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

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

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)

Let z=9+ai, where i=sqrt(-1) and a be non-zero real. If Im(z^(2))=Im(z^(3)) , sum of the digits of a^(2) is

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

(i)If z is a complex no.such that |z|=1; prove that (z-1)/(z+1) is purely imaginary.what will be the conclusion if z=1 (ii) Find real theta such that (3+2i sin theta)/(1-2i sin theta) is purely real

The equation z^(2)-i|z-1|^(2)=0, where i=sqrt(-1), has.

Find real theta such that (3+2i sin theta)/(1-2i sin theta) is purely real.

If z^(3)+(3+2i)z+(-1+ia)=0, " where " i=sqrt(-1) , has one real root, the value of a lies in the interval (a in R)