How do you use demoivre's theorem to simplify 4(1−3i)3?

DeMoivre's theorem for exponents says that any complex number z can be written as r(cos⁡θ+isin⁡θ), or rcis(θ) for short.It continues to say that when raising an imaginary number to a certain power, the result is:zn=rncis(nθ)To simplify this, let's first calculate (1−i3)3.We find r by using Pythagoras' theorem:r=12+32=4=2We find θ by taking the inverse tangent of Im(z)Re(z)θ=tan−1(−31)=−π3So (1−i3)=2cis(−π3). Therefore,(1−i3)3=23cis(3(−π3))=8cis(π)Finally, we expand cis(π)8cis(π)=8cos⁡π+8isin⁡π=8(−1)+8i(0)=−8And don't forget to multiply by 4!−8⋅4=−32So 4(1−i3)3=−32

"How do you use demoivre's theorem to simplify..." was answered by our community

Nylah Church

Answered question

2022-01-31

How do you use demoivre's theorem to simplify

4 {(1 - \sqrt{3} i)}^{3}

Answer & Explanation

cevarnamvu

Beginner2022-02-01Added 12 answers

DeMoivre's theorem for exponents says that any complex number z can be written as

r (\cos θ + i \sin θ), or r c i s (θ)

for short.
It continues to say that when raising an imaginary number to a certain power, the result is:

z^{n} = r^{n} c i s (n θ)

To simplify this, let's first calculate

{(1 - i \sqrt{3})}^{3}

.
We find r by using Pythagoras' theorem:

r = \sqrt{1^{2} + {\sqrt{3}}^{2}} = \sqrt{4} = 2

We find

θ

by taking the inverse tangent of

\frac{I m (z)}{R e (z)}

θ = \tan^{- 1} (\frac{- \sqrt{3}}{1}) = - \frac{π}{3}

(1 - i \sqrt{3}) = 2 c i s (- \frac{π}{3})

. Therefore,

{(1 - i \sqrt{3})}^{3} = 2^{3} c i s (3 (- \frac{π}{3})) = 8 c i s (π)

Finally, we expand

c i s (π)

8 c i s (π) = 8 \cos π + 8 i \sin π = 8 (- 1) + 8 i (0) = - 8

And don't forget to multiply by 4!

- 8 \cdot 4 = - 32

4 {(1 - i \sqrt{3})}^{3} = - 32

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Precalculus

Didn't find what you were looking for?