Home
Class 12
MATHS
If |z|=1 and w=(z-1)/(z+1) (where z!=-1)...

If `|z|=1` and `w=(z-1)/(z+1)` (where `z!=-1),` then `R e(w)` is 0 (b) `1/(|z+1|^2)` `|1/(z+1)|,1/(|z+1|^2)` (d) `(sqrt(2))/(|z|1""|^2)`

A

0

B

`(-1)/(|z+1|^(2))`

C

`|(z)/(z=1)|*(1)/(|z+1|^(2))`

D

`(sqrt(2))/(|z+1|^(2))`

Text Solution

Verified by Experts

The correct Answer is:
a
Promotional Banner

Topper's Solved these Questions

  • COMPLEX NUMBERS

    ARIHANT MATHS ENGLISH|Exercise EXAMPLE|10 Videos
  • COMPLEX NUMBERS

    ARIHANT MATHS ENGLISH|Exercise EXAMPLE(Single integer answer type questions)|1 Videos
  • CIRCLE

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|16 Videos
  • CONTINUITY AND DIFFERENTIABILITY

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|20 Videos

Similar Questions

Explore conceptually related problems

If |z|=1 and z'=(1+z^(2))/(z) , then

If z > 0 ? (1) (z + 1)(z)(z - 1) (2) |z| < 1

If |z_(1)|=|z_(2)| and arg (z_(1)//z_(2))=pi, then find the of z_(1)z_(2).

If Real ((2z-1)/(z+1)) =1, then locus of z is , where z=x+iy and i=sqrt(-1)

If Real ((2z-1)/(z+1)) =1, then locus of z is , where z=x+iy and i=sqrt(-1)

If |z|=1 and let omega=((1-z)^2)/(1-z^2) , then prove that the locus of omega is equivalent to |z-2|=|z+2|

If |z_1|=1a n d|z_2|=2,t h e n Max (|2z_1-1+z_2|)=4 Min (|z_1-z_2|)=1 |z_2+1/(z_1)|lt=3 Min (|z_1=z_2|)=2

If |z_1/z_2|=1 and arg (z_1z_2)=0 , then a. z_1 = z_2 b. |z_2|^2 = z_1*z_2 c. z_1*z_2 = 1 d. none of these

If |z_(1)|= |z_(2)|= ….= |z_(n)|=1 , prove that |z_(1) + z_(2) + …+ z_(n)|= |(1)/(z_(1)) + (1)/(z_(2)) + …(1)/(z_(n))|

if z_(1) = 3i and z_(2) =1 + 2i , then find z_(1)z_(2) -z_(1)