Find all values of $(-1)^{1/3}$ I used the identity's and such and got a part where I got $e^{1/3\log(-1)}$, and I'm not sure how to do the next step and get to the answer. Can anyone send in the right direction ?
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3Do you know about the nth roots of unity? – Hiten May 21 '15 at 07:39
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Yes I used it, and i got the principle axis which equals = 1/2 + sqrt(3)i/2 But the answer to the question is = e^ipi/3 , e^ipi and e^(i*5pi/3) and Im not sure how they got that – Vishwjit May 21 '15 at 07:42
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just write them in the form of e^ix where x is the angle.... The answers do match. – Hiten May 21 '15 at 07:45
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@Vishwjit, See http://math.stackexchange.com/questions/192742/how-to-solve-x3-1 – lab bhattacharjee May 21 '15 at 08:15
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$$-1 = e^{i\pi}$$
so
$$\sqrt[3]{-1} = e^{i\pi/3}$$
is your first solution. The other two are obtained by adding integer multiples of $2\pi/3$ in the exponential ($e^{ix}$ is $2\pi$-periodic). The general solution is then
$$\sqrt[3]{-1} = e^{i\pi/3 + ik 2\pi/3}$$
for $k = 0,1,2$.
GDumphart
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A fast but non-rigorous answer could go as follows: Use the polar representation of -1. In general $-1=e^{i(2k+1)\pi}$ for every integer k. Assuming that the properties of exponentiation hold also for complex numbers, you would have $(-1)^{1/3}=e^{i(2k\pi/3)+i\pi/3}$. Now the only thing that is left is to identify the different values that the phase $2k\pi/3$ can take, considering that multiples of $2\pi$ produce the same number.
I hope that helps.
