While I am reading Rudin's proof on the completeness of complex field,
I am stuck in two steps.
The first one is in the derivation of (57). How can $Q(z)$ be represented with $1+b_kz^k+\dots+b_nz_n$ ($1\geq k\geq n$)? I tried to compute some examples but it seems the first $1$ does not appear in the expanded result and the $k$ does not make sense to me.
And the second one is in the derivation of $|1+b_kr^ke^{ik\theta}|=1-r^k|b_k|$. How is this derived? Here's my try: $$ \begin{align*} e^{ik\theta}b_k&=-|b_k|\\ b_kr^ke^{ik\theta}&=-r^k|b_k|\\ 1+b_kr^ke^{ik\theta}&=1-r^k|b_k|\\ |1+b_kr^ke^{ik\theta}|&=|1-r^k|b_k||\\ \end{align*} $$ Then use the triangle inequality of complex modulus, $$ |1-r^k|b_k||\leq1+|-r^k|b_k||=1+r^k|b_k|. $$ And now I have a result that is different from Rudin's. But suprisingly my way works. With $|1+b_kr^ke^{ik\theta}|=1+r^k|b_k|$, $$ |Q(re^{i\theta})|\leq1+r^k\{|b_k|+r|b_{k+1}|+\dots+r^{n-k}|b_n|\}. $$ And with sufficiently small $r$, the expression in the brace is still positive and can be sufficiently small, so is $r^k$, thus $|Q(re^{i\theta})|\leq1+\varepsilon$, and $|Q(re^{i\theta})|\leq1$. The desired contradiction still appears. Is this new argument valid?
By the way, here's my attempt computing an example of (57):
Note that $Q(z)=\dfrac{P(z+z_0)}{P(z)}$ and that therefore $Q(0)=\dfrac{P(z_0)}{P(z_0)}=1$. In other words, the constant term of $Q(z)$ is $1$.
Concerning the other doubt, remember that $e^{ik\theta}b_k=-\lvert b_k\rvert$ and that therefore $1+b_kr^ke^{ik\theta}=1-r^k\lvert b_k\rvert$.