In the Argand plane,the values of $z$ satisfying the equation $|z-1|=|i(z+1)|$ lie on

If $z_1 = 10 + 6i$,$z_2 = 4 + 6i$ and $z$ is any complex number such that the argument of $\frac{z - z_1}{z - z_2}$ is $\frac{\pi}{4}$,then

The area of the triangle formed by the complex numbers $z$,$iz$,and $z+iz$ as vertices in the Argand diagram is:

Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as:
$S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$
$S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$
$S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$
Then the set $S_{1} \cap S_{2} \cap S_{3}$

If $\left|\frac{z}{1+i}\right|=2$,where $z=x+iy$ and $i=\sqrt{-1}$ represents a circle,then the centre $C$ and radius $r$ of the circle are:

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,