= Solution
{c}
<Conformality> implies $f'(0)\neq0$, so $a_1\neq0$. Apply the stated <Green's theorem> with $F=z^n\bar z$:
$$
M_n=\frac1{2i}\oint_\Gamma z^n\bar z\,dz
=\frac1{2i}\oint_{|\zeta|=1}f(\zeta)^n
\left(\sum_{j=1}^N\bar a_j\zeta^{-j}\right)f'(\zeta)\,d\zeta.
$$
Here $\bar\zeta=\zeta^{-1}$ on the <unit circle>. The lowest power in $f^nf'$ is $\zeta^n$, while the most negative power in the conjugate <polynomial> is $\zeta^{-N}$. For $n\geq N$ their product has no $\zeta^{-1}$ term. The <residue theorem> gives \b[$M_n=0$ for all $n\geq N$].
When $N=2$, the residue for $n=0$ is $|a_1|^2+2|a_2|^2$. For $n=1$, expand
$$
ff'=a_1^2\zeta+3a_1a_2\zeta^2+2a_2^2\zeta^3.
$$
Only $a_1^2\zeta\,\bar a_2\zeta^{-2}$ contributes. Therefore
$$
\boxed{M_0=\pi(|a_1|^2+2|a_2|^2),\qquad M_1=\pi a_1^2\bar a_2.}
$$
The first is the area; the second records its complex first moment, and need not be real.
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