Let $S = \{z \in \mathbb{C} : |z-1|=1 \text{ and } (\sqrt{2}-1)(z+\bar{z}) - i(z-\bar{z}) = 2\sqrt{2}\}$. Let $z_1, z_2 \in S$ be such that $|z_1| = \max_{z \in S} |z|$ and $|z_2| = \min_{z \in S} |z|$. Then $|\sqrt{2}z_1 - z_2|^2$ equals:

If ${z_1}$ and ${z_2}$ are complex numbers such that ${z_1} \neq {z_2}$ and $|{z_1}| = |{z_2}|$. If ${z_1}$ has a positive real part and ${z_2}$ has a negative imaginary part,then $\frac{{z_1 + z_2}}{{z_1 - z_2}}$ may be

$A$ 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 $Z_2 =$

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:

The locus of $z=x+iy$, such that $\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)=0$ is

If a complex number $z=x+iy$ represents a point $P$ on the Argand plane and $\operatorname{Arg}\left(\frac{z-(3-2i)}{z-(-2+3i)}\right)=\frac{\pi}{4}$,then the locus of $P$ is a