In chiral perturbation theory with complex $\phi$-s, would the next lo leading order renormalization $\gamma$-s change?

Question

The Lagrangian of chiral perturbation theory (with two quark flavors) is written using the following matrix $U$
$$U=e^{i\sigma^i\phi_i/f}$$
where $\sigma^i$ are the Pauli matrices, $\phi_i$ are three scalar fields and $f$ is a constant with mass dimension. $U$ is unitary, which makes the $\phi_i$ fields real. 

The Lagrangian at $O(p^2)$ order is 
$$\mathcal{L}_2=\frac{f^2}{4}tr(D_{\mu}U^{\dagger}D^{\mu}U)$$
the Lagrangian at next to leading order $O(p^4)$ is
$$\mathcal{L}_4=\frac{l_1}{4}tr{}(D_{\mu}U^{\dagger}D^{\mu}U)tr(D_{\nu}U^{\dagger}D^{\nu}U)+\frac{l_2}{4}tr(D_{\mu}U^{\dagger}D_{\nu}U)tr(D^{\mu}U^{\dagger}D^{\nu}U)\\
+\frac{l_3+l_4}{16}\big[tr(\chi{}U^{\dagger}+U\chi^{\dagger})\big]^2+\frac{l_4}{8}tr(D_{\mu}UD^{\mu}U^{\dagger})tr(\chi{}U^{\dagger}+U\chi^{\dagger})\\
+l_5tr(U^{\dagger}F_R^{\mu\nu}UF_{L\mu\nu})+i\frac{l_6}{2}tr(F_R^{\mu\nu}D_{\mu}UD_{\nu}U^{\dagger}+F_L^{\mu\nu}D_{\mu}U^{\dagger}D_{\nu}U)\\
-\frac{l_7}{16}\big[tr(\chi{}U^{\dagger}-U\chi^{\dagger})\big]^2+\frac{h_1+h_3-l_4}{4}tr(\chi^{\dagger}\chi)\\
+\frac{h_1-h_3-l_4}{16}\bigg(\big[tr(\chi{}U^{\dagger}+U\chi^{\dagger})\big]^2+\big[tr(\chi{}U^{\dagger}-U\chi^{\dagger})\big]^2\\
-2tr(\chi{}U^{\dagger}\chi{}U^{\dagger}+U\chi^{\dagger}U\chi^{\dagger})\bigg)-\frac{4h_2+l_5}{2}tr(F_{R\mu\nu}F_R^{\mu\nu}+F_{L\mu\nu}F_L^{\mu\nu})$$

at next to leading order the coefficients $l_i$ and renormalize with loops with vertices coming only from the $O(p^2)$ Lagrangian like

$$l_i=l_i^r(\mu)+\frac{\gamma_i}{32\pi^2}\big(-\frac{1}{\epsilon}-\ln4\pi+\gamma_e-1\big)$$

where the gammas are

$$\gamma_1=1/3\qquad\gamma_2=2/3\qquad\gamma_3=-1/2\qquad\gamma_4=2\qquad\gamma_5=-1/6\qquad\gamma_6=-1/3\qquad\gamma_7=0$$

Now comes the question. Let's consider the same theory with the same Lagrangian but allowing the $\phi$ fields in $U$ to be complex.

My question is, would the $\gamma_i-s$ change?