$f(z)= (s-1) \zeta(s)$. Prove that $\log |f(0)|=0$

Question

I am reading Balazard, Saias and Yor Paper on the RH where they consider a function$$ f(z)= (s-1) \zeta(s)$$ where $ \ s=\frac{1}{1-z}$ and $\zeta $ denotes the Zeta function

$$\text{Claim:} \qquad \log|f(0)|=0$$ My Try-

$$\lim_{s \to 1} (s-1)\zeta(s)=1$$ $$\lim_{z \to 0} f(z)=1$$ $f$ is continuous at $0$ as it is analytic near $0$ $$f(0)=\lim_{z \to 0} f(z)=1$$ $$\log|f(0)|=0$$ Thank you for your help.

Answer 1

No mistake, the reasoning is fine.

$(s-1)\zeta(s)$ is analytic at $s=1$ and $1/(1-z)$ is analytic at $0$ whence the composition is analytic at $z=0$.