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I'm working through an exercise in Chapter 4.2 of the PDE book by Robert McOwen. I have derived that when $n=2$ and $a=1$, the Poisson integral formula can be represented by

$$u(r,theta)=frac1-r^22piint_0^2pifracg(phi)dphi1+r^2-2rcos(theta-phi)$$

Now, I'm trying to verify that

$$r^kcos ktheta=frac1-r^22piint_0^2pifraccos (kphi)dphi1+r^2-2rcos(theta-phi)$$

where $k$ is an integer and $0leq r < 1$.

I've previously derived that we can find $u(r,theta)$ by the Fourier series representation

$$u(r,theta)=a_0+sum_n=1^inftyr^n(a_n cos ntheta + b_nsin ntheta)$$

But, I'm not sure why this implies that

$$r^kcos ktheta=frac1-r^22piint_0^2pifraccos (kphi)dphi1+r^2-2rcos(theta-phi)$$

Is there a property of Fourier series that I can reference to show that

$$g(phi)=cos (kphi)$$

and that

$$u(r,theta)=frac1-r^22piint_0^2pifraccos (kphi)dphi1+r^2-2rcos(theta-phi)$$

Remember that $g(theta)$ is the boundary condition for your PDE: that is, you want to find a function $u(r,theta)$ solving
$$left{
beginarraycc
Delta u = 0& textin B(0,1)\
u(1,theta) = g(theta)&\
endarrayright.tag*$$
The function $u$ given by the Poisson formula is the unique solution to the problem, so if we have any function solving $(*)$, then that function must be equal to the Poisson integral.

In particular, we notice that the function $u(r,theta) = r^kcos(ktheta)$ solves $(*)$ with $g(theta) = cos(ktheta)$ (this is just a computation). But, since the solution is unique, this means that
$$r^kcos(ktheta) = frac1-r^22piint_0^2pifraccos(kphi)1+r^2-2rcos(theta-phi);dphi$$

Notice in particular that we aren't declaring $g(theta) = cos (ktheta)$ always. What it is varies from problem to problem. However, if $g(theta) = cos(ktheta)$, then $u(r,theta) =r^k cos(ktheta)$ is the only solution to the PDE $(*)$.

Note that $g(theta)$ is the function satisfying the following property: $Delta_2 u = u_rr + frac1ru_r + frac1r^2u_thetatheta = 0$ and $u(1,theta) = g(theta) = cos(ktheta)$.

Substituting $u(r,theta) = r^k cos(ktheta)$ we see that it satisfies the hypothesis above. So we can let $u$ and $g$ be as such.