Order of zero of $(e^z - 1)^2$ and $f(z) = (z^2 - 4z + 4)^3$ Maybe it's just that I need to sleep, but despite reading the definition of order of a zero and looking at another example on this website, I'm still having trouble fully understanding the concept.
(a) $f(z) = (e^z - 1)^2$. My intuition is that the order of zero is $2$, but I'm not sure how to show that. I know that $e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}$.
(b) $f(z) = (z^2 - 4z + 4)^3$. Here, I think the order of zero is $6$ because the sixth derivative is not zero. 

Your answers are correct. To prove that $(e^{z}-1)^{2}$ has a zero of order $2$ at $z=0$ use the fact that $\frac {e^{z}-1} z \to 1$ as $z \to 0$ or the fact that $e^{z}-1=z(1+\frac z {2!}+ \frac {z^{2}} {3!}+...)$. 
