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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.29(c),7.30

Please point out errors.

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Exer 7.29

(c) $M_k = r^k$ because $$\frac{|z|^k}{r^k} \le 1 \le |z^k + 1| $$

Elaboration: I believe that $|\frac{z^k}{z^k+1}| \le r^k$.

Pf: For $|z| \le r,$ we have $$|z|^k \le r^k \implies \frac{|z|^k}{r^k} \le 1. \tag{1}$$

But for $|z| \in [0,\infty),$ $$|z^k+1| \ge ||z^k|-|1|| = ||z|^k-1| \ge |0-1| = 1 \implies 1 \le |z^k+1| \tag{2}$$

Thus,

$$(1) \wedge (2) \implies \frac{|z|^k}{r^k} \le 1 \le |z^k+1| \implies \frac{|z|^k}{|z^k+1|} \le r^k$$

$$\therefore, |\frac{z^k}{z^k+1}| = \frac{|z^k|}{|z^k+1|} = \frac{|z|^k}{|z^k+1|} \le r^k \ \text{QED}$$


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Exer 7.30

Weierstrass M-Test: $|\frac{z}{w}|^k \le \frac{|z|^k}{r^k} = |\frac{z}{r}|^k =: M_k$

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In exercise 7.29 (c) $$\left\lvert \frac{z^k}{z^k+1} \right\rvert \leq r^k$$ doesn't hold for all of the $z \in \overline{D}[0,r]$. According to the maximum modulus principle, the maximum occurs on the boundary where $z=r \exp{i \theta}$. The functions to be maximized then become $$\frac{r^k}{|r^k \exp{i k \theta}+1|},$$ and the maxima occur when $\exp{i k \theta}=-1$. Hence $M_k$ really should be $$M_k:=\frac{r^k}{1-r^{k}} .$$

Exericse 7.30 looks ok to me.

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  • $\begingroup$ Thanks user1337, but what exactly is wrong with my argument please? $\endgroup$ – BCLC Aug 4 '18 at 2:26
  • $\begingroup$ @BCLC Your choice of $M_k$. $\endgroup$ – user1337 Aug 4 '18 at 17:22
  • $\begingroup$ lol I mean why? I believe it's an upper bound whose series is convergent $\endgroup$ – BCLC Aug 4 '18 at 20:11
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    $\begingroup$ @BCLC It is not an upper bound. Consider $z=-r$ when $k$ is odd, for instance. $\endgroup$ – user1337 Aug 4 '18 at 21:02
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    $\begingroup$ @BCLC $||z|^k-1|\ge|0-1|$ is not necessarily true, consider $z=0.1$ $\endgroup$ – user341124 Aug 5 '18 at 14:08
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In re user1337's answer:

$$\frac{r^k}{1-r^{k}} \le \frac{r^k}{1-r}$$

Without user1337's answer, we can use zhw.'s answer to directly show that:

$$|\frac{z^k}{1+z^k}| \le \frac{r^k}{1-r}$$

Note: Upper bound is a my proposed (albeit incorrect) upper bound multiplied by $\frac{1}{1-r}$. Note that $\frac{1}{1-r}$ does not depend on $k$.

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