# Thread: Show that function is bounded

1. ## Show that function is bounded

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.

2. 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 $K$ .

3. 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.

4. L'Hopital rule.

5. 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?

6. 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.

7. 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.