# Math Help - Complex number identity

1. ## Complex number identity

Show that $(-1)^{\frac{1}{\pi}}=e^{(2n+1)i} (n=0, \pm1, \pm2, ...)$

2. Originally Posted by davesface
Show that $(-1)^{\frac{1}{\pi}}=e^{(2n+1)i} (n=0, \pm1, \pm2, ...)$

$(-1)^{\frac{1}{\pi}}=e^{\frac{1}{\pi}Ln(-1)}=e^{\frac{1}{\pi}[\ln 1+i\arg(-1)]}=e^{\frac{1}{\pi}(2n+1)\pi i}=$ ...

Tonio

3. These sign errors get me every time. Thanks for the help.