$POQ$ is a straight line through the origin $O$. $P$ and $Q$ represent the complex numbers $z_1 = a + ib$ and $z_2 = c + id$ respectively. If $OP = OQ$,then:

If $P(x)=0$ is a polynomial equation of least degree with integer coefficients and $\sqrt{2}+\sqrt{3} i$ is one of its roots,then that equation is

Let the point $P$ represent $z=x+iy$, where $x, y \in \mathbb{R}$, in the Argand plane. Let the curves $C_1$ and $C_2$ be the loci of $P$ satisfying the conditions $(i)$ $\frac{2z+i}{z-2}$ is purely imaginary and $(ii)$ $\operatorname{Arg}\left(\frac{z+i}{z+1}\right)=\frac{\pi}{2}$, respectively. Then the point of intersection of the curves $C_1$ and $C_2$, other than the origin, is

Let $s, t, r$ be non-zero complex numbers and $L$ be the set of solutions $z = x + iy$ $(x, y \in \mathbb{R}, i = \sqrt{-1})$ of the equation $sz + t\bar{z} + r = 0$,where $\bar{z} = x - iy$. Then,which of the following statement$(s)$ is (are) $TRUE$?
$(A)$ If $L$ has exactly one element,then $|s| \neq |t|$
$(B)$ If $|s| = |t|$,then $L$ has infinitely many elements
$(C)$ The number of elements in $L \cap \{z : |z - 1 + i| = 5\}$ is at most $2$
$(D)$ If $L$ has more than one element,then $L$ has infinitely many elements

The equation $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$ represents a circle with:

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