The locus of $z$ given by $\left| \frac{z - 1}{z - i} \right| = 1$ is

If $Z \neq \pm 1$ is a complex number and $\operatorname{Arg}\left(\frac{Z-1}{Z+1}\right)=\frac{\pi}{4}$,then the locus of $Z$ in the Argand plane is

Let $z_1$ and $z_2$ be two complex numbers and roots of the equation $z^2 + az + b = 0$. If $O$ is the origin such that $OA = OB$ and $a^2 = \lambda b \cos^2 \frac{\alpha}{2}$,where $\alpha$ is the angle $\angle AOB$,then $\lambda$ is equal to:

$A$ particle $P$ starts from the point $Z_0 = 1 + 2i$ where $i = \sqrt{-1}$. It moves first horizontally away from the origin by $5$ units and then vertically upwards parallel to the positive $y$-axis by $3$ units to reach a point $Z_1$. From $Z_1$,the particle moves $\sqrt{2}$ units in the direction of vector $\hat{i} + \hat{j}$ and then it moves through an angle $\frac{\pi}{2}$ in the anticlockwise direction on a circle with the centre at the origin to reach point $Z_2$. Then $Z_2 =$

If the vertices $A, B$ and $C$ of an isosceles $\triangle ABC$ are respectively $z_1, z_2$ and $z_3$ and if $\angle C=90^{\circ}$,then

If $z = x + iy$ is a complex number,then the equation $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$ represents the