Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • How to solve this problem- - Complex numbers and quadratic equations - JEE Main-6

Letw\left ( Im\: w\neq 0 \right ) be a complex number. Then the set of all complex numbers z satisfying the equation w-\bar{w}z=k\left ( 1-z \right ) for some real number k, is :

  • Option 1)

    \left \{ z:\left | z \right | =1\right \}

  • Option 2)

    \left \{ z:z=\bar{z} \right \}

  • Option 3)

    \left \{ z:z\neq 1 \right \}

  • Option 4)

    \left \{ z:\left | z\right |= 1,z\neq 1 \right \}

 

Answers (2)

best_answer

As we have learned

Property of Modulus of z(Complex Number) -

\left |\frac{z_{1}}{z_{2}} \right |=\frac{\left |z_{1} \right |}{\left |z_{2} \right |}

- wherein

|.| denotes modulus of complex number

 

 

Definition of Modulus of z(Complex Number) -

\left | z \right |=\sqrt{a^{2}+b^{2}} is the distance of z from origin in Argand plane

- wherein

Real part of z = Re (z) = a & Imaginary part of z = Im (z) = b

 

 \omega -\bar{\omega }z= k(1-z)

\Rightarrow \omega -k =\bar{\omega }z-kz

\Rightarrow z= \frac{\omega -k}{\omega -k}\neq1

And 

\Rightarrow |z|= \frac{|\omega -k|}{|\omega -k|}

= \frac{|x+iy-k |}{|x-iy-k|}

= \frac{\sqrt{(x-k)^2+y^2 }}{\sqrt{(x-k)^2+y^2}}= 1

 

 


Option 1)

\left \{ z:\left | z \right | =1\right \}

Option 2)

\left \{ z:z=\bar{z} \right \}

Option 3)

\left \{ z:z\neq 1 \right \}

Option 4)

\left \{ z:\left | z\right |= 1,z\neq 1 \right \}

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question