# How does Mathematica apply laws of exponents? [duplicate]

When computing the value of -1 raised to the 2/3 power, Mathematica returns a complex number instead of the value 1.

• The correct result is -(1/2) + (I Sqrt[3])/2, not 1. – bbgodfrey Apr 15 '17 at 0:37
• @bbgodfrey No, the correct result is $e^{2/3 \ln(-1)} = e^{2/3 i \pi (1 + 2n)}$, where the value of the integer $n$ depends on your choice of branch cut for the log function. The OP's choice of $n = 1$ is just as valid as your implicit choice of $n = 0$. I think the OP is asking which branch of the log function Mathematica uses. I assume it's the principle branch, whose imaginary part lies between $-i \pi$ and $i \pi$. – tparker Apr 15 '17 at 0:48
• (CubeRoot[-1])^2 =1 – PRG Apr 15 '17 at 3:57
• For Mathematica to compute the real value of (-1)^(2/3), as one would traditionally think of this, you have to write it in Mathematica as (Surd[-1,3])^2=1. Since this is a cube root, there is a Mathematica function (shown above) known as CubeRoot; however, if you want the real value of (say) (-1)^(2/5) then you can use (Surd[-1,5])^2 = 1 (as it should), – PRG Apr 15 '17 at 4:06