Question

Solve x^7=1 by using De Moivre's theorem.​

Answer 1

Answer:

For an odd-order polynomial, there must be at least one real root. For (1), a real root is x=−1, and for (2) a real root is x=+1. You may divide out those roots if you wish.

For complex roots, x=|x| (cosθ±i sinθ).

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

z=1andz=cos

7

2kπ

+isin

7

2kπ

,k=1,2,3,4,5,6

z

7

=1

⇒z=1

7

1

=(cos0+isin0)

7

1

⇒z=cos

7

2kπ

+isin

7

2kπ

...{ De Moivre's Theorem}

Where k=0,1,2,3,4,5,6

For k=0

z=1

And z=cos

7

2kπ

+isin

7

2kπ

for k=1,2,3,4,5,6