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

Let $z$ be a complex number such that the real part of $\frac{z-2i}{z+2i}$ is zero. Then,the maximum value of $|z-(6+8i)|$ is equal to:

Let $z=x+iy$ and $P(x, y)$ be a point on the Argand plane. If $z$ satisfies the condition $\operatorname{Arg}\left(\frac{z-3i}{z+2i}\right)=\frac{\pi}{4}$, then the locus of $P$ is:

$P$ is a point denoting $z$ in the Argand diagram. If $\frac{z-i}{z-1}$ is always purely imaginary,then the locus of $P$ is

If $|z^2 - 1| = |z|^2 + 1$,then $z$ lies on

Let $a, b \in \mathbb{R}$ and $a^2+b^2 \neq 0$. Suppose $S = \{z \in \mathbb{C} : z = \frac{1}{a+ibt}, t \in \mathbb{R}, t \neq 0\}$,where $i = \sqrt{-1}$. If $z = x+iy$ and $z \in S$,then $(x, y)$ lies on: