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

The number of points $z$ on the Argand plane which satisfy the conditions $\operatorname{Re}\left(\frac{z-2}{z-4i}\right)=0$ and $\operatorname{Im}\left(\frac{z-2}{z-4i}\right)=1$ simultaneously is

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}$ and $|z_{1}|=|z_{2}|$. If $\operatorname{Re}(z_{1}) > 0$ and $\operatorname{Im}(z_{2}) < 0$, then $\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is

In the Argand diagram,if $O, P$ and $Q$ represent the origin,the complex number $z$ and the complex number $z + iz$ respectively,then the angle $\angle OPQ$ is

Let $z = x + iy$ be a non-zero complex number such that $z^{2} = i|z|^{2},$ where $i = \sqrt{-1}.$ Then $z$ lies on the:

Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1+\theta_2+\ldots+\theta_{10}=2 \pi$. Define the complex numbers $z_1=e^{i \theta_1}, z_k=z_{k-1} e^{i \theta_k}$ for $k=2,3, \ldots, 10$,where $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ given below:
$P: |z_2-z_1|+|z_3-z_2|+\ldots+|z_{10}-z_9|+|z_1-z_{10}| \leq 2 \pi$
$Q: |z_2^2-z_1^2|+|z_3^2-z_2^2|+\ldots+|z_{10}^2-z_9^2|+|z_1^2-z_{10}^2| \leq 4 \pi$
Then,