Exercise 27 (iii): $u:U \rightarrow \mathbb{R}$ is a real polynomial, $\sum _{n,m \ge 0 ; n+m \le d } a_{n,m}x^ny^m $ in $x$ and $y$ for some real coefficients $a_{n,m}$: Show $u$ is harmonic iff it is real part of polynomial $f(z)$ of one complex variable. I am stuck on $(\Rightarrow)$.
For $(\Rightarrow)$,We can substitute $x=\frac{z+\bar{z}}{2}, \, y = \frac{z-\bar{z}}{2i}$, to obtain $$ u(z) = \sum_{n,m \ge 0 , n+m \le d} c_{n,m} z^n \bar{z}^m \stackrel{*}= \sum_{0 \le k \le d} a_k z^k +b_k \bar{z}^k , \quad c_{n,m}, a_k,b_k \in \mathbb{C} $$ where $(*)$ follows by Exercise 27(ii) as $u(z)$ is harmonic.
EDIT: I end the proof as follows: $u(z) = \bar{u}(z)$ for all $z$, so $$ \sum_{0 \le k \le d} (a_k - \bar{b_k}) z^k + (b_k-\bar{a}_k)\bar{z}^k = 0 $$ for all $z$. Regarding this as a polynoial in $x,y$ in both real and complex components, we have $a_k - \bar{b}_k = 0 $ for all $k$. So, $f(z) := \sum_{0 \le k \le d } 2 a_k z^k $, yields $u = \frac{f+\bar{f}}{2} = Re\, f(z)$.
Is above proof correct?