" Tif "(v)z+(1)/(z)=3,z!=0

Modulus of a Complex Number & its properties If z;z_1;z_2inCC then (i)|z|=0hArrz=0 i.e. Re(z)=Im(z)=0 (ii)|z|=|barz|=|-z| (iii) -|z|leRe(z)le|z|;-|z|leIm(z)le|z| (iv) zbarz=|z|^2 (v)|z_1z_2|=|z_1||z_2| (vi)|(z_1)/(z_2)|=|z_1|/|z_2|; z_2!=0

(v) |(z_1)/(z_2)|=|z_1| /|z_2|

If |u|=|v|=1uv!=-1 and z=(u-v)/(1+uv) then (a) |z|=1( b) Re(z)=0 (c) Im(z)=0 (d) Re(z)=Im(z)

The centre of a square ABCD is at z_(0). If A is z_(1) ,then the centroid of the ABC is 2z_(0)-(z_(1)-z_(0))(b)(z_(0)+i((z_(1)-z_(0))/(3))(z_(0)+iz_(1))/(3) (d) (2)/(3)(z_(1)-z_(0))

Show that if z_(1)z_(2)+z_(3)z_(4)=0 and z_(1)+z_(2)=0 ,then the complex numbers z_(1),z_(2),z_(3),z_(4) are concyclic.

Show that if z_(1)z_(2)+z_(3)z_(4)=0 and z_(1)+z_(2)=0 ,then the complex numbers z_(1),z_(2),z_(3),z_(4) are concyclic.

If z_(1);z_(2) and z_(3) are the vertices of an equilateral triangle; then (1)/(z_(1)-z_(2))+(1)/(z_(2)-z_(3))+(1)/(z_(3)-z_(1))=0