (1 - i)⁸ ]

=  (1/16) × 2(1 + 28i² + 70i⁴ + 28i⁶ + i⁸)

=  (1/8) × (1 - 28 + 70 - 28 + 1)

=  (1/8) × 16

= 2

Answer 2

Answer:

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