How do I use DeMoivre's theorem to solve z3−1=)?

Step 1 z3−1=0 z3=1 We know that any complex number, a+bi can be written in modulus-argument form, r(cos⁡x+isin⁡x), where r=a2+b2 and x satisfies sin⁡x=br and cos⁡x=ar ∴1=1(cos⁡0+isin⁡0) So z3=cos⁡(0+2kπ)+isin⁡(0+2kπ) Since the solutions to trig equations arent

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High School
Calculus and Analysis
Precalculus
Complex numbers

elbluffz1

Answered question

2022-01-30

How do I use DeMoivre's theorem to solve

z^{3} - 1 =)

Answer & Explanation

egowaffle26ic

Beginner2022-01-31Added 7 answers

Step 1
If

z^{3} - 1 = 0

, then we are looking for the cubic roots of unity, i.e. the numbers such that

z^{3} = 1

If youre

Maritza Mccall

Beginner2022-02-01Added 17 answers

Step 1

z^{3} - 1 = 0

z^{3} = 1

We know that any complex number,

a + b i

can be written in modulus-argument form,

r (\cos x + i \sin x)

, where

r = \sqrt{a^{2} + b^{2}}

and x satisfies

\sin x = \frac{b}{r}

and

\cos x = \frac{a}{r}

∴ 1 = 1 (\cos 0 + i \sin 0)

z^{3} = \cos (0 + 2 k π) + i \sin (0 + 2 k π)

Since the solutions to trig equations arent

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