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 $N:=\frac{-\ln(\varepsilon)}{\ln(2)}$.

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


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


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.


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


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