## 2 Mathematical Background
[PERSON] and [PERSON] [3] provided the following estimate: \[|\lambda|\leq\sqrt{1+\sum_{k=1}^{n}\left|a_{k}\right|^{2}}.\] (7)[PERSON] [3] proved the following estimate:  
\[|\lambda|\leq\max\Biggl{\{}1,\sum\limits_{k=1}^{n}{{{{\left|{{a_{k}}}\right|}}}} \Biggr{\}}. \tag{8}\]
4. [PERSON] and [PERSON] [16] have shown that \[|\lambda|\leq\cos\biggl{(}\frac{\pi}{n+1}\biggr{)}+\frac{1}{2}\Biggl{(}|{{a_{n }}}|+\sum\limits_{k=1}^{n}{{{{\left|{{a_{k}}}\right|}^{2}}}}\biggr{)}.\] (9)
5. [PERSON] [17] provided the following estimate: \[|\lambda|\leq\frac{1}{2}\Biggl{(}|{{a_{n}}}|+\cos\Bigl{(}\frac{\pi}{n}\Bigr{)} +\sqrt{\left({{{{\left|{{a_{n}}}\right|}}}-\cos\Bigl{(}\frac{\pi}{n}\Bigr{)} }\right)^{2}+\left({1+\sum\limits_{k=1}^{n-1}{{{{\left|{{a_{k}}}\right|}^{2}} }}}\right)^{2}}\Biggr{)}.\] (10) It seems that [PERSON] and [PERSON] [18] did not notice the [PERSON] result where they provided the same estimate. 6. [PERSON] [19] provided the following estimate: \[|\lambda|\leq\frac{|{{a_{n}}}|}{n}+\sqrt{\frac{n-1}{n}\[PERSON]{(}n-1+\sum \limits_{k=1}^{n}{{{{\left|{{a_{k}}}\right|}^{2}}}-\frac{{{{{\left|{{a_{n}}} \right|}^{2}}}}}{n}}\[PERSON]{)}}.\] (11)
7. [PERSON] [14] improved the [PERSON] estimate by proving that \[|\lambda|\leq\frac{1}{2}\Biggl{(}|{{a_{n}}}|+\cos\Bigl{(}\frac{\pi}{n}\Bigr{)} +\sqrt{\left({{{{\left|{{a_{n}}}\right|}}}-\cos\[PERSON]{(}\frac{\pi}{n}\Bigr{)} }\right)^{2}+\left({{{{\left|{{a_{n-1}}}\right|}}-1}}\right)^{2}+\sum\limits_{ j=1}^{n-2}{{{{\left|{{a_{j}}}\right|}^{2}}}}}\Biggr{)}.\] (12)
8. [PERSON] and [PERSON] [20] introduced the following estimate: \[|\lambda|\leq\frac{1}{2}\Biggl{(}\frac{|{{a_{n}}}|+\alpha}{2}+\cos\biggl{(} \frac{\pi}{n+1}\biggr{)}+\sqrt{\left({\frac{{{{\left|{{a_{n}}}\right|}+\alpha }}}{2}-\cos\biggl{(}\frac{\pi}{n+1}\biggr{)}}\right)^{2}+4\beta}\Biggr{)},\] (13) where \(\alpha=\sqrt{\sum\limits_{k=1}^{n}{{{{\left|{{a_{k}}}\right|}^{2}}}}}\) and \(\beta=\sqrt{\sum\limits_{k=1}^{n-1}{{{{\left|{{a_{k}}}\right|}^{2}}}}}\). 9. [PERSON] et al. [21] provided the following estimate: \[|\lambda|\leq\frac{1}{2}\Biggl{(}|{{a_{n}}}|+2\cos\Bigl{(}\frac{\pi}{n}\Bigr{)} +\sqrt{{t^{2}{{{\left|{{a_{n}}}\right|}^{2}}}}+\sum\limits_{k=1}^{n-1}{{{{ \left|{{a_{k}}}\right|}^{2}}}}}+\sqrt{1+(1-t)^{2}{{{\left|{{a_{n}}}\right|}^{2} }}}\Biggr{)}\] (14) for \(t\in[0,1]\). In fact, the upper bound above should be rewritten by taking'min' over \(t\in[0,1]\), which gives the best value for this estimate. No single known algorithm has been used in the literature to bound the zeros of the polynomial \(p(z)\) using the Frobenius companion matrix partitioned by the Cartesian decomposition method, as far as we are aware. To apply the numerical radius inequalities established in the previous section to \(C(p)\), we note that we have a little partition challenge in applying our obtained results because the main diagonal in the presented results requires a square sub-matrix. As a result, the usual well-known methods of partitioning the companion matrix \(C(p)\) are useless, as demonstrated by the inequalities presented in the previous section, for example in (3). Hence,  
\[C(q):=\text{Re}(C(q))+i\text{Im}(C(q)).\]  
Moreover, it is easy to observe that \(A_{kj}=P_{kj}+iQ_{kj}\) for \(k,j=1,2\). This observation may not hold true in general for operator matrices, i.e., the Cartesian decomposition of operator matrices is not equal to the Cartesian decomposition of their sub-matrices. However, because we construct a special type of \(C(q)\) partition, it appears that such equality holds true only for this construction. It is not easy to apply (3) to a general algorithm because it has the norms \(\||P_{11}|+|Q_{11}|\|\) and \(\||P_{22}|+|Q_{22}|\|\), which are difficult to evaluate for general \(n\times n\) matrices. In fact, it will be easier as long as we have numeric entries with specific \(n\), as explored in the presented examples below.