Limits of complex functions

**Richmond** · March 22nd 2009, 10:14 PM

lim log(sinh(z))
z->pi i/2

**Jhevon** · March 22nd 2009, 10:24 PM

Originally Posted by Richmond

lim log(sinh(z))
z->pi i/2

Hint: $\sinh z = \frac {e^z - e^{-z}}2 = - i \sin iz$

**Chris L T521** · March 22nd 2009, 10:30 PM

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?

**Richmond** · March 22nd 2009, 10:43 PM

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.

$<br /> \frac{\left(4k+1\right)\pi}{2}i<br />$

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 $- i \sin iz$

Thread: Limits of complex functions

LinkBack

Thread Tools

Display

Limits of complex functions

Similar Math Help Forum Discussions

Complex limits and addition

Complex Sequence Limits

Limits on Complex Numbers

Complex Limits

limits in the complex domain

Search Tags