If `|z| =1 ` and `w=(z-1)/(z+1)` (where `z != -1`) then `Re(w)` is (A) 0 (B) `-1/|z+1|^2` (C) `|z/(z+1)| 1/|z+1|^2` (D) `sqrt2/|z+1|^2`

Similar Questions

Explore conceptually related problems

If |z|=1 and omega=(z-1)/(z+1) (where z in -1 ), then Re (omega) is

If |z|=1 and w=(z-1)/(z+1) (where z!=-1) then Re(w) is (A) 0(B)-(1)/(|z+1|^(2))(C)|(z)/(z+1)|(1)/(|z+1|^(2))(D)(sqrt(2))/(|z+1|^(2))

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

If w=(z)/(z-(1)/(3)i) and |w|=1, then z lies on

If |(z-1)/(z-4)|=2 and |(w-4)/(w-1)|=2 , (where z,w in C ) .Then the value of |z-w|_(max)+|z-w|_(min)

If |u|=|v|=1uv!=-1 and z=(u-v)/(1+uv) then (a) |z|=1( b) Re(z)=0 (c) Im(z)=0 (d) Re(z)=Im(z)

If |z|=1 and |omega-1|=1 where z, omega in C ,then the range of values of |2z-1|^(2)+|2 omega-1|^(2) equals

Prove that Re (z_ (1) z_ (2)) = Re z_ (1) Rez_ (2) -Im z_ (1) Im z_ (2)

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