I have the following question:
Let $n\in \mathbb{N}\setminus \{0\}$ be an arbitrary, but fixed natural number, let $f:\mathbb{R}^{+}\rightarrow \mathbb{C}:\ t\mapsto n^{-b} - t^{-b}$, when $b\in \mathbb{C}$ denotes a fixed complex number with Re$(b)>0$.
How can I prove that $\sup\limits_{n\leq t \leq n+1} \{|f(t)|\}\leq \frac{|b|}{n^{1+ \text{Re}(b)}}$ ?
I know that the derivative of $f$ is given by $\frac{b}{t^{b+1}}$, but I'm not sure, what theorems stay valid, because we have a complex valued function $f$ here.
Thanks for the help!
Let $f\colon(0,\infty)\to\mathbb{C}$ be given by $$ f(x)=x^{-b}=e^{-b\log x}, $$ where $\log$ is the principal branch of the logarithm, that is, $\log x$ is real and $\log 1=0$. Then $f$ is $C^\infty$ and $$ f'(x)=-b\,e^{-b\log x}\,\frac1x=-b\,e^{-(b+1)\log x}=-b\,x^{-(b+1)}. $$ Now $$ |t^{-b}-t^{-b}|=|f(t)-f(n)|=\Bigl|\int_n^tf'(x)\,dx\Bigr|\le |b|\int_n^{n+1}|x^{-(b+1)}|\,dx\le\frac{|b|}{n^{\operatorname{Re}(b)+1}}. $$ Note: $f(t)-f(n)=\int_n^tf'(x)\,dx$ can be justified integrating the real and imaginary parts.