onesun0000
Junior Member
- Joined
- Dec 18, 2018
- Messages
- 83
I just thought of this and tried to simplify. [MATH]{(-1)}^\frac{1}{6}[/MATH]
[MATH]{(-1)}^\frac{1}{6}={[{(-1)}^\frac{1}{3}]}^{\frac{1}{2}}[/MATH]
Since the cube root of [MATH]-1[/MATH] is [MATH]-1[/MATH]
[MATH]{(-1)}^{\frac{1}{6}}=\sqrt{-1}\to{(-1)}^{\frac{1}{6}}=i[/MATH]
Is this a correct conclusion? If so, does this work to every roots with even index, where the radicand is negative?
[MATH]{(-1)}^\frac{1}{6}={[{(-1)}^\frac{1}{3}]}^{\frac{1}{2}}[/MATH]
Since the cube root of [MATH]-1[/MATH] is [MATH]-1[/MATH]
[MATH]{(-1)}^{\frac{1}{6}}=\sqrt{-1}\to{(-1)}^{\frac{1}{6}}=i[/MATH]
Is this a correct conclusion? If so, does this work to every roots with even index, where the radicand is negative?
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