Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

If $z_1, z_2, z_3 \in \mathbb{C}$ are the vertices of an equilateral triangle,whose centroid is $z_0$,then $\sum_{k=1}^3 (z_k - z_0)^2$ is equal to

Let the complex numbers $\alpha$ and $\left(\frac{1}{\bar{\alpha}}\right)$ lie on circles $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ and $\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2$ respectively. If $z_0=x_0+i y_0$ satisfies the equation $2|z_0|^2=r^2+2$,then $|\alpha|=$

The complex number $z$ satisfying the equation $|z-i|=|z+1|=1$ is

If the complex number $z=x+iy$,where $i=\sqrt{-1}$,satisfies the condition $|z+1|=1$,then $z$ lies on

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,