I have a confused with this problem? I calculate this by 2 ways: $$j^3 = jj^2 = j(-1) = -j$$ $$j^3 = j^{\frac{12}{4}} = (j^{12})^{0.25} = 1^{0.25} = 1$$ Why does it have different result?
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3$1^{0.25}=1^{\frac14}=1,-1,j,-j$. The roots other than $-j$ are extraneous. – GoodDeeds Apr 23 '16 at 16:08
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Same as why $(-2)^3=-8$ and $(-2)^3 = (-2)^{\frac{12}{4}} = ((-2)^{12})^{1\over4} = 4096^{1\over4} = 8$ seem to give different results. – Aritra Das Apr 23 '16 at 16:13
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Related: For which complex $a$, $b$, $c$ does $(a^{b})^{c} = a^{bc}$ hold?. – Andrew D. Hwang Apr 23 '16 at 17:12
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In the complex number field $x^{\frac{12}{4}}$ is not a single valued function because we have four fourth roots of a complex number, so we cannot write $j^3=j^{\frac{12}{4}}$.
Emilio Novati
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Thank for your reply Emilio, Like Aritra calculate (-2)^3 it also have 2 solutions. It make me confused :'( – Cuong Bui Apr 23 '16 at 16:21
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So if i calculate by pocket calculator, I press (4096)^0.25 it has only equal = 8; not have result -8 (assumed we use R field, not complex field) – Cuong Bui Apr 23 '16 at 16:36
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Generally a pocket calculator give only the principal root, that , in the case of a fourth root is the positive real value. See: http://mathworld.wolfram.com/PrincipalSquareRoot.html – Emilio Novati Apr 23 '16 at 16:43