Any hint to solve such a complex limit

Question

What would be the limit of the following function: $\lim\limits_{z\rightarrow m\pi i} \dfrac{(z-m\pi i) \exp(z)}{\sin z}$ . Can we apply L'Hospital's rule?

Answer 1

I'd do it this way:

$$\lim_{z \to m \pi i} \frac{ (z-m \pi i) \exp(z) }{ \sin z } = \lim_{z \to m \pi i} \frac{\exp(z)}{ \frac{\sin z - \sin(m \pi i)}{z - m \pi i} } = \frac{\exp( m \pi i )}{\cos(m \pi i )} = \frac{(-1)^m}{\cosh( m \pi )}.$$

If you prefer, de l'Hospital rule can also be applied.

Edit: As pointed out in the comments, I made a mistake assuming $\sin( m \pi i ) = 0$. The correct solution:

$$\lim_{z \to m \pi i} \frac{ (z-m \pi i) \exp(z) }{ \sin z } = \frac{0 \cdot \exp( m \pi i )}{\sin( m \pi i )} = 0.$$

This also means the l'Hospital rule cannot be applied since we don't have a $\frac{0}{0}$.