## 5 Numerical Radius of Real and Imaginary Parts of \(C(p)\)
\begin{array}{cc}w(\left|\mathrm{Im}(a_{n}) \right|)&\left\|u\right\|\\ \left\|u\right\|&0\end{array} \right]\bigg{)}\] \[=\frac{1}{2}\bigg{(}\left|\mathrm{Im}(a_{n})\right|+\sqrt{\left| \mathrm{Im}(a_{n})\right|^{2}+4\left\|u\right\|^{2}}\bigg{)}\] \[=\frac{1}{2}\Bigg{(}\left|\mathrm{Im}(a_{n})\right|+\sqrt{\left| \mathrm{Im}(a_{n})\right|^{2}+\left|1+a_{1}\right|^{2}+\sum_{k=2}^{n-1}\left|a_{ k}\right|^{2}}\bigg{)}\,. \tag{23}\]  
In addition, by (1) and Lemma 1, we have  
\[\omega(P_{22})=\cos\biggl{(}\frac{\pi}{n+1}\biggr{)}. \tag{24}\]Combining all above inequalities and equalities (21)-(24) in (20), and following the same steps for \(\omega(\text{Im}[C(q)])\), we obtain the required result in Theorem 3. Thus, the proof of Theorem 3 is established. The example in Table 2 shows that our estimated rectangle given in Theorem 3 might be better than the one given in Theorem 2. **Example 1**. : _Consider \(q_{2}(z)=z^{6}+2 iz^{5}+4 iz^{4}+\frac{1}{4}z+\frac{1}{16}\); then, the real and imaginary parts of the zeros of \(q_{2}(z)\) are bounded as obtained in Table 2._  
**Remark 1**. : _Several particular cases of Theorem 3, which are of great interest, could be deduced. Among others, we note the following cases:_  
* \(a_{2n}=a_{n}=0\)_,_ \(a_{2n-1}=\pm 1\) _and_ \(a_{k}=0\) _for all_ \(k=2,3,\cdots,n-1\)_. In particular, take_ \(a_{1}=1\)_._
* \(a_{2n}=a_{n}=0\)_,_ \(a_{2n-1}=\pm 1\) _and_ \(a_{k}=0\) _for all_ \(k=n+1,n+2,\cdots,2n-2\)_. In particular, take_ \(a_{1}=1\)_._  
**Theorem 4**. : _Under the assumptions of Theorem 3, we have_  
\[\omega(C(q))\leq\frac{1}{2}\left(L+\cos\biggl{(}\frac{\pi}{n+1} \biggr{)}+\sqrt{\left(L-\cos\biggl{(}\frac{\pi}{n+1}\biggr{)}\right)^{2}+ \left(D_{1}+D_{2}\right)^{2}}\right), \tag{25}\]  
_where_  
\[L:=\frac{1}{2}\left(\sqrt{\sum_{k=n+2}^{2n}|a_{k}|^{2}}+\sqrt{ \sum_{k=n+2}^{2n}|a_{k}|^{2}+\left(|a_{n+1}|+1\right)^{2}}\right),\]  
\[D_{1}=\frac{1}{2}\left(|a_{n}|+\sqrt{|a_{n}|^{2}+|1-a_{1}|^{2}+ \sum_{k=2}^{n-1}|a_{k}|^{2}}\right)\]  
_and_  
\[D_{2}=\frac{1}{2}\left(|a_{n}|+\sqrt{|a_{n}|^{2}+|1+a_{1}|^{2}+ \sum_{k=2}^{n-1}|a_{k}|^{2}}\right),\]  
**Proof. ** Applying (20) to \(C(q)\) given in (16) by setting \(A=A_{11}\), \(B=A_{12}\), \(C=A_{21}\) and \(D=A_{22}\). So that, we obtain  
\[\omega(C(q))\leq\] \[\frac{1}{2}\biggl{(}\omega(A_{11})+\omega(A_{22})+\sqrt{\left( \omega(A_{11})-\omega(A_{22})\right)^{2}+\left(\omega(A_{12}+A_{21})+\omega(A _{12}-A_{21})\right)^{2}}\biggr{)}. \tag{26}\]  
Let us observe that \begin{table}
\begin{tabular}{c c c} \hline
**Result** & **Upper Bound of \(|\text{Re}(\lambda)|\)** & **Upper bound of \(|\text{Im}(\lambda)|\)** \\ \hline Kittaneh [14] & 3.999737494 & 3.576384821 \\ Theorem 3 & 2.476786336 & 2.585204772 \\ \hline \end{tabular}
\end{table} 
Table 2: Comparisons of upper bounds based on the considered polynomial \(q_{2}(z)\). Because \(\omega(A_{12}-A_{21})=\cos\bigl{(}\frac{\pi}{n+1}\bigr{)}\), by substituting (27)-(29) into (26), we obtain the required result in (25). The following example illustrates that the upper bound given in Theorem 4 is better than some famous and recent upper bounds obtained in the literature. Any zero of \(q_{3}(z)=z^{6}+\frac{1}{2}z^{5}+\frac{1}{16}z^{2}+1\) is bounded by any values given in Table 3 and shows that our presented results could be much better than all compared upper bounds listed in Table 2.  
Under the assumptions of Theorem 4. If \(a_{k}=0\) for all \(k=2,3,\cdots,n\) and \(a_{1}=1\) (or \(a_{1}=-1\)), then, we have  
\[\omega(C(q))\leq\frac{1}{2}\Biggl{(}L+\cos\biggl{(}\frac{\pi}{n+1}\biggr{)}+ \sqrt{\biggl{(}L-\cos\biggl{(}\frac{\pi}{n+1}\biggr{)}\biggr{)}^{2}+1}\Biggr{)}. \tag{30}\]  
where \(L\) is defined in Theorem 4. Proof. : From (20), we have \(\omega(A_{12}+A_{21})=0\) and \(\omega(A_{12}-A_{21})=1\), if \(a_{1}=1\). Thus, we have  
\[(\omega(A_{12}+A_{21})+\omega(A_{12}-A_{21}))^{2}=1.\]  
In addition, we always have \(\omega(A_{22})=\cos\bigl{(}\frac{\pi}{n+1}\bigr{)}\). Employing (20), we obtain the desired result in (30).