## 2 Mathematical Background
}^{m}\Big{|}\Big{\langle}A_{kj}x_{j},x_{k}\Big{\rangle} \Big{|}^{2}\] \[\leq\frac{1}{m}\sum_{k=1}^{m}\Big{(}\langle P_{kk}x_{k},x_{k} \rangle^{2}+\langle Q_{kk}x_{k},x_{k}\rangle^{2}\Big{)}+\frac{1}{m}\sum_{ \begin{subarray}{c}j=1\\ j\
eq k\end{subarray} }^{m}\Big{[}\Big{\langle}f^{2}\Big{(}\Big{|}P_{kj}\Big{|} \Big{)}x_{j},x_{k}\Big{\rangle}^{\frac{1}{2}}\Big{\langle}g^{2}\Big{(}\Big{|} P_{kj}\Big{|}\Big{)}x_{j},x_{k}\Big{\rangle}^{\frac{1}{2}}\] \[\qquad\qquad\qquad+\Big{\langle}f^{2}\Big{(}\Big{|}Q_{kj}\Big{|} \Big{)}x_{j},x_{k}\Big{\rangle}^{\frac{1}{2}}\Big{\langle}g^{2}\Big{(}\Big{|} Q_{kj}\Big{|}\Big{)}x_{j},x_{k}\Big{\rangle}^{\frac{1}{2}}\Big{]}^{2}\] \[\leq\frac{1}{m}\sum_{k=1}^{m}\Big{(}\Big{\langle}P_{kk}^{2}x_{k},x_{k}\Big{\rangle}+\Big{\langle}Q_{kk}^{2}x_{k},x_{k}\Big{\rangle}\Big{)}\] \[\qquad+\frac{1}{4m}\sum_{ \begin{subarray}{c}j=1\\ j\
eq k\end{subarray} }^{m}\Big{\|}f^{2}\Big{(}\Big{|}P_{kj}\Big{|}\Big{)}+g^{2} \Big{(}\Big{|}P_{kj}\Big{|}\Big{)}+f^{2}\Big{(}\Big{|}Q_{kj}\Big{|}\Big{)}+g^{ 2}\Big{(}\Big{|}Q_{kj}\Big{|}\Big{)}\Big{\|}^{2}\|x_{j}\|^{2}\|x_{k}\|^{2}\] \[\leq\frac{1}{m}\sum_{k=1}^{m}\Big{\langle}\Big{(}P_{kk}^{2}+Q_{kk }^{2}\Big{)}x_{k},x_{k}\Big{\rangle}\] \[\qquad+\frac{1}{4m}\sum_{ \begin{subarray}{c}j=1\\ j\
eq k\end{subarray} }^{m}\Big{\|}f^{2}\Big{(}\Big{  
Proof. : From Theorem 1 and by employing Lemma 3, we have  
\[\omega\bigg{(}\bigg{[} \begin{array}{cc}A_{11}&A_{12}\\ A_{21}&A_{22}\end{array} \bigg{]}\bigg{)}\] \[\leq\sqrt{2}\cdot\omega^{\frac{1}{2}}\Bigg{(}\Bigg{[} \begin{array}[ ]{cc}\omega\big{(}P_{11}^{2}+Q_{11}^{2}\big{)}&\||P_{12}|+|Q_{12}|\|\\ \||P_{21}|+|Q_{21}|\|&\omega\big{(}P_{22}^{2}+Q_{22}^{2}\big{)}\end{array}  \bigg{]}\Bigg{)}\] \[=\sqrt{2}\cdot r^{\frac{1}{2}}\Bigg{(}\Bigg{[} \begin{array}{cc }\omega\big{(}P_{11}^{2}+Q_{11}^{2}\big{)}&\frac{\||P_{12}|+|Q_{12}|\|+|Q_{21} |\|}{2}\\ \frac{\||P_{12}|+|Q_{12}|\|+\|P_{21}|+|Q_{21}|\|}{2}&\omega\big{(}P_{22}^{2}+Q _{22}^{2}\big{)}\end{array} \bigg{]}\Bigg{)}\] \[=\sqrt{\omega\big{(}P_{11}^{2}+Q_{11}^{2}\big{)}+\omega\big{(}P_{ 22}^{2}+Q_{22}^{2}\big{)}+\sqrt{\big{(}\omega\big{(}P_{11}^{2}+Q_{11}^{2} \big{)}-\omega\big{(}P_{22}^{2}+Q_{22}^{2}\big{)}\big{)}^{2}+N^{2}}},\]  
which proves the result. **Corollary 2**. : _If \(\mathbf{A}=\bigg{[} \begin{array}{cc}A_{11}&0\\ 0&A_{22}\end{array} \bigg{]}\) in \(\mathscr{M}_{n}\bigoplus\mathscr{M}_{n}\), then,_  
\[\omega\bigg{(}\bigg{[} \begin{array}{cc}A_{11}&0\\ 0&A_{22}\end{array} \bigg{]}\bigg{)}\leq\max\Big{(}\omega\Big{(}P_{11}^{2}+Q_{1 1}^{2}\Big{)},\omega\Big{(}P_{22}^{2}+Q_{22}^{2}\Big{)}\Big{)},\]  
_where \(P_{kk}+iQ_{kk}\) is the Cartesian decomposition of \(A_{kk}\)._  
Proof. : Setting \(A_{12}=0=A_{21}\) in Corollary 1 and using the fact that \(\omega\bigg{(}\bigg{[} \begin{array}{cc}A&0\\ 0&D\end{array} \bigg{]}\bigg{)}\leq\max(\omega(A),\omega(D))\), see [15]. **4. Applications for Bounding Zeros of Polynomials**  
One of the most interesting and useful applications of the numerical radius inequalities is to bound zeros of complex polynomials using a suitable partition of the well-known Frobenius companion matrix. Let  
\[p(z)=z^{n}+a_{n}z^{n-1}+\cdots+a_{2}z+a_{1}, \tag{4}\]  
be any polynomial with \(a_{1}\
eq 0\). The general corresponding companion matrix is defined as  
\[C(p):=\left[ \begin{array}{ccccc}-a_{n}&-a_{n-1}&\cdots&-a_{2}&-a_{1}\\ 1&0&\cdots&0&0\\ 0&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&0\end{array} \right]. \tag{5}\]  
It is well known that the eigenvalues of \(C(p)\) are exactly the zeros of \(p(z)\), see [3], p. 316. Based on some numerical radius estimations of \(C(p)\), several authors paid serious attention to finding various upper bounds of the zeros of \(p(z)\). Some famous upper bounds are listed as follows:  
If \(\lambda\) is a zero of \(p(z)\), then,  
1. [PERSON] [3] obtained the following upper bound: \[|\lambda|\leq 1+\max\{|a_{k}|:k=1,2,,\cdots,n\}.\] (6)
2.