Let $f(x) = x^4 + ax^3 + bx^2 + c$ be a polynomial with real coefficients such that $f(1) = -9$. Suppose that $i\sqrt{3}$ is a root of the equation $4x^3 + 3ax^2 + 2bx = 0$,where $i = \sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$,and $\alpha_4$ are all the roots of the equation $f(x) = 0$,then $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2$ is equal to:

Let $S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}$. Then $\sum_{z \in S} |z + \sqrt{3}i|^2$ is equal to:

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:

If $\sqrt{5}-i \sqrt{15}=r(\cos \theta+i \sin \theta)$ where $-\pi < \theta < \pi$,then find the value of $r^2(\sec \theta+3 \operatorname{cosec}^2 \theta)$.

If $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is purely real,then $\cos ^3 \theta+\sin ^2 \theta+\cos \theta+1=$

Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $|z-z_0|^2=4$ and $|z-z_0|^2=16$ respectively,where $z_0=1+i$. Then,the value of $100|\alpha|^2$ is.