Show that $\displaystyle (-1)^{\frac{1}{\pi}}=e^{(2n+1)i} (n=0, \pm1, \pm2, ...)$
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Originally Posted by davesface Show that $\displaystyle (-1)^{\frac{1}{\pi}}=e^{(2n+1)i} (n=0, \pm1, \pm2, ...)$ $\displaystyle (-1)^{\frac{1}{\pi}}=e^{\frac{1}{\pi}Ln(-1)}=e^{\frac{1}{\pi}[\ln 1+i\arg(-1)]}=e^{\frac{1}{\pi}(2n+1)\pi i}=$ ... Tonio
These sign errors get me every time. Thanks for the help.
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