Consider the annulus $G = \{2 \leq |z| \leq 3 \}$ (this is basically a ring with inner radius $2$ and outer radius $3$). Find a harmonic function which vanishes on $C[0, 2] $(this is a circle with center $0$ and radius $2$) and satisfies $u(3\mathrm{e}^{i \theta}) = 65 \cos(2 \theta)$ on $C[0,3].$
Hint: consider functions of the form $z^n$ for suitable $n$.
Since $z^n$ is holomorphic for $n \neq 1$ and therefore harmonic, I've tried $n = 2, 3, 4$. But it doesn't satisfy the condition. I'm not sure if it's worth it to keep trying more $n$s because the expansions get large after this point and it might be a waste of time if it's a big $n$.
How else can I approach this problem? Thank you for your help!
Let's take a look at holomorphic functions of the form $$f(z)=\sum_{n=-2}^2 a_n z^n\; a_n \in \mathbf{R}. $$
When $|z|=2$ we can write $z=2 \mathrm{e}^{i \theta}$ for some real theta, and then $$f(2 \exp(i \theta))=4a_{-2} \mathrm{e}^{-2i \theta}-2a_{-1} \mathrm{e}^{-i \theta}+a_0+2a_1 \mathrm{e}^{i \theta}+4 a_2 \mathrm{e}^{2 i \theta} $$ The real part of this is $$4a_{-2} \cos 2 \theta-2a_{-1} \cos \theta+a_0+2a_1 \cos \theta+4a_2 \cos 2 \theta. $$ In order for it to vanish, we need $$a_{-2}=-a_2 \\ a_{-1}=a_1 \\a_0=0, $$ so that $$f(z)=a_2(z^2-z^{-2})+a_1(z-z^{-1}). $$ Now, for the other boundary condition, substitute $z=3 \exp( i \theta)$ to get $$\Re f(3 \exp(i \theta)) =a_2(9 \cos 2 \theta-1/9 \cos 2 \theta)+a_1 (3 \cos \theta-1/3 \cos \theta).$$ Can you find the values of constants $a_1,a_2$?