How do you find the 3rd root of $8e^(30i)$ ?

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Answer 1 · 2016-09-16T23:14:39

$root(3)(8e^(30i)) = 2e^((10+(2pi)/3)i)$

Note that $2e^(10i)$ is a third root of $8e^(30i)$ , but it may or may not be the principal one...

$(2e^(10i))^3 = 2^3*e^((10i)*3) = 8e^(30i)$

To determine what the principal cube root is, we first need to determine which quadrant $8e^(30i)$ is in.

$30/(pi/2) ~~ 19.1 -= 3.1" "$ modulo $4$

This places $8e^(30i)$ in Q4, so its principal cube root is also in Q4.

$10/(pi/2) ~~ 6.4 -= 2.4" "$ modulo $4$

So $2e^(10i)$ is in Q3 and is not the principal cube root.

To get the principal cube root we can multiply by the primitive Complex cube root of $1$ :

$omega = -1/2+sqrt(3)/2 = e^((2pi)/3i)$

This will result in a number in Q4

$2e^(10i) * omega = 2e^(10i) e^((2pi)/3i) = 2e^((10+(2pi)/3)i)$

How do you find the 3rd root of 8e^(30i)8e^(30i)?