Prove $z^n$ converges uniformly to $0$ for $|z| < \frac 1 2$ by definition.

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Eg 7.24

Note in the following example (Eg 7.24): $D[0,\frac 1 2]$ the disc with centre 0 and radius $\frac 1 2$, i.e. $|z| < \frac 1 2$

Question: In the following example (Eg 7.24), why do we choose $N:=-\ln2\ln(\varepsilon)$?

I think this should be $N:=\frac{-\ln(\varepsilon)}{\ln(2)}$.

(Well strictly speaking the proof chooses $N>-\ln2\ln(\varepsilon)$. I take this to mean the same thing as choosing $n > N$ and $N:=-\ln2\ln(\varepsilon)$ and as choosing $ N > -\ln2\ln(\varepsilon))$

Here is my pf:

Let $\varepsilon > 0$ and $z$ be any complex number s.t. z's modulus is less than $\frac 1 2$, i.e. $z$ is in the disc $\{|z| < \frac12\}$. Suppose $n > N:=\frac{-\ln(\varepsilon)}{\ln(2)}$. Then $$n \ln(\frac 1 2) < \ln \varepsilon \implies \ln((\frac 1 2)^n) < \ln \varepsilon \implies (\frac 1 2)^n < \varepsilon. \tag{*}$$

Therefore, $$|f_n(z)-f(z)| = |z^n| = |z|^n < (\frac 1 2)^n \stackrel{\text{by} \ (*)}{<} \varepsilon$$

or

$$|f_n(z)-f(z)| = |z^n| = |z|^n < (\frac 1 2)^n < (\frac 1 2)^N \stackrel{\text{by} \ (*)}{=} \varepsilon.$$

QED

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Note: Please don't use Vitali convergence, Lebesgue dominated convergence, Exer 7.19, etc or Power series, Taylor series, etc. Please just help me find the right N.

Answer 1

You are right. The argument depends on choosing $\varepsilon$ so that $$\varepsilon>\frac1{2^n}.$$ This inequality is exactly $$n>-\frac{\log\varepsilon}{\log 2}.$$