2

How can one prove that $|n^s|= n^{Re(s)}$, where $n \in \mathbb{N}$ and $s \in \mathbb{C}$, such that $Re(s) > 1$ (I found this on the link Convergence of $\zeta(s)$ on $\Re(s)> 1$)? I tried: $|n ^s|= |e^{ s log n}| = | e^{ s( ln(n) + i Arg (n) + 2 k \pi i) } |= |e^{ s( ln(n) + 2 k \pi i) }| = | e^{ (x + iy ) (ln (n) + 2 k \pi i) } | = | e^{x ln (n) + i y ln(n) +2 k \pi x i - 2 k \pi y )} | = | e^{x ln (n) - 2 k \pi y } | = n^x e^{- 2 k \pi y } $.

How do we obtain $n^x$?

user122
  • 151
  • @AnneBauval Whic logarithm is used on that link? Is it the principal branch? – user122 Dec 13 '22 at 17:00
  • Your definition of $n^s$ is wrong. It is simply $e^{s\ln(n)}$, there is no branch nor $2k\pi i.$ – Anne Bauval Dec 13 '22 at 17:02
  • @AnneBauval Here it says that it is multivalued function http://mathonline.wikidot.com/complex-power-functions , there is $2k \pi i$ – user122 Dec 13 '22 at 17:10

3 Answers3

1

Try using the definition $|z| = \sqrt{\overline{z}z}$, for $s = x+iy$ \begin{align*} |n^s| & = \sqrt{\overline{n^s}\cdot n^s} \\ & = \sqrt{\overline{n^{x+iy}}\cdot n^{x+iy}} \\ & = \sqrt{n^{x-iy}\cdot n^{x+iy}} \\ & = \sqrt{n^{x}n^{-iy} n^{x}n^{iy}} \\ & = \sqrt{n^{2x}n^{iy-iy}} \\ & = \sqrt{n^{2x}n^{0}} \\ & = \sqrt{n^{2x}} \\ & = n^{x} \\ \end{align*}

ineffable-clanging
  • 11
1

Assuming that $s = x + iy$, and following this link, then:

$$n^s = n^{x+iy} = n^x \cdot n^{iy} = n^x \cdot e^{iy\log(n)}.$$

Recall that:

$$e^{iy\log(n)} = \cos(y \log(n)) + i \sin(y \log(n)),$$

and hence:

$$|e^{iy\log(n)}| = \sqrt{\cos^2(y \log(n)) + \sin^2(y \log(n))} = 1.$$

As a consequence:

$$|n^s| = |n^x| \cdot |e^{iy \log(n)}| = n^x \cdot 1 = n^x.$$

But $x = \text{Re}(s)$, and hence:

$$|n^s| = n^{\text{Re}(s)}.$$

the_candyman
  • 14,064
  • 4
  • 35
  • 62
  • Thank you! Just one question: could you also please answer what is wrong in my reasoning? – user122 Dec 13 '22 at 17:14
  • @the_candyman Better not duplicate an answer to a duplicate question. – Anne Bauval Dec 13 '22 at 19:14
  • @user122 : K.defaoite and I told you what is wrong. It is to add to $\ln(n)$ unwanted terms. Even zwim, though more tolerant, told you that when $b>0,$ the usual definition is $b^z=\exp(z\ln b).$ – Anne Bauval Dec 13 '22 at 19:21
0

On one hand we could have this reasoning:

$n^s=n^{a+ib}=n^a\times n^{ib}$

And since $n^{ib}=\exp(ib\ln n)=e^{i\theta}\quad$ with $\theta=b\ln n$ so it is of module $1$.

Yet actually the premises $n^{a+ib}=n^a\cdot n^{ib}$ is false in general if you consider complex exponentiation.

But we generally assume the definition $b^z=\exp(z\ln b)$ when $b$ is a positive real, where $\exp$ is defined by its power series.

See this Definition of exponential function, single-valued or multi-valued?

Therefore it is really a matter of conventions, the result $|n^z|=n^a$ is fine according to this latter convention, but yours is the correct one assuming the complex exponentiation convention.

zwim
  • 28,563