How do you use demoivre's theorem to simplify (1−i)12?

Step 1 z=1−i will be in 4th quadrant of argand diagram. Important to note for when we find the argument. r=12+(−1)2=2 θ=2π−tan−1(1)=7π4=−π4 z=r(cos⁡θ+isin⁡θ) zn=rn(cos⁡nθ+isin⁡nθ) Step 2 z12=(2)12(cos⁡(−12π4)+isin⁡(−12π4)) z12=212⋅12(cos⁡(−3π)+isin⁡(−3π)) z12=26(cos⁡(3π)−isin⁡(3π)) cos⁡(3π)=cos⁡(π)=−1 sin⁡(3π)=sin⁡(π)=0 z12=−26=−64

Solved: "How do you use demoivre's theorem to simplify..."

pozicijombx

Answered question

2022-01-31

How do you use demoivre's theorem to simplify

{(1 - i)}^{12}

Answer & Explanation

search633504

Beginner2022-02-01Added 16 answers

Step 1

z = 1 - i

will be in 4th quadrant of argand diagram. Important to note for when we find the argument.

r = \sqrt{1^{2} + {(- 1)}^{2}} = \sqrt{2}

θ = 2 π - \tan^{- 1} (1) = \frac{7 π}{4} = - \frac{π}{4}

z = r (\cos θ + i \sin θ)

z^{n} = r^{n} (\cos n θ + i \sin n θ)

Step 2

z^{12} = {(\sqrt{2})}^{12} (\cos (- 12 \frac{π}{4}) + i \sin (- 12 \frac{π}{4}))

z^{12} = 2^{\frac{1}{2} \cdot 12} (\cos (- 3 π) + i \sin (- 3 π))

z^{12} = 2^{6} (\cos (3 π) - i \sin (3 π))

\cos (3 π) = \cos (π) = - 1

\sin (3 π) = \sin (π) = 0

z^{12} = - 2^{6} = - 64

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