Given that the equation $z^2 + (p + iq)z + r + is = 0$,where $p, q, r, s$ are real and non-zero,has a real root,then:

Let $S$ be the set of all $(\alpha, \beta)$ such that $\pi < \alpha, \beta < 2\pi$,for which the complex number $\frac{1-i \sin \alpha}{1+2i \sin \alpha}$ is purely imaginary and $\frac{1+i \cos \beta}{1-2i \cos \beta}$ is purely real. Let $Z_{\alpha \beta} = \sin 2\alpha + i \cos 2\beta$ for $(\alpha, \beta) \in S$. Then $\sum_{(\alpha, \beta) \in S} \left(i Z_{\alpha \beta} + \frac{1}{i \bar{Z}_{\alpha \beta}}\right)$ is equal to:

$\cosh (\alpha + i\beta ) - \cosh (\alpha - i\beta )$ is equal to

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

If $i = \sqrt{-1}$, then $\frac{e^{xi} + e^{-xi}}{2} = $

For a complex number $Z = a + ib$,let $\hat{Z} = b + ia$. If $Z_1$ and $Z_2$ are such complex numbers,then $\widehat{Z_1 Z_2} = $