Embibe Experts Solutions for Chapter: Complex Numbers, Exercise 1: JEE Main - 10th April 2016

Let $\mathrm{z}$ be a complex number such that $\left|\frac{z-2i}{z+i}\right|=2, z\ne -i$. Then $z$ lies on the circle of radius $2$ and centre

For two non-zero complex number $z_1$ and $z_2$, if $Re\left(z_1{z}_2\right)=0$ and $Re\left(z_1+z_2\right)=0$, then which of the following are possible?

(A) $Im\left({\mathrm{z}}_1\right)>0$ and $Im\left({\mathrm{z}}_2\right)>0$
(B) $Im\left({\mathrm{z}}_1\right)<0$ and $Im\left({\mathrm{z}}_2\right)>0$
(C) $Im\left({\mathrm{z}}_1\right)>0$ and $Im\left({\mathrm{z}}_2\right)<0$
(D) $Im\left({\mathrm{z}}_1\right)<0$ and $Im\left({\mathrm{z}}_2\right)<0$

Choose the correct answer from the options given below:

Let $\alpha =8-14\mathrm{i}, \mathrm{A}=\left\{\mathrm{z}\in \mathrm{ℂ}:\frac{\alpha \mathrm{z}-\overline{\alpha }\overline{\mathrm{z}}}{{\mathrm{z}}^2-(\overline{\mathrm{z}}{)}^2-112\mathrm{i}}=1\right\}$ and $B=\left\{z\in ℂ:\left|z+3i\right|=4\right\}$

Then, $\sum _{z\in A\cap B}(\mathit{Re}z-Imz)$   is equal to  ________

Let $z=1+i$ and $z_1=\frac{1+\mathrm{i}\overline{\mathit{z}}}{\overline{\mathit{z}}(1-z)+\frac{1}{z}}·$ Then $\frac{12}{\pi }$ $\arg \left(z_1\right)$ is equal to

For all $z\in C$ on the curve $C_1:\left|z\right|=4$, let the locus of the point $\mathrm{z}+\frac{1}{\mathrm{z}}$ be the curve $C_2$. Then

The complex number $z=\frac{i-1}{\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}}$ is equal to:

If the center and radius of the circle $\left|\frac{z-2}{z-3}\right|=2$ are respectively $\left(\alpha ,\beta \right)$ and $\gamma$, then $3\left(\alpha +\beta +\gamma \right)$ is equal to

Let $a,b$ be two real numbers such that $ab<0$. If the complex number $\frac{1+ai}{b+i}$ is of unit modulus and $a+ib$ lies on the circle $\left|z-1\right|=\left|2z\right|$, then a possible value of $\frac{1+\left[a\right]}{4b}$, where $\left[t\right]$ is greatest integer function, is :