Suppose $\Omega=\{z\in\mathbb{C}:2<|z|<5\}$ and $K=\{z\in\mathbb{C}: 3\leqslant|z|\leqslant4\}$. Let $f(z)=\dfrac{\cos z}{z}$. Clearly $f\in Hol(\Omega)$ Find an rational function $r$ such that
$ |f(z)-r(z)|\leqslant1\qquad(z\in K)$
And more over
$ P(r)\subset \{1,i6\}$
My attemp:
We can write $\displaystyle\dfrac{1}{z}=\dfrac{1}{(z-1)+1}=\dfrac{1}{z-1}\cdot\dfrac{1}{1+\frac{1}{z-1}}=\dfrac{1}{z-1}\sum_{k=0}^{\infty}\dfrac{1}{(z-1)^k}= \sum_{k=0}^{\infty}(\dfrac{1}{z-1})^{k+1}$
and we have
$\Big|\dfrac{\cos z}{z}-\dfrac{1}{z}\big(1-\dfrac{z^2}{2}+\dfrac{z^4}{4!}\big) \Big|= \Big|\dfrac{\cos z}{z}-\big(\dfrac{1}{z}-\dfrac{z}{2}+\dfrac{z^3}{4!}\big) \Big|=\Big|\sum_{k=3}^{\infty}\dfrac{z^{2k}}{(2k)!} \Big|<?$
Now I don't know how can I continue it?