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

Prove that z=i^i ,w h e r ei=sqrt(-1) , is purely real.

If z=i^(i) where i=sqrt(-)1 then |z| is equal to

If Z=(i^i)^i" where "i=sqrt(-1) then

If z=i^(i^(i)) where i=sqrt-1 then |z| is equal to

If z=i^(i^(i)) where i=sqrt-1 then |z| is equal to

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

Prove that if the ratio (z - i)/( z - 1) is purely imaginary then the point z lies on the circle whose centre is at the point 1/2(1+i)and " radius is " 1/sqrt(2)

If z=x+iyandw=(1-iz)/(z-i) such that |w|=1 , then show that z is purely real.

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