Show that if `iz^3+z^2-z+i=0`, then `|z|=1`

If 8iz^3+12z^2-18z+27i=0, then (a). |z|=3/2 (b). |z|=2/3 (c). |z|=1 (d). |z|=3/4

If z_1, z_2 and z_3 , are the vertices of an equilateral triangle ABC such that |z_1 -i = |z_2 -i| = |z_3 -i| .then |z_1 +z_2+ z_3| equals:

If z_1, z_2, z_3 are three complex numbers such that 5z_1-13 z_2+8z_3=0, then prove that [(z_1,(bar z )_1, 1),(z_2,(bar z )_2 ,1),(z_3,(bar z )_3 ,1)]=0

If z is a complex number satisfying z^4+z^3+2z^2+z+1=0 then the set of possible values of z is

If z= re^(i theta)" then "|e^(iz)|=

If |z| = 1, show that 2 le |z^(3) - 3| le 4.

If z_1 = 4 - 7i , z_2 = 2 + 3i and z_3 = 1 + i show that z_1 + (z_2 + z_3) = (z_1 + z_2)+z_3