# Limits of complex functions

• March 22nd 2009, 11:14 PM
Richmond
Limits of complex functions
lim log(sinh(z))
z->pi i/2
• March 22nd 2009, 11:24 PM
Jhevon
Quote:

Originally Posted by Richmond
lim log(sinh(z))
z->pi i/2

Hint: $\sinh z = \frac {e^z - e^{-z}}2 = - i \sin iz$
• March 22nd 2009, 11:30 PM
Chris L T521
Quote:

Originally Posted by Richmond
lim log(sinh(z))
z->pi i/2

Note that $\sinh z=\frac{e^z-e^{-z}}{2}$

So, $\lim_{z\to i\frac{\pi}{2}}\log\left(\sinh z\right)=\log\left(\frac{e^{i\frac{\pi}{2}}-e^{-i\frac{\pi}{2}}}{2}\right)=\log\left(i\right)$

Since $\log z=\ln\left|z\right|+i\left(\text{Arg}\,z+2k\pi\rig ht) \,z\,;\, -\pi<\text{Arg}\,z\leqslant\pi$, we see that $\log\left(z\right)=\ln \left|i\right|+i\left(\text{Arg}\,i+2k\pi\right)= \ln 1+i\left(\frac{\pi}{2}+2k\pi \right)=\frac{\left(4k+1\right)\pi}{2}i$

However, I think you're dealing with the Principal Logarithm.

So, we should say that $\lim_{z\to i\frac{\pi}{2}}\log\left(\sinh z\right)=\boxed{\frac{\pi}{2}i}$

Does this make sense?
• March 22nd 2009, 11:43 PM
Richmond
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.

$
\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 $- i \sin iz$