- #1
Undoubtedly0
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It can be shown from Euler's formula that
[tex] \left(-1\right)^x = \cos(\pi x) + i\sin(\pi x) [/tex]
However, consider [itex] x = 2/3 [/itex]. The left expression gives
[tex] \left(-1\right)^\frac{2}{3} = \left(\left(-1\right)^2\right)^\frac{1}{3} = \left(1\right)^\frac{1}{3} = 1 [/tex]
while the right expression gives
[tex] \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i [/tex]
What has gone wrong here? Thanks all.
[tex] \left(-1\right)^x = \cos(\pi x) + i\sin(\pi x) [/tex]
However, consider [itex] x = 2/3 [/itex]. The left expression gives
[tex] \left(-1\right)^\frac{2}{3} = \left(\left(-1\right)^2\right)^\frac{1}{3} = \left(1\right)^\frac{1}{3} = 1 [/tex]
while the right expression gives
[tex] \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i [/tex]
What has gone wrong here? Thanks all.