If `(z-1)/(z+1)` is purely imaginary `(zne-1)`, then show that |z|=1

If z_1,z_2 are comples numbers such that (2 z_1)/(3z_2) is purely imaginary number, then find |(z_1-z_2)/ (z_1+ z_2)| .

If (z+1)/(z+i) is a purely imaginary number (where (i=sqrt(-1) ), then z lies on a

If iz^3+z^2-z+i = 0 , then show that |z|=1.

If |z|=1, prove that (z-1)/(z+1) (z ne -1) is purely imaginary number. What will you conclude if z=1?

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 the complex number z_1 and z_2 be such that arg(z_1)-arg(z_2)=0 , then show that |z_1-z_2|=|z_1|-|z_2| .

If |z|=1 and w=(z-1)/(z+1) (where z!=-1), then R e(w) is

If |z-1|=1, where z is a point on the argand plane, show that (z-2)/(z)=i tan (argz),where i=sqrt(-1).

If z = x + iy and w=(1-iz)/(z-i) , show that : |w|=1 implies z is purely real.

Given that |z-1|=1, where z is a point on the argand planne , show that (z-2)/(z)=itan (arg z), where i=sqrt(-1).