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Which of the following is a solution of z^6 = 64i?

2(cos105° + isin105°)
2(cos120° + isin120°)
2(cos135° + isin135°)
8(cos15° + isin15°)

2 Answers

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Answer:

2(cos135° + isin135°)

Explanation:

C

User Opal
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5.8k points
5 votes

Answer:

2(cos135° + isin135°)

Explanation:

z^6 = 64i

We need to change i to polar form

cos (90) + isin (90) = i

x^6 = 64 cos (90) + isin (90)

Now we need to take the sixth root of each side

(x^6) ^ 1/6 = ((64)( cos (90) + isin (90)) ^ 1/6

(x^6) ^ 1/6 = ((64) ^ 1/6 * cos (90) + isin (90)) ^ 1/6

(x^6) ^ 1/6 = 2( * cos (90) + isin (90)) ^ 1/6

We we take the roots of the trig functions, we have 6 roots

360/n means the roots are 60 degrees are apart

take 90 /6 = 15 degrees

The first root is at 2 (cos (15) + isin (15))

The second root is at 2 (cos (15+60) + isin (15+60))

2 (cos (75) + isin (75))

The third root is at 2 (cos (75+60) + isin (75+60))

2 (cos (135) + isin (135))

and so on

User Cardoso
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5.9k points