11/20/2012, 11:01 PM
(This post was last modified: 11/20/2012, 11:03 PM by sheldonison.)
(11/20/2012, 10:53 PM)tommy1729 Wrote:if \( \text{2sinh}^{[z]}=n\pi i \) then \( \text{2sinh}^{[z+1]}=e^{n\pi i}-e^{-n\pi i}=0 \)(11/20/2012, 12:01 AM)sheldonison Wrote: \( \text{2sinh}^{[z]}=n\pi i \)
What is wrong with that ?
We can compute log of that ?
and therefore there is a singularity.
\( \log(\text{2sinh}^{[z+1]})=\log(0) \)
- Sheldon