Home
Class 12
MATHS
If w=alpha+ibeta where beta ne 0 and z ...

If `w=alpha+ibeta` where `beta ne 0 ` and `z ne 1` satisfies the condition that `((w- bar wz)/(1-z))` is purely real then the set of values of z is

Promotional Banner

Similar Questions

Explore conceptually related problems

If w=alpha+ibeta , where beta ne 0 and z ne 1 , satisfies the condition that ((w-barwz)/(1-z)) is purely real, then the set of values of z is

If w=alpha+ibeta, where beta!=0 and z!=1 , satisfies the condition that ((w- bar w z)/(1-z)) is a purely real, then the set of values of z is |z|=1,z!=2 (b) |z|=1a n dz!=1 (c) z=bar z (d) None of these

If w=alpha+ibeta, where beta!=0 and z!=1 , satisfies the condition that ((w- bar w z)/(1-z)) is a purely real, then the set of values of z is |z|=1,z!=2 (b) |z|=1a n dz!=1 (c) z=bar z (d) None of these

If w=alpha+ibeta, where beta!=0 and z!=1 , satisfies the condition that ((w- bar w z)/(1-z)) is a purely real, then the set of values of z is |z|=1,z!=2 (b) |z|=1a n dz!=1 (c) z=bar z (d) None of these

If w=alpha+ibeta, where beta!=0 and z!=1 , satisfies the condition that ((w- barw z)/(1-z)) is a purely real, then the set of values of z is |z|=1,z!=2 (b) |z|=1a n dz!=1 z= z (d) None of these

If w=alpha+i beta, where beta!=0 and z!=1 satisfies the condition that ((w-wz)/(1-z)) is a purely real,then the set of values of z is |z|=1,z!=2( b) |z|=1 and z!=1z=z(d) None of these

If (w- bar(w)z)/(1-z) is purely real where w= alpha + ibeta, beta ne 0 and z ne 1 , then the set of the values of z is

Let z ne 1 be a complex number and let omeg= x+iy, y ne 0 . If (omega- bar(omega)z)/(1-z) is purely real, then |z| is equal to

If ((z-1)/(z+1)) is purely an imaginary number and z ne -1 then find the value of |z|.