De moirve's (√3-i ÷ √3+i)^6 = 1

1 answer:

7 0

(√3 - i ) / (√3 + i ) × (√3 - i ) / (√3 - i ) = (√3 - i )² / ((√3)² - i ²)

… = ((√3)² - 2√3 i + i ²) / (3 - i ²)

… = (3 - 2√3 i - 1) / (3 - (-1))

… = (2 - 2√3 i ) / 4

… = 1/2 - √3/2 i

… = √((1/2)² + (-√3/2)²) exp(i arctan((-√3/2)/(1/2))

… = exp(i arctan(-√3))

… = exp(-i arctan(√3))

… = exp(-iπ/3)

By DeMoivre's theorem,

[(√3 - i ) / (√3 + i )]⁶ = exp(-6iπ/3) = exp(-2iπ) = 1

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Let's go through the steps of factoring that Venita should take.

1.) Find the greatest common factor (GCF). We only have two terms, so that makes it pretty easy.
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The greatest common factor of 32 and 8 is 8. We can also factor out a b since that term appears in each part of the original expression. The GCF and variable should go on the outside of the parentheses.
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2.) Now let's figure out what should go in the middle of the parentheses. To do this, use the original expression and divide each term. This is written in the parentheses.

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This would then result in the factored expression 8b(4a - 1). You can always check this by using the distributive property. Distribute 8b out to both expressions:
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Now that we went through the steps to solve the factored expression, let's check her answer. The only difference between Venita's and ours is that she has 0 as the second term while we have a 1. It seems that she had subtracted the GCF from the second term instead of dividing.

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