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

$A$ function $f$ is defined on the complex numbers by $f(z) = (a + ib)z$,where $a, b \in \mathbb{R}^+$. This function has the property that the $f$-image of any point in the complex plane is equidistant from that point and the origin. If $|a + bi| = 10$ and $b^2 = \frac{p}{q}$,where $p, q \in \mathbb{Z}$ and $\text{gcd}(p, q) = 1$,then $p + q$ is:

The locus of the points represented by $|z+3|-|z-3|=6$,where $z$ is a complex number,is ....

The locus of the complex number $z$ such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$ is:

Let $z \in \mathbb{C}$ be such that $|z| < 1$. If $w = \frac{5 + 3z}{5(1 - z)}$,then

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a