How do you solve for x^6 = i?

It is easiest to take roots of complex numbers by first representing them in polar form.

We start with i = exp(i * pi/2) which is the polar representation of i.

In fact, we can add any multiple of 2*pi to the angle and get the same answer, so we can write:

i = exp(i * (pi/2 + 2*k*pi))

where k is an integer. The important thing is that even though k can take different values, the resulting complex number is always the same because the angle is just increasing by 2*pi every time.

To take the sixth root, we can apply the usual method of applying an exponent to an exponential.

i^(1/6) = exp(i * (pi/2 + 2*k*pi)) ^ 1/6

= exp(1/6 * i * (pi/2 + 2*k*pi))

= exp(i * (pi/12 + k/3*pi))

Which is a polar representation of the sixth roots of i.

Note that now the different values of k are no longer just different ways of representing the same number, but now they actually represent different numbers. To rhyme off a few:

k = 0 gives exp(i*pi/12)

k = 1 gives exp(i * 5/12 * pi)

k = 2 gives exp(i * 9/12 * pi)

k = 3 gives exp(i * 13/12 * pi)

k = 4 gives exp(i * 17/12 * pi)

k = 5 gives exp(i * 21/12 * pi)

k = 6 gives exp(i * 25/12 * pi) = exp(i * 1/12 * pi)

Note that due to the periodicity of angle, the value for k = 6 is the same as the value for k = 0. In fact all values for k >= 6 are the same as one of the values for k=0..5.

Each of these numbers taken to the sixth power will give you i.

there aren't 6 answers, just 2.

x=+-i^(1/6)

thats pos/neg sixth root of i or 12th root of -1

read the applications in the article about moivre theorem

http://en.wikipedia.org/wiki/De_Moivre%27s_formula