How can we calculate the complex limit $\frac{1 - e^{iz}}{z}$ when $z$ approach zero 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  $$\frac{1 - e^{iz}}{z}$$
 A: 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 $\lim\limits_{z \to 0} \frac{1 - e^{iz}}{z} = - \lim\limits_{z \to 0} \frac{e^{iz} - 1}{z} = - \lim\limits_{z \to 0} \frac{f(z) - f(0)}{z} = - f'(0) = -i e^{i \cdot 0} = -i$.
A: Notation: $B(a, r) = \{z \in \Bbb C : |z - a| < r\}$ for $a \in \Bbb C$ and $r > 0$.

Complex differentiable functions have the following very nice property:

Suppose $a \in \Bbb C$, $r > 0$, and $f : B(a, r) \to \Bbb C$ is complex differentiable.
Then, we there exists a complex sequence $(a_n)_{n = 0}^\infty$ such $$f(z) = \sum_{n = 0}^\infty a_n(z - a)^n$$ for all $z \in B(a, r)$.

Moreover, the above summation enjoys nice properties, including differentiation term-by-term on $B(a, r)$.
Thus, if $f, g$ are differentiable on a neighbourhood of $0$, then we can write $$\frac{f(z)}{g(z)} = \frac{a_0 + a_1 z + a_2 z^2 \cdots}{b_0 + b_1 z + b_2 z^2 + \cdots}$$ on a neighbourhood of $0$.
From the above, depending on how many of the beginning $a_k$ and $b_k$ are $0$, we can calculate the limit. Moreover, we have $$a_k = \frac{1}{k!} f^{(k)}(0)$$ which shows you how L'Hôpital is valid.
In fact, the proof of L'Hôpital is much easier for the case of complex differentiable functions since you have continuity of derivatives.
