The locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$,is

Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$. Then

If ${z_1} = 10 + 6i$,${z_2} = 4 + 6i$ and $z$ is a complex number such that $\text{amp}\left( \frac{z - z_1}{z - z_2} \right) = \frac{\pi}{4}$,then the value of $|z - 7 - 9i|$ is equal to

If $z = x + iy$ and $|z - zi| = 1,$ then

If $|z - 3 - 4i| = 4$,where $i = \sqrt{-1}$,then the maximum possible value of $|z|$ is:

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: