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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?

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