Solve for `z : z^2-(3-2i)z=(5i-5)dot`

Solve the equation z^3= z (z!=0)dot

If the maximum value of |3z +9 - 7i | if |z+2-i|=5 is 5K, then find k

The locus of z such that arg[(1-2i)z-2+5i]= (pi)/(4) is a

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

In the Argands plane what is the locus of z(!=1) such that a rg{3/2((2z^2-5z+3)/(2z^2-z-2))}=(2pi)/3dot

For all complex numbers z_1,z_2 satisfying |z_1|=12 and |z_2-3-4i|=5 , find the minimum value of |z_1-z_2|

A(z_1),B(z_2),C(z_3) are the vertices of the triangle A B C (in clockwise). If /_A B C=pi//4 and A B=sqrt(2)(B C) , then prove that z_2=z_3+i(z_1-z_3)dot

Find the radius and centre of the circle z bar(z) - (2 + 3i)z - (2 - 3i)bar(z) + 9 = 0 where z is a complex number