Explanation Calculation: Consider the following equation, x 5 − 243 = 0 Rewrite the above equation, x 5 = 243 The modulus of z is, r = 243 2 + 0 2 = 243 = 243 And, The argument θ is, tan θ = 0 243 = 0 Then, θ = 0 ° The solution is the fifth root of 243 is, z = 243 cos 0 + i sin 0 Now, using n th the root of a complex number z = r cos θ + i sin θ has exactly n distinct n th the root is given by z k = r n cos θ + 2 π k n + i sin θ + 2 π k n where, k = 0 , 1 , 2 , ⋯ n − 1 . Here, n = 5 Therefore, z k = 243 5 cos 0 + 2 π k 5 + i sin 0 + 2 π k 5 …… (1) Where, k = 0 , 1 , 2 , 2 , 4 . For k = 0 , Now, substitute the value of k = 0 in equation (1). Therefore, z 0 = 243 5 cos 2 π 0 5 + i sin 2 π 0 5 = 3 cos 0 + i sin 0 = 3 For k = 1 , Now, substitute the value of k = 1 in equation (1). Therefore, z 1 = 243 5 cos 2 π 1 5 + i sin 2 π 1 5 = 3 cos 2 π 5 + i sin 2 π 5 = 0

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ISBN: 9781337278461

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 4.5, Problem 64E

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To calculate: The solution of the equation x5−243=0 and represent the solution graphically.

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Chapter 4.5, Problem 63E

Chapter 4.5, Problem 65E

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Ch. 4.1 - Perform each operation and write the result in...Ch. 4.1 - Prob. 2ECP Ch. 4.1 - Prob. 3ECP Ch. 4.1 - Prob. 4ECP Ch. 4.1 - Prob. 5ECP Ch. 4.1 - Prob. 6ECP Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E

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The solution of the equation x 5 − 243 = 0 and represent the solution graphically.

Solution Summary: The author explains how to calculate the solution of the equation x5-243=0 and represent it graphically.

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