Show that `|(2z + 5)(sqrt2-i)|=sqrt(3) |2z+5|`, where z is a complex number.

Solve |z|+z= 2+ i , where z is a complex number

Solve the equation 2z=|\z|+2i , where z is a complex number.

Solve the equation z^2 +|z|=0 , where z is a complex number.

Number of solutions of Re(z^(2))=0 and |Z|=a sqrt(2) , where z is a complex number and a gt 0 , is

Complex numbers z satisfy the equaiton |z-(4//z)|=2 Locus of z if |z-z_(1)| = |z-z_(2)| , where z_(1) and z_(2) are complex numbers with the greatest and the least moduli, is

Statement-1 : IF |z+1/z| =a , where z is a complex number and a is a real number, the least and greatest values of |z| are (sqrt(a^(2)+4-a))/2 and (sqrt(a^(2)+ 4)+a)/2 and Statement -2 : For a equal ot zero the greatest and the least values of |z| are equal .

If |z_(1)|=|z_(2)|=|z_(3)|=1 and z_(1)+z_(2)+z_(3)=sqrt(2)+i , then the complex number z_(2)barz_(3)+z_(3)barz_(1)+z_(1)barz_(2) , is

The point z_1=3+sqrt(3)i and z_2=2sqrt(3)+6i are given on a complex plane. The complex number lying on the bisector of the angel formed by the vectors z_1a n dz_2 is z=((3+2sqrt(3)))/2+(sqrt(3)+2)/2i z=5+5i z=-1-i none of these

If z_(1),z_(2)andz_(3) are the vertices of an equilasteral triangle with z_(0) as its circumcentre , then changing origin to z^(0) ,show that z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=0, where z_(1),z_(2),z_(3), are new complex numbers of the vertices.

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.