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⇒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