If $z = (\lambda + 3) + i\sqrt{5 - \lambda^2}$,then the locus of $z$ is a

If $z=x+iy$, where $x$ and $y$ are real numbers and $i=\sqrt{-1}$, then the points $(x, y)$ for which $\frac{z-1}{z-i}$ is real, lie on

The equation $|z - 5i| / |z + 5i| = 12,$ where $z = x + iy,$ represents a/an

Let $\arg(z)$ represent the principal argument of the complex number $z$. The curves $|z|=3$ and $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ intersect:

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:

Let $z=x+iy$ and a point $P$ represent $z$ in the Argand plane. If the real part of $\frac{z-1}{z+i}$ is $1$,then a point that lies on the locus of $P$ is