How do I find the imaginary root of 216?

Question

I need help for an upcoming test, and can't seem to find help anywhere. Help is appreciated. thank you in advance

Answer 1

I am going to assume by root you mean nth root.
you can express 216 using argument and angle as follows:

216 = 216e^(i*2*pi*t)

where t = 0,1,2,3,... and i = sqrt(-1)

When you take the nth root of 216 you take the real root of the argument and divide the angle into n;

216^(1/n) = Re(216^(1/n) ) * e^(i*2*pi*t/n)

so if n = 5 we'll have 5 distinct answers (where the angle is between 0 and 2*pi). These angles are:

2*pi*0/ 5, 2*pi*1/ 5, 2*pi*2/ 5, 2*pi*3/ 5, 2*pi*4/ 5
= 0, 1.2566, 2.5133, 3.7699, 5.0265

216^(1/n) = 2.93 or 2.93 e^i*1.2566 or 2.93 e^i*2.5133 or,...

/m

Answer 2

216 = 6^3, so I assume you are looking for the two complex cube roots, which are represented in polar coordinates on the complex plane as 6@120° and 6@240° (or 6@- 120°.

The magnitude of each nth root is the same as the magnitude of the "real" nth root, and the angular separations are 360/n where n is the number of roots.

6@120° = - 3 + i3√3 = 3(- 1 + i√3)
[3(- 1 + √3)]^3 =
27(- 1 + i√3)(- 1 + i√3)(- 1 + i√3) =
27(- 1 + i√3)(- (- 1 + i√3) + i√3(- 1 + i√3)) =
27(- 1 + i√3)(1 - i√3 + - i√3 - 3) =
27(- 1 + i√3)(- 2 - i2√3) =
54(1 - i√3)(1 + i√3) =
54(1 + 3) = 54*4 = 216

The other root proves out similarly.

Answer 3

216, since it is a positive number, has no imaginary roots. So then, the roots of 216 are:
sqrt (216) and -sqrt(216).

If you were looking for the roots of -216, then youd have:
sqrt (-216) = sqrt (-1 x 216) = sqrt (-1) x sqrt(216) = i x sqrt (216), or i x -sqrt(216)

So basicall, you can separte a negative number to -1 and a positive number, and then find the square root of each. This would then give you the roots of your number.

Hope this helped.

Answer 4

not sure what u want?
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like this form

216 = cos (x) + j sin(x)