If $z_1$ and $z_2$ are two roots of the equation $z^2+az+b=0$ with $a^2 < 4b$, then the origin, $z_1$, and $z_2$ form an equilateral triangle if

The set of all real values of $c$ for which the equation $z \bar{z} + (4 - 3i) \bar{z} + (4 + 3i) z + c = 0$ represents a circle is

$z_1$ and $z_2$ are the roots of $3z^2 + 3z + b = 0$. If the origin,$A(z_1)$,and $B(z_2)$ form an equilateral triangle,then the value of $b$ is equal to

Let $S = \{z \in \mathbb{C} - \{i, 2i\} : \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \}$. If $\alpha - \frac{13}{11}i \in S$ and $\alpha \in \mathbb{R} - \{0\}$,then $242\alpha^2$ is equal to

Let $z = x + iy$, where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$ is purely imaginary, lie on

If $z = x + iy$ and $\omega = \frac{1 - iz}{z - i}$,then $|\omega| = 1$ shows that in the complex plane: