The number of solutions of `sqrt(2)|z-1|=z-i, where z=x+iy` is (A) 0 (B) 1 (C) 2 (D) 3

The number of solutions of z + sqrt2 (| z+ 1|) + i=0

If z+sqrt(2)|z+1|+i=0 , then z= (A) 2+i (B) 2-i (C) -2-i (D) -2+i

If z+sqrt(2)|z+1|+i=0 , then z= (A) 2+i (B) 2-i (C) -2-i (D) -2+i

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

If B: {z: |z-3-4i|}=5 and C={z:Re[(3+4i)z]=0} then the number of elements in the set B intesection C is (A) 0 (B) 1 (C) 2 (D) none of these

Solve the equation |z|+z=2+i, where z = x + iy.

Maximum value of |z+1+i|, where z in S is (a) sqrt(2) (b) 2 (c) 2sqrt(2) (d) 3sqrt(2)

The locus of z satisfying the inequality |(z+2i)/(2z+1)|<1, where z=x+iy, is :

If z=2-i sqrt(3) then z^(4)-4z^(2)+8z+35 is (A) 6(B)0(C)1 (D) 2

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