Question

The complex number 16 has the following roots calculated. 1st root (k = 0): 2 left-bracket cosine (StartFraction 0 pi Over 4 EndFraction) + I sine (StartFraction 0 pi Over 4 EndFraction right-bracket 2nd root (k = 1): 2 left-bracket cosine (StartFraction 2 pi Over 4 EndFraction) + I sine (StartFraction 2 pi Over 4 EndFraction right-bracket 3rd root (k = 2): 2 left-bracket cosine (StartFraction 4 pi Over 4 EndFraction) + I sine (StartFraction 4 pi Over 4 EndFraction right-bracket What is the 4th root (k = 3) of 16?

Answer 1

The 4th root (k = 3) of 16 is calculated using De Moivre's Theorem resulting in -2i, which involves evaluating the trigonometric functions cosine and sine at 3π/2.

Step-by-step explanation:

The 4th root (k = 3) of 16 in complex numbers can be found using De Moivre's Theorem, which states that the nth roots of a complex number can be found by taking the nth root of the modulus and dividing the argument by n. Since 16 has a modulus of 16 and an argument of 0 (because it's a real number), we can calculate its roots. For the 4th root, we take the fourth root of the modulus, which is 2, and divide the argument by 4, applying an increment of 2π/4 for each successive k.

Thus, the 4th root for k=3 is calculated as follows:

• Modulus: √16 = 2
• Argument: (3⋅(2π/4)) = 3π/2
• Complex root: 2[cos(3π/2) + i sin(3π/2)]

Evaluating the trigonometric functions: cos(3π/2) = 0 and sin(3π/2) = -1. Thus, the 4th root when k=3 is 2(0 - i) = -2i.