Given equation is z+√2|z+1|+i=0 put z=x+iy in the given equation. (x+iy)+√2|x+iy+1|+i=0 ⇒x+iy+√2[√(x+1)2+y2]+i=0 Now, equating real and imaginary part, we get x+√2√(x+1)2+y2=0 and y+1=0⇒y=−1 ⇒x+√2√(x+1)2+(−1)2=0(∵y=−1) ⇒√2√(x+1)2+1=−x ⇒2[(x+1)2+1]=x2 ⇒x2+4x+4=0 ⇒x=−2 Thus, z=−2+i(−1)⇒|z|=√5