if z+√2|z+1|+i=0 and z=x+iythen
A)x=-2
B)x=2
C)y=-2
D)y=1

z+2z+1+i=0x+iy+2x+1+iy+i = 0x+2x+12+y2+iy+1=0+i0Npw comparing the real and imaginary part we get, x+2x+12+y2=0....1or y+1=0  y = -1Putting this value in equation 1 we get, x +2x+12+1=0x = -2x+12+1Squaring we get, x2=2x+12+2x2=2x2+4x+2+2x2+4x+4=0x+22=0x +2=0x = -2

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