If $z=x+iy$ and $z^{1/3}=p+iq$,where $x, y, p, q \in R$ and $i=\sqrt{-1}$,then the value of $\left(\frac{x}{p}+\frac{y}{q}\right)$ is

If $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $\arg(z) - \arg(w) = \frac{\pi}{2}$,then

If $e^{i x}$ is a solution of the equation $z^n+p_1 z^{n-1}+p_2 z^{n-2}+\ldots+p_n=0$,where $p_i$ are real $(i=1, 2, \ldots, n)$,then $p_n \sin nx + p_{n-1} \sin(n-1)x + \ldots + p_1 \sin x + \sin(0) = $ (Note: The constant term in the equation is $p_n$ and the coefficient of $z^0$ is $1$ if we normalize,but here the equation is given as $z^n + p_1 z^{n-1} + \ldots + p_n = 0$. Let us assume the constant term is $p_n$. The expression to evaluate is $p_n \sin nx + p_{n-1} \sin(n-1)x + \ldots + p_1 \sin x + \sin(0)$). Given the standard form,find the value of $p_n \sin nx + p_{n-1} \sin(n-1)x + \ldots + p_1 \sin x$.

Let the complex number $z = x + iy$ be such that $\frac{2z - 3i}{2z + i}$ is purely imaginary. If $x + y^2 = 0$,then $y^4 + y^2 - y$ 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 $z = x - iy$ and $z^{1/3} = p + iq$,then $\left( \frac{x}{p} + \frac{y}{q} \right) / (p^2 + q^2)$ is equal to