lim log(sinh(z))
z->pi i/2
Note that $\displaystyle \sinh z=\frac{e^z-e^{-z}}{2}$
So, $\displaystyle \lim_{z\to i\frac{\pi}{2}}\log\left(\sinh z\right)=\log\left(\frac{e^{i\frac{\pi}{2}}-e^{-i\frac{\pi}{2}}}{2}\right)=\log\left(i\right)$
Since $\displaystyle \log z=\ln\left|z\right|+i\left(\text{Arg}\,z+2k\pi\rig ht) \,z\,;\, -\pi<\text{Arg}\,z\leqslant\pi$, we see that $\displaystyle \log\left(z\right)=\ln \left|i\right|+i\left(\text{Arg}\,i+2k\pi\right)= \ln 1+i\left(\frac{\pi}{2}+2k\pi \right)=\frac{\left(4k+1\right)\pi}{2}i$
However, I think you're dealing with the Principal Logarithm.
So, we should say that $\displaystyle \lim_{z\to i\frac{\pi}{2}}\log\left(\sinh z\right)=\boxed{\frac{\pi}{2}i}$
Does this make sense?
I've got till Log(i)
I'm not sure what is the Principal Logarithm but I'll try search around the net about it.
$\displaystyle
\frac{\left(4k+1\right)\pi}{2}i
$
I think this is what I'm looking for. Thanks again for your help. Greatly appreciated. and thanks to Jhevon too. Prefer to work with$\displaystyle - i \sin iz$