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:

If $z = x + iy$ is a complex number,then the equation $\left|\frac{z+i}{z-i}\right| = \sqrt{3}$ represents the

If $z=x+iy, x, y \in R$ and the imaginary part of $\frac{\bar{z}-1}{\bar{z}-i}$ is $1$,then the locus of $z$ is

Let $\arg(z)$ represent the principal argument of the complex number $z$. The curves $|z|=3$ and $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ intersect:

$z_1$ 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:

The area (in sq units) of the triangle whose vertices are the points represented by the complex numbers $0, z$,and $z e^{i \alpha}$ $(0 < \alpha < \pi)$ is: