Let $z \in \mathbb{C}$. We set $z = a + ib$, where $a, b \in \mathbb{R}$.
Deduce that $$\lim_{n \rightarrow \infty} \left(1 + \frac{z}{n}\right)^n = e^z$$

Let $z$ be a complex number, with real part $x$ and imaginary part $y$, such that $(x,y) \notin \mathbb{R}^{-} \times \{0\}$. We denote $$\theta(z) = 2\arctan\left(\frac{y}{x + \sqrt{x^2 + y^2}}\right) \quad \text{and} \quad R(z) = \frac{z + |z|}{\sqrt{2(\operatorname{Re}(z) + |z|)}}$$ Determine $[R(z)]^2$, $\theta \circ R(z)$ and $|z|^{1/2}\mathrm{e}^{\mathrm{i}\theta(z)/2}$ as functions of $z$, $R(z)$ and $\theta(z)$.

Let $A = \left\{ \theta \in ( 0,2 \pi ) : \frac { 1 + 2 i \sin \theta } { 1 - i \sin \theta } \right.$ is purely imaginary $\}$ Then the sum of the elements in $A$ is
(1) $4 \pi$
(2) $3 \pi$
(3) $\pi$
(4) $2 \pi$

If $\alpha$ and $\beta$ are the roots of the equation $2z^2 - 3z - 2i = 0$, where $i = \sqrt{-1}$, then $16 \cdot \operatorname{Re}\left(\frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}}\right) \cdot \operatorname{Im}\left(\frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}}\right)$ is equal to
(1) 441
(2) 398
(3) 312
(4) 409