`|z|-z=1+2i`

| z | -z = 1 + 2i

Solve :z+|z|=1+2i , where z=x+iy,x,yinRR .

If |z - 1| = |z + 1| = |z - 2i|, then value of |z| is

Locus of z in the following curves: Im(z)=|z-(1+2i)| and 4-Im(z)=|z-(1+2i)| represent A and B respectively. If locus of z in arg (z-(1+2i))=theta intersect A and B at points P(z_(1)) and Q(z_(2)) respectively, then minimum value of |z_(1)-(1+2i)||z_(2)-(1+2i)| is: (Re(z_(1))+Re(z_(2))!=2)

locus of the point z satisfying the equation |z-1|+|z-i|=2 is

locus of the point z satisfying the equation |z-1|+|z-i|=2 is

locus of the point z satisfying the equation |z-1|+|z-i|=2 is

which of the foolowing satisfies |z+1|=z+2+2i