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)#