The proof proceeds by constructing a near-identity coordinate change that eliminates all non-resonant monomials up to cubic order. **Step 1: Setup.** After a real change of basis using the eigenvectors of $A$ (see the [Neimark-Sacker Bifurcation](/page/Neimark-Sacker%20Bifurcation) page), the map in complex coordinates $z = u + iv$ takes the form: \begin{align*} \bar{z} = e^{i\theta_0}z + \sum_{j+k \ge 2} c_{jk}\,z^j\bar{z}^k. \end{align*} **Step 2: Eliminating a non-resonant monomial.** To remove a monomial $c_{jk}\,z^j\bar{z}^k$, try the substitution $z = w + h_{jk}\,w^j\bar{w}^k$. At leading order $\bar{w} \approx e^{i\theta_0}w$, so $\bar{z} \approx e^{i\theta_0}w + h_{jk}\,e^{i(j-k)\theta_0}\,w^j\bar{w}^k$ on the left side, and $e^{i\theta_0}z + c_{jk}\,z^j\bar{z}^k \approx e^{i\theta_0}w + (e^{i\theta_0}h_{jk} + c_{jk})\,w^j\bar{w}^k$ on the right. Matching coefficients: \begin{align*} h_{jk}\left(e^{i(j-k)\theta_0} - e^{i\theta_0}\right) = c_{jk}. \end{align*} This has a solution $h_{jk} = c_{jk}/(e^{i(j-k)\theta_0} - e^{i\theta_0})$ provided the **divisor** $d_{jk} := e^{i(j-k)\theta_0} - e^{i\theta_0}$ is nonzero. **Step 3: Check all divisors.** For the six quadratic and non-resonant cubic monomials, the divisor $d_{jk}$ vanishes when $e^{i(j-k-1)\theta_0} = 1$. The values of $|j - k - 1|$ across all six monomials are $1, 1, 3$ (quadratics) and $2, 2, 4$ (non-resonant cubics). The non-resonance hypothesis $e^{im\theta_0} \neq 1$ for $m = 1, 2, 3, 4$ ensures all six divisors are nonzero, so all six monomials can be eliminated. **Step 4: The resonant term.** The monomial $z^2\bar{z}$ has $j - k = 1$, giving divisor $d_{21} = e^{i\theta_0} - e^{i\theta_0} = 0$. This term **cannot** be eliminated and remains in the normal form. Its coefficient after the transformation is $(L_1 + i\Omega_1)e^{i\theta_0}$, where $L_1 = \operatorname{Re}(c_{21}^{\mathrm{new}})$ accounts for both the original $c_{21}$ and the indirect contributions from eliminating the quadratic terms. **Step 5: Polar form.** Factor out $e^{i\theta_0}$: $\bar{w} = e^{i\theta_0}w(1 + (L_1 + i\Omega_1)|w|^2 + O(|w|^4))$. Taking the modulus: \begin{align*} r_{n+1} = r_n\left|1 + (L_1 + i\Omega_1)r_n^2 + O(r_n^4)\right| = r_n\sqrt{(1 + L_1 r_n^2)^2 + \Omega_1^2 r_n^4 + O(r_n^6)} = r_n(1 + L_1 r_n^2 + O(r_n^4)). \end{align*} The $\Omega_1$ term is absorbed into the $O(r_n^4)$ correction of the modulus because $(1 + L_1 r^2)^2 + \Omega_1^2 r^4 = 1 + 2L_1 r^2 + (L_1^2 + \Omega_1^2)r^4$, and $\sqrt{1 + 2L_1 r^2 + O(r^4)} = 1 + L_1 r^2 + O(r^4)$.