Embibe Experts Solutions for Chapter: Complex Numbers, Exercise 1: Assam CEE 2017

If $\left|z-i\right|\le 2$ and $z_0=5+3i$, then the maximum value of $\left|iz+z_0\right|$ is

If the equation $a{z}^2+2bz+a=0$, where $a,b$ are complex numbers, has both roots purely real, then

If $z=1+i\cot \theta ,\frac{3\pi }{2}<\theta <2\pi$ then $\left|-\frac{4z}{i}\right|$ is equal to

Let $z_1$ and $z_2$ be two complex numbers such that ${\overline{z}}_1+i{z}_2=0$ and $\arg \left(z_1{\overline{z}}_2\right)=\pi$, then $\arg \left({\overline{z}}_2\right)$ is equal to

If $z^2+z+1=0$, then the value of  ${\left(z+\frac{1}{z}\right)}^2+{\left(z^2+\frac{1}{z^2}\right)}^2+{\left(z^3+\frac{1}{z^3}\right)}^2+\dots +{\left(z^{21}+\frac{1}{z^{21}}\right)}^2$

If $\mathrm{z}=x+iy$ and $\omega =\frac{1-iz}{z-i}$, then $\left|\omega \right|=1$ implies that, in the complex plane

If $\omega$ is an imaginary cube root of unit, then ${\left(1+\omega -{\omega }^2\right)}^7$ is equal to

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