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

Let the curve $z(1+i)+\bar{z}(1-i)=4, z \in \mathbb{C}$,divide the region $|z-3| \leq 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha-\beta|$ equals :

$z_1$ and $z_2$ are two complex numbers such that $\left|z_1-z_2\right| < k$. If a complex number $z$ satisfies the condition $\left|z-z_1\right|+\left|z-z_2\right|=k$,then $z$ lies on:

If $z=x+iy$ and the point $P$ in the Argand plane represents $z$,then the locus of $z$ satisfying the equation $|z-2|+|z-2i|=4$ is

$A$ circle whose radius is $r$ and centre is $z_0$,then the equation of the circle 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