Geometrically `Re(z^(2)-i)=2,where i=sqrt(-1)` and Re is the real part, represents.

Prove that z=i^(i), where i=sqrt(-1), is purely real.

The real part of (1-i)^(-i), where i=sqrt(-1) is

if |z-i Re(z)|=|z-Im(z)| where i=sqrt(-1) then z lies on

Prove that z=i^i, where i=sqrt-1, is purely real.

Prove that z=i^i, where i=sqrt-1, is purely real.

Prove that z=i^i, where i=sqrt-1, is purely real.

For a complex number z ,let Re(z) denote the real part of z . Let S be the set of all complex numbers z satisfying z^(4)-|z|^(4)=4iz^(2) ,where i=sqrt(-1) . Then the minimum possible value of |z_(1)-z_(2)|^(2) ,where, z_(1),z_(2) in S with Re(z_(1))>0 and Re(z_(2))<0, is

If Re((z-1)/(2z+i))=1, where z=x+iy ,then the point (x,y) lies on a