Embibe Experts Solutions for Chapter: Complex Numbers, Exercise 1: Exercise 1

If a complex number $z$ lies on a circle of radius $3$ and centre at $z_0=0$ then the complex number $\left(-3+9z\right)$ lies on a circle of radius

Given that the equation $z^2+(a+ib)z+c+id=0$ where $a, b, c, d$ are non-zero, has a real root then

It is given that the equation $|z{|}^2-2iz+2\alpha (\lambda +i)=0$ possesses a solution for all $\alpha \in R,$ then the number of integral value(s) of ' $\lambda$ ' for which it is true is

The complex number $z_1,z_2$ satisfies the equation $z+1+8 i=\left|z\right|\left(1+i\right)$, where $i=\sqrt{-1}$ then the equation whose roots are $\left|z_1\right|$ and $\left|z_2\right|$ is

The complex number $z$ and $\omega$ are such that $\left|z\right|=\left|\omega \right|=1$. Then, the locus of $\frac{z+\omega }{1+z\omega }$ is

Let $\omega =\cos 3°+i\sin 3°$ then $\sum _{r=1}^{10}Re\left({\omega }^{2r-1}\right)$ equals

If the points $A,B$ and $C$ are the affixes of the complex number $z_1,z_2$ and $z_3$ in the argand plane $z$ in any complex number such that

$\left|z-z_1\right|=\frac{1}{2}\left(\left|z-z_2\right|+\left|z-z_3\right|\right)$,

$\left|z-z_2\right|=\frac{1}{2}\left(\left|z-z_3\right|+\left|z-z_1\right|\right)$ and

$\left|z-z_3\right|=\frac{1}{2}\left(\left|z-z_1\right|+\left|z-z_2\right|\right)$, then the affix of $z$ is

The locus of $z=i+2\exp \left(i\left(\theta +\frac{\pi }{4}\right)\right)$, (where $\theta$ parameter) is