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

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

A particle P starts from the point z_0=1+2i , where i=sqrt(-1) . It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z_1dot From z_1 the particle moves sqrt(2) units in the direction of the vector hat i+ hat j and then it moves through an angle pi/2 in anticlockwise direction on a circle with centre at origin, to reach a point z_2dot The point z_2 is given by (a) 6+7i (b) -7+6i (c) 7+6i (d) -6+7i

If x=a+bi is a complex number such that x^2=3+4i and x^3=2+11i , where i=sqrt-1 , then (a+b) equal to

let z= (-1+sqrt(3i))/2, where i=sqrt(-1) and r,s epsilon P1,2,3}. Let P= [((-z)^r, z^(2s)),(z^(2s), z^r)] and I be the idenfity matrix or order 2. Then the total number of ordered pairs (r,s) or which P^2=-I is

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

If z=x+i ya n dw=(1-i z)//(z-i) , then show that |w|=1 z is purely real.

If z=(3+7i)(a+ib) , where a , b in Z-{0} , is purely imaginery, then minimum value of |z|^(2) is

Prove that the equation Z^3+i Z-1=0 has no real roots.