If $z$ is a complex number such that $\frac{z-1}{z-i}$ is purely imaginary and the locus of $z$ represents a circle with centre $(\alpha, \beta)$ and radius $r$,then $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=$

The area (in sq units) of the triangle whose vertices are the points represented by the complex numbers $0, z$,and $z e^{i \alpha}$ $(0 < \alpha < \pi)$ is:

Let $a$ be a complex number such that $|a| < 1$ and $z_1, z_2, \dots$ be vertices of a polygon such that $z_k = 1 + a + a^2 + \dots + a^{k-1}$. Then the vertices of the polygon lie within a circle:

Let point $P = \alpha + i\beta$,where $\alpha, \beta > 0$,undergo the following three transformations successively on the Argand plane:
$(I)$ Reflection about $\text{amp}(z) = \frac{\pi}{4}$
$(II)$ Transformation through a distance $\beta$ units along the positive direction of the real axis
$(III)$ Rotation through an angle $\frac{\pi}{4}$ about the origin in the counter-clockwise direction
If the final position of the point is given by $Q = -\sqrt{2} + i\sqrt{6}$,then:

$PQ$ and $PR$ are two infinite rays. $QAR$ is an arc. $A$ point lying in the shaded region,excluding the boundary,satisfies:

Let $x_1, x_2, x_3, x_4$ be the roots of the equation $4x^4 + 8x^3 - 17x^2 - 12x + 9 = 0$. If $(4+x_1^2)(4+x_2^2)(4+x_3^2)(4+x_4^2) = \frac{125}{16}m$,then the value of $m$ is: