Properties of Conjugate, Modulus and Argument of Complex Numbers

$z_1,z_2,z_3$ are three points lying on the circle $\left|z\right|=1$, then maximum value of ${\left|z_1-z_2\right|}^2+{\left|z_2-z_3\right|}^2+{\left|z_3-z_1\right|}^2$ is equal to -

Let $S=\left\{z\in C : z\left(i{z}_1-1\right)=z_1+1, \left|z_1\right|<1\right\}$. Then, for all $z\in S,$ which one of the following is always true?

If $\left|z_1\right|=\left|z_2\right|$ and $\arg \left(\frac{z_1}{z_2}\right)=\pi$, then $z_1+z_2$ is equal to

If $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1$ and $z_1+z_2+z_3=\sqrt{2}+i$, then the number $z_1{\overline{z}}_2+z_2{\overline{z}}_3+z_3{\overline{z}}_1$ is

If $z_1, z_2$ and $z_3$ are any three distinct complex numbers such that $\left|z_1\right|=1, \left|z_2\right|=2, \left|z_3\right|=4, \arg z_2=\arg z_1-\pi$ and $\arg z_3=\arg z_1+\frac{\pi }{2},$ then $z_2{z}_3$ is equal to

If $\frac{3+2i\sin \theta }{1-2i\sin \theta }$ is purely real, then $\theta$ is equal to

If the number $\frac{z-1}{z+1}$ is purely imaginary, then

The modulus of the complex number $Z=\frac{\left(1-i\sqrt{3}\right)\left(\cos \theta +i\sin \theta \right)}{2\left(1-i\right)\left(\cos \theta -i\sin \theta \right)}$ is

If $\left(a+ib\right)\left(c+id\right)\left(e+if\right)\left(g+ih\right)=A+iB$, then $\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)$ is equal to

For all complex numbers $z_1,z_2$ satisfying $\left|z_1\right|=12$ and $\left|z_2-3-4i\right|=5,$ then the minimum value of $\left|z_1-z_2\right|$ is

If $\left|z_1\right|=\left|z_2\right|=\dots .=\left|z_n\right|=1,$ then $\left|\frac{1}{z_1}+\frac{1}{z_2}+\dots \dots +\frac{1}{z_n}\right|$.

If $\left|z-\frac{4}{z}\right|=2$, then the maximum value of $\left|z\right|$ is equal to

If $\left|z_1-1\right|<1, \left|z_2-2\right|<2, \left|z_3-3\right|<3,$ then $\left|z_1+z_2+z_3\right|$

If $\left|z_1\right|=2, \left|z_2\right|=3, \left|z_3\right|=4$ and $\left|z_1+z_2+z_3\right|=2,$ then the value of $\left|4z_2{z}_3+9z_3{z}_1+16z_1{z}_2\right|$

Let arg$\left(z_k\right) = \frac{\left(2k+1\right)\pi }{n}$ where $k=1, 2, 3, 4...,n$. If arg$\left(z_1{z}_2{z}_3...z_n\right)=\pi$, then $n$ must be of form $\left(m\in \mathrm{I}\right)$

The region represented by the inequality $\left|2z-3i\right|<\left|3z-2i\right|$ is :

$a,b,c$ are three complex number on the unit circle $\left|z\right|=1$, such that $abc =a +b+c$. Then $|ab+bc+ca|$ is equal to 

Let $z$ be a complex number satisfying $\left|z+16\right|=4\left|z+1\right|$, then

Let $\left|z_1\right|=3,\left|z_2\right|=2$ and $z_1+z_2+z_3=3+4i$. If the real part of $\left(z_1\stackrel{-}{z_2}+z_2\stackrel{-}{z_3}+z_3\stackrel{-}{z_1}\right)$ is equal to $4$, then $\left|z_3\right|$ is equal to (where, $i^2=-1$)