“…where A, B, C, D, H, K, L, M, N, P, Q ∈ C.For H = Q = 0, the solution of the center-focus problem can be found in[1].Letω 1 (x) = Nx 5 (H + Qx) 3 − x 3 (C + Mx)(H + Qx) 2 (1 + Dx + P x 2 ) + 3x 2 (B + Lx)(H + Qx)(1 + Dx + P x 2 ) 2 + (1 − Ax − Kx 2 )(1 + Dx + P x 2 ) 3 , ω 2 (x) = Nx 3 (H + Qx) 2 − x(2C + D + 2(M + P )x)(H + Qx)(1 + Dx + P x 2 )/3 + ((3B + 2H)/3 + (L + Q)x)(1 + Dx + P x 2 ) 2 .…”