Find a sequence of complex numbers $\{z_k\}$ such that $\lim\limits_{k\to \infty} z_k = 0$ and $\lim\limits_{k\to \infty} e^{1/z} =2+i$.
Text book doesnot really explain this assignment based question. $$ e^{1/z} = \sum_{k=0}^\infty \frac{1}{k!} z^{-k}$$
All you have to do is to find 'a' logarithm of $2+i$. Verify, using basic trigonometry, that $$e^{\ln \sqrt 3 +i \tan ^{-1} (1/2)}=2+i$$ Now take $$z_k=\frac 1 {2\pi i k+\ln \sqrt 3 +i \tan ^{-1} (1/2)}$$ Not only does $e^{1/z_k} \to 2+i$ but we actually have $e^{1/z_k} = 2+i$ for all $k$.