Show that function is bounded

**EinStone** · April 9th 2011, 04:39 PM

Hi,

I want to show that $f(z) = \dfrac{e^z - 1}{\cos z + \sin z - 1}$ is bounded on some neighborhood of zero in the complex plane, for example on the unit circle. Any bound suffices, I have

$f(z) = \dfrac{e^z - 1}{\frac{1}{2}(e^{iz}+e^{-iz}) + \frac{1}{2i}(e^{iz}-e^{-iz}) - 1} = \dfrac{e^z - 1}{e^{iz}(\frac{1}{2} + \frac{1}{2i}) + e^{-iz}(\frac{1}{2} - \frac{1}{2i}) - 1}$
but I don't know what to do now.

**FernandoRevilla** · April 9th 2011, 10:36 PM

We have $\lim_{z\to 0}f(z)=1$ so, defining $f(0)=1$ the function $|f(z)|$ is continuous on a closed unit disk $K$ centered at $0$ . As $K$ is a compact set, $|f(z)|$ as an absolute maximum on $K$ .

**EinStone** · April 10th 2011, 02:53 AM

The problem is showing exactly $\lim_{z\to 0}f(z)=1$ . If I can prove that its bounded on a punctured neighborhood of zero then this would follow.

**FernandoRevilla** · April 10th 2011, 04:07 AM

L'Hopital rule.

**EinStone** · April 10th 2011, 09:00 AM

Does L'Hopital rule also apply for holomorphic quotients? I always thought its a real method. And if I use it, how do I get to the result?

**roninpro** · April 10th 2011, 11:53 AM

You could also try to write down the Laurent series:

$\displaystyle \dfrac{e^z - 1}{\cos z + \sin z - 1}=\frac{z + z^2/2 + z^3/6 + z^4/24+\ldots}{z - z^2/2 - z^3/6 + z^4/24-\ldots}$

You can factor out the $z$ and then evaluate the limit.

**FernandoRevilla** · April 10th 2011, 12:13 PM

Originally Posted by EinStone

Does L'Hopital rule also apply for holomorphic quotients?

Yes, it does.

And if I use it, how do I get to the result?

Trying it.

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