Find the exact value of (1/(2^1/2) - 1/(2^1/2)i)^100

Question

What is the exact value of the above equation in polar form? Please show the steps in how to get to the answer. Thanks.

Answer 1

1/(2^1/2) - 1/(2^1/2)i = r(cos(θ) + i sin(θ)) for some r, θ.  [I'm assuming "1/(2^1/2) - 1/(2^1/2)i" means "1/√2 - i 1/√2".]

 

Matching real and imaginary parts, you get equations r cos(θ) = 1/√2  &  r sin(θ) = -1/√2 .

 

We can compute r = √((1/√2)²+(-1/√2)²) = 1, making equations

 

cos(θ) = 1/√2  &  sin(θ) = -1/√2 .

 

Solve them for θ.  On the unit circle chart, θ = 315°.  This gives us the polar form before raising to the 100th power.

 

Using deMoivre's theorem, (1/√2 - i 1/√2)¹⁰⁰ = (cos(θ) + i sin(θ))¹⁰⁰ = cos(100θ) + i sin(100θ).