If $|z|=1$ and $w=\frac{z-1}{z+1}$ (where $z \neq -1$),then $\operatorname{Re}(w)$ is

Let ${z_1}$ and ${z_2}$ be two roots of the equation ${z^2 + az + b = 0}$,where ${z}$ is a complex number. Further,assume that the origin,${z_1}$,and ${z_2}$ form an equilateral triangle. Then:

Let the complex numbers $\alpha$ and $\left(\frac{1}{\bar{\alpha}}\right)$ lie on circles $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ and $\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2$ respectively. If $z_0=x_0+i y_0$ satisfies the equation $2|z_0|^2=r^2+2$,then $|\alpha|=$

The maximum distance from the origin of coordinates to the point $z$ satisfying the equation $\left| z + \frac{1}{z} \right| = a$ is

The locus of the point $z=x+iy$ satisfying the equation $\left|\frac{z-1}{z+1}\right|=1$ is given by :

The equation $|z - i| = |z - 1|$,where $i = \sqrt{-1}$,represents: