= Solution
{c}
Write $H=\phi_1$ and $P=\phi_2$, both <Hermitian matrices>. Through quadratic order around the chosen <vacuum expectation value>,
$$
X=\Phi^\dagger\Phi=C^2\mathbf1+\sqrt2 CH+\frac12(H^2+P^2+i[H,P]).
$$
The <trace of a commutator> vanishes. The needed <matrix traces> are therefore
$$
\begin{aligned}
\operatorname{tr}X&=NC^2+\sqrt2 C\operatorname{tr}H+\frac12\operatorname{tr}(H^2+P^2),\\
\operatorname{tr}X^2&=NC^4+2\sqrt2 C^3\operatorname{tr}H+C^2\operatorname{tr}(3H^2+P^2)+O((H,P)^3),\\
(\operatorname{tr}X)^2&=N^2C^4+2\sqrt2 NC^3\operatorname{tr}H+NC^2\operatorname{tr}(H^2+P^2)+2C^2(\operatorname{tr}H)^2+O((H,P)^3).
\end{aligned}
$$
Using $m^2=-(\alpha+N\beta)C^2$, all linear terms cancel. The quadratic <scalar potential> reduces to
$$
\boxed{V^{(2)}=\alpha C^2\operatorname{tr}H^2+\beta C^2(\operatorname{tr}H)^2,}
$$
with no $P^2$ term. The kinetic term is
$$
\mathcal L_{\mathrm{kin}}^{(2)}=\frac12\operatorname{tr}[(\partial H)^2+(\partial P)^2],
$$
since its mixed term is the <trace of a commutator>. A <Hermitian matrix> has $N^2$ independent real components, so $P$ describes exactly $N^2$ massless scalar modes.
To determine the other <masses>, take $T_0=\mathbf1/\sqrt N$ and traceless Hermitian $T_A$ with $\operatorname{tr}(T_AT_B)=\delta_{AB}$, and expand $H=h_0T_0+\sum_{A=1}^{N^2-1}h_AT_A$. These are canonically normalized real <scalar fields>. The <scalar mass matrix> is diagonal in this basis:
$$
\boxed{m_{h_0}^2=2C^2(\alpha+N\beta)=-2m^2,\qquad m_{h_A}^2=2\alpha C^2\quad(A=1,\ldots,N^2-1),\qquad m_{P_A}^2=0\quad(A=0,\ldots,N^2-1).}
$$
Thus $H$ contains one massive singlet and a degenerate multiplet of $N^2-1$ massive fields in the <Adjoint representation> of the unbroken diagonal $SU(N)$. Both squared <masses> are strictly positive for the specified parameters. The $P$ fields are the $N^2$ <Goldstone bosons> associated with the broken generators: $N^2-1$ traceless relative left-right rotations and one overall phase rotation.
<Goldstone's theorem> guarantees those massless directions. It does not alone forbid additional accidental massless modes; the positive <scalar mass matrix> computed here supplies the further conclusion that every $H$ component is massive. The fields describe a global symmetry, so these <Goldstone bosons> are physical rather than being absorbed into massive <gauge bosons>.
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