= Solution
Use the supplied oscillator expansion, with the boundary-term convention for <momentum> stated in part (b). Use a real <polarization vector> basis, as in the supplied unconjugated <polarization completeness relation>. Raise its second field index to obtain
$$
\sum_{\lambda,\lambda'}\epsilon_\mu^\lambda\epsilon^{\nu\lambda'}\eta^{\lambda\lambda'}=\delta_\mu{}^\nu.
$$
The mixed <commutator> contains only the annihilation-creation and creation-annihilation terms. Their signs are both positive after combining the minus sign in the <momentum> expansion with the negative <Minkowski metric> oscillator <commutator>. Setting $\mathbf r=\mathbf x-\mathbf y$ and using the <momentum> <Dirac delta distribution> gives
$$
[A_\mu(\mathbf x),\pi^\nu(\mathbf y)]
=\frac i2\delta_\mu{}^\nu\int\frac{d^3p}{(2\pi)^3}\bigl(e^{i\mathbf p\cdot\mathbf r}+e^{-i\mathbf p\cdot\mathbf r}\bigr)
=i\delta_\mu{}^\nu\delta^3(\mathbf r).
$$
Here the last equality is the <Fourier representation of the Dirac delta function>. Similarly,
$$
[A_\mu(\mathbf x),A_\nu(\mathbf y)]
=-\eta_{\mu\nu}\int\frac{d^3p}{(2\pi)^3}\frac{e^{i\mathbf p\cdot\mathbf r}-e^{-i\mathbf p\cdot\mathbf r}}{2|\mathbf p|}=0,
$$
because the integrand is odd under $\mathbf p\mapsto-\mathbf p$. The momentum-momentum <commutator> is proportional to the same odd difference, now weighted by $|\mathbf p|/2$, and also vanishes. Therefore the equal-time <canonical commutation relations> are
$$
\boxed{[A_\mu(\mathbf x),\pi^\nu(\mathbf y)]=i\delta_\mu{}^\nu\delta^3(\mathbf x-\mathbf y),\quad[A_\mu,A_\nu]=[\pi^\mu,\pi^\nu]=0.}
$$
These are identities of <operator-valued distributions>, understood after smearing. The extra indices in the TeX polarization relation are transcription defects; the PDF has the ordinary two-polarization completeness contraction used above. The direct momenta of part (b) also satisfy the <canonical commutation relations> after their boundary-generated shift, although they do not have the unmodified mode expansion printed here.
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