I'm trying to find the limit of $z_n$ as $n\to \infty$

where $z_n = n\left[1-\cos\left(\frac{\theta}{n}\right) -i\sin\left(\frac{\theta}{n}\right)\right]$

Here's what I have so far: $$\begin{align}\lim_{n \to \infty} z_n & = \lim \limits_{n \to \infty} \left[n -n\cos\left(\frac{\theta}{n}\right) -in\sin\left(\frac{\theta}{n}\right)\right] \\ &= 0 -i\lim_{n \to \infty}n\sin\left(\frac{\theta}{n}\right) \\ &= -i\lim_{n \to \infty} \frac{\sin\left(\frac{\theta}{n}\right)}{1/n} \\ &= -i\theta \end{align}$$
(Last equality is by L'hospital)

Is this correct? It looks correct to me but a bit hand-wavy.

  • 2
    $\begingroup$ $\lim_{z\to 0}\frac{e^z-1}{z}=1$, either if $z$ is a real number or a complex one. $\endgroup$ – Jack D'Aurizio Jan 23 '18 at 20:53

Note that

$$n\left[1-\cos\left(\frac{\theta}{n}\right) -i\sin\left(\frac{\theta}{n}\right)\right]=n\left[ 1-e^{ i\left( \frac{\theta}{n} \right) } \right]=-i\theta\frac{ \left[ e^{ i\left( \frac{\theta}{n} \right) } -1\right]}{i\frac{\theta}n}\to -i\theta$$

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