Recent content by Jacopo

Oh yes :D Thanks

I know why I can add 2 \pi i to log_{e} \omega ( log_{e} \omega = log_{e}z + i \theta if \omega = z(cos \theta +i sen \theta ) ) but I can't understand why I can add \frac{2 \pi i}{log_{e}b} to log_{b} \omega . Does anyone have the answer for me? Thanks!