How can we calculate the complex limit $\frac{1 – e^{iz}}{z}$ when $z$ approach zero

Question

I would like for an explanation how one should approach limit's from the form "$\frac{0}{0}$" in the complex field. Untill now I just used L'hopital, but recently my teacher told us that we do not talk about L'hopital rule in our course (so I did not understand if this rule is true in the complex field). for example, how I can show then without L'hopital that the following limit is equal to $-i$ when $z$ approach zero $$\frac{1 – e^{iz}}{z}$$

Answer 1

You don't need L'Hopital's rule. By definition, $\lim\limits_{z \to 0} \frac{f(z) - f(0)}{z} = f'(0)$. So letting $f(z) = e^{iz}$, we have $\lim\limits_{z \to 0} \frac{1 - e^{iz}}{z} = - \lim\limits_{z \to 0} \frac{e^{iz} - 1}{z} = - \lim\limits_{z \to 0} \frac{f(z) - f(0)}{z} = - f'(0) = -i e^{i \cdot 0} = -i$.