An inequality involving a function of complex variables Let $n$ be a positive integer. Let $z$ be a complex number and assume $|z|\le C.$
How to prove that there exists an absolute constant $A$ (independent of $n$ and $z$) such that
$\bigg|\dfrac 1{n^z} - \dfrac 1 {x^z} - \dfrac  z{x^{z+1}}\bigg|\le A x^{-\Re z -2}, \forall x\in (n,n+1)$, where $\Re z$ denotes the real part of $z$ ?
I tried considering the function $f:[n,n+1]\to \mathbb C$ as $f(x)=\dfrac 1{n^z} - \dfrac 1 {x^z} - \dfrac  z{x^{z+1}}$ whose derivative is  $f'(x)=\dfrac z {x^{z+1}} + \dfrac  {(z+1)z}{x^{z+2}}$, so $|f'(x)|\le \dfrac C {x^{\Re z+1}} + \dfrac  {(C+1)C}{x^{\Re z+2}}$ , and then Mean value estimate along with triangle inequality only gives
$|f(x)|\le |f(x)-f(n)|+|f(n)|\le \dfrac C {x^{\Re z+1}} + \dfrac  {(C+1)C}{x^{\Re z+2}}+\dfrac  C{n^{\Re z+1}} $ which does not seem to give me what I want.
Please help.

Something doesn't look right. If the inequality were true, it would hold for $x=n$ by continuity. Suppose $x=n$ and $z=1.$ The inequality simplifies to $n^{-2}\le An^{-3},$ which has no chance of holding.
