The inequality $|z - 4| < |z - 2|$ represents the region given by

If a complex number $z = x + iy$ is taken such that the amplitude of the fraction $\frac{z - 1}{z + 1}$ is always $\frac{\pi}{4}$,then:

$z_1$ 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

If complex numbers ${z_1}, {z_2}, \text{and } {z_3}$ represent the vertices $A, B, \text{and } C$ respectively of an isosceles triangle $ABC$ of which $\angle C$ is a right angle,then the correct statement is:

For any two complex numbers $z_1, z_2$,if $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$,then:

If $|8 + z| + |z - 8| = 16$,where $z$ is a complex number,then the point $z$ will lie on