If `u+iv=(2+i)/(z+3) and z=x+iy` find u, v

If u=(2z+i)/(z-ki) where z=x+iy and k gt 0 Curve Re(u)+Im(u)= cuts y-axis at two point P and Q such that PQ=5 then value of k is

Given that abs(z+i)=abs(z-i)=abs(z-1) and z=x+iy find number of pairs (x,y)

If u+iv=(3i)/(x+iy+2), then y in terms of u,v is

If u + iv = (3i)/(x + iy + 2) , then y in terms of u , v is

If z+sqrt(2)|z+1|+i=0 and z=x+iy then

The locus of z satisfying the inequality |(z+2i)/(2z+i)|lt1 where z = x + iy , is :

If U=(y)/(z)+(z)/(x), then find x(del u)/(del x)+y(del u)/(del y)+z(del u)/(del z) .

If (x+iy)^(3) = u + iy , then show that u/x + v/y = 4(x^(2)-y^(2))