Hint: you might start with wely Bound which is defined as :$t\in\mathbb{R}$, then $|\zeta(\frac{1}{2}+it)|\leq c(|t|+1)^{1/6}$ this means that $\log|\zeta(\frac{1}{2}+it)|\leq \frac16 \log(c|t|+1)$, you can easly deduce the representation series of $\log(c|t|+1)$ which it is defined as : $\log(c|t|+1)=\log(c|t|)-\sum_{k=1}^{\infty} \frac{(-1^k)}{k c^k|t|^k},|ct|>1$

Hint: you might start with wely Bound which is defined as :$t\in\mathbb{R}$, then $|\zeta(\frac{1}{2}+it)|\leq c(|t|+1)^{1/6}$ this means that $\log|\zeta(\frac{1}{2}+it)|\leq \frac16 \log(c|t|+1)$, you can easly deduce the representation series of $\log(c|t|+1)$ which it is defined as : $\log(c|t|+1)=\log(c|t|)-\sum_{k=1}^{\infty} \frac{(-1^k)}{k c^k|t|^k},|ct|>1$,And if you want really to find such non trivial series representation of $\zeta(0.5+i t)$ in general , you may look to find such polynomial of degree N using local trigonometric approximation I recomond you to check this nice post by Terence Tao