Determine analytic function outside the unit circle from value of real part at the boundary. I'm addressing the following problem.
Let $G(z)$ be an analytic function outside of the unit circle with decay at infinity and Re $G(z)$ at $|z|=1$ some known Laurent polynomial (with poles only at $z=0$, not on the circle). Then it should be possible to fully determine $G(z)$ everywhere in the complex plane.
I know the answer is that $G(z)$ is twice the sum of all negative degree terms of the given Laurent polynomial, but don't get to the proof.
One attempt of solution is the following. As $G(z)$ is analytic outside the circle (let's denote the oriented contour bounding this region $C_1$), it follows that
$\frac{1}{2\pi i}\oint_{-C_1} dz'\:z' \:G(z-\frac{1}{z'})=G(z)=-\frac{1}{2\pi i}\oint_{C_1} dz' z'\left[\text{Re}\: G(z-\frac{1}{z'})+i\text{Im}\:G(z-\frac{1}{z'})\right]$,
with $-C_1$ the contour surrounding the whole region of the plane outside the unit circle. Then I would think of use the known value of the real part on the circle and somehow find the imaginary part and perform the integral. But this strategy seems to be leading nowhere.
To my knowledge Poisson and other formulas from potential theory don't solve this problem.
Thanks forward.
 A: Let $u(e^{it})=\Re G(e^{it})=\sum_{n \ge 1}(2a_n \cos nt + 2b_n \sin nt), a_n, b_n \in \mathbb R$
(clearly there is no free term if one assumes $G(z) \to 0, z \to \infty$)
We have that $u=\sum_{n \ge 1}(a_n(e^{int}+e^{-int})-ib_n(e^{int}-e^{-int}))$
Here we may have a finite or infinite sum with some convergence conditions but the result is the same as coefficients go.
We need to find the harmonic conjugate for $u$ on the outside of the unit circle, so 
$v(t)=\sum_{n \ge 1}(2c_n \cos nt + 2d_n \sin nt), a_n, b_n \in \mathbb R$ s.t $u+iv$ has no positive powers of $e^{int}$ since $G=u+iv$ is given to be bounded at infinity.
Identifying coefficients we get $a_n-ib_n+d_n+ic_n=0, n \ge 1$ so $b_n=c_n, d_n=-a_n$ which gives $G(z)=\sum_{n \ge 1}(2a_n+2ib_n)z^{-n}$ and obviously $u(z)=\sum_{n \ge 1}(a_n+ib_n)z^{-n}+\sum_{n \ge 1}(a_n-ib_n)z^n$, hence the answer is exactly as noted by the OP (twice the coefficients of the negative powers of $z$)
Edit later - per comments one can give an argument with convolution that doesn't explicit the coefficients (though it's essentially the same as above if you dig in); 
Let $u=u_P+u_N, v=v_P+v_N, v=\Im G, u_p,u_N$ the positive/negative power of $u$, same for $v_P, v_N$
Since $u, v$ are real, $u=\bar u, v =\bar v$ which means that on the unit circle (only!) where $\bar z=z^{-1}$, we have $\bar u_P=u_N, \bar v_P=v_N$
Let $H=1/z+1/z^2+...$ be the convolution identity of the space of analytic functions outside the unit disc that vanish at infinity;  by definition, we have $G*H=(u+iv)*H=G=u+iv$; but $u*H=u_N, v*H=v_N$ so we get $u_P+iv_P=0$ on the outside of the unit circle; by continuity, this must hold on the unit circle and conjugating we get $u_N-iv_N=0$ on the unit circle, which again must hold on the outside of the unit circle by analytic continuation as both $u_N,v_N$ are analytic functions there(!); substituting back $G=G*H=u_N+iv_N=2u_N$ as required!
