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证明 . we have $$ \begin{aligned} \frac{1}{1+\cos t+i \sin t} &=\frac{1+\cos t-i \sin t}{(1+\cos t)^{2}+\sin ^{2} t} \\ &=\frac{1+\cos t-i \sin t}{2(1+\cos t)} \\ &=\frac{1}{2}-i \frac{\sin t}{2(1+\cos t)} \end{aligned} $$ Thus, for \(t \neq \pi(\bmod 2 \pi)\) we have \(\operatorname{Re} \frac{1}{1+\cos t+i \sin t}=\frac{1}{2} \quad\) and \(\quad \operatorname{Im} \frac{1}{1+\cos t+i \sin t}=-\frac{1}{2} \tan \frac{t}{2}\) As for the polar decomposition, recall the formulas $$ 1+\cos t=2 \cos ^{2}(t / 2) \text { and } \sin t=2 \cos (t / 2) \sin (t / 2) $$ Thus, for \(t\) not an odd multiple of \(\pi\) we have $$ z=\frac{1}{\cos (t / 2)} \frac{1}{\cos (t / 2)+i \sin (t / 2)}=\frac{1}{\cos (t / 2)}(\cos (t / 2)-i \sin (t / 2)) $$ For \(t \in(0, \pi) \cup(3 \pi, 4 \pi)(\bmod 4 \pi),\) we have \(\cos (t / 2)>0\) and the polar representation of \(z\) is $$ z=\frac{1}{\cos (t / 2)}(\cos (t / 2)-i \sin (t / 2)) $$ For \(t \in(\pi, 3 \pi)(\bmod 4 \pi),\) we have \(\cos (t / 2)<0\) and the polar representation of \(z\) is $$ z=\frac{-1}{\cos (t / 2)}(\cos ((t / 2)+\pi)-i \sin ((t / 2)+\pi)) $$

证明 . We have $$ \left|2+z^{n}+z^{5 n}\right| \geq|2-| z^{n}+z^{5 n}|| $$ For \(|z|<1\) we have $$ \left|z^{n}+z^{5 n}\right| \leq 2|z|<2 $$ and so $$ |2-| z^{n}+z^{5 n}||=2-\left|z^{n}+z^{5 n}\right| \geq 2-2|z|=2(1-|z|) $$ and hence the result.

证明 . We have $$ \begin{aligned} 1-b_{w}(z) \overline{b_{w}(v)} &=1-\frac{(z-w)(\bar{v}-\bar{w})}{(1-z \bar{w})(1-\bar{v} w)} \\ &=\frac{(1-z \bar{w})(1-\bar{v} w)-(z-w)(\bar{v}-\bar{w})}{(1-z \bar{w})(1-\bar{v} w)} \\ &=\frac{(1-z \bar{v})\left(1-|w|^{2}\right)}{(1-z \bar{w})(1-\bar{v} w)} \end{aligned} $$ and hence we obtain the required identity.

证明 . We have $$ \sin z_{1}-\sin z_{2}=2 \cos \left(\frac{z_{1}+z_{2}}{2}\right) \sin \left(\frac{z_{1}-z_{2}}{2}\right) $$ and hence $$ \sin z_{1}=\sin z_{2} \Longleftrightarrow\left\{\begin{array}{ll} \frac{z_{1}+z_{2}}{2} \in \frac{\pi}{2}+\pi \mathbb{Z}, & \text { or } \\ \frac{z_{1}-z_{2}}{2} \in \pi \mathbb{Z} \end{array}\right. $$

证明 . The idea behind the four factorizations is that the polynomials are real, and hence their non-real roots appear in pairs, which lead to second-degree real polynomials: $$ \left(z-z_{0}\right)\left(z-\overline{z_{0}}\right)=z^{2}-2\left(\operatorname{Re} z_{0}\right) z+\left|z_{0}\right|^{2} $$ We focus on the first and third equalities, and leave to the reader the proofs of the other two. The roots of the polynomial \(z^{2 n}+1\) are \(z_{k}=e^{i \theta_{k}},\) with $$ \theta_{k}=\frac{\pi}{2 n}+\frac{k \pi}{n}, \quad k=0, \ldots, 2 n-1 $$ The roots corresponding to \(k=0, \ldots, n-1\) are not conjugate to each other; indeed a pair of indices \(\left(k, k^{\prime}\right)\) corresponds to conjugate roots if $$ \frac{\pi}{2 n}+\frac{k \pi}{n}=-\frac{\pi}{2 n}-\frac{k^{\prime} \pi}{n} \quad(\bmod 2 \pi) $$ that is $$ \frac{1}{n}+\frac{k+k^{\prime}}{n}=0 \quad(\bmod 2) $$ which cannot hold if both \(k\) and \(k^{\prime}\) are between 0 and \(n-1\). Thus $$ z^{2 n}+1=\prod_{k=0}^{2 n-1}\left(z-z_{k}\right)=\prod_{k=0}^{n-1}\left(z-z_{k}\right)\left(z-\overline{z_{k}}\right) $$ But $$ \left(z-z_{k}\right)\left(z-\overline{z_{k}}\right)=z^{2}-2 z \cos \theta_{k}+1 $$ which concludes the proof of the first equality since $$ \cos \theta_{k}=\cos \left(\frac{(2 k+1) \pi}{2 n}\right) $$ We now prove the third equality. The roots of order \(2 n\) of the unity are $$ z_{k}=\exp i \frac{2 k \pi}{2 n}=\exp i \frac{k \pi}{n}, \quad k=0, \ldots, 2 n-1 $$ We have \(z_{0}=1\) and \(z_{n}=-1 .\) The other roots are not real, and appear in pairs since \(p(z)=z^{2 n}-1\) has real coefficients (and thus, \(p(w)=0 \Longrightarrow p(\bar{w})=0 ;\) see Exercise 1.5 .3\() .\) The roots from \(k=1\) to \(k=n-1\) are all different and so the roots of \(p(z)\) are, besides 1 and -1 , $$ z_{k} \quad \text { and } \quad \overline{z_{k}}, \quad k=1, \ldots, n-1 . $$ Thus $$ \begin{aligned} p(z) &=(z+1)(z-1) \prod_{k=1}^{n-1}\left(z-z_{k}\right)\left(z-\overline{z_{k}}\right) \\ &=(z+1)(z-1) \prod_{k=1}^{n-1}\left(z^{2}-2 z \operatorname{Re} z_{k}+1\right) \end{aligned} $$ which concludes the proof of the third equality since \(\operatorname{Re} z_{k}=\cos \left(\frac{k \pi}{n}\right)\) Using previous formula for the sum of a geometric series we obtain $$ p(z)=\sum_{k=0}^{n-1} z^{2 k}=\frac{1-z^{2 n}}{1-z^{2}} $$ and hence, using the previous arguments to prove the third equality and also using the third equality itself we have $$ p(z)=\prod_{k=0}^{n-1}\left(z-\exp \frac{i k \pi}{n}\right)\left(z-\exp \frac{-i k \pi}{n}\right)=\prod_{k=1}^{n-1}\left(z^{2}-2 z \operatorname{Re} z_{k}+1\right) $$ We now prove the previous formula. Setting \(z=1\) in the above equality we have $$ n=\prod_{k=1}^{n-1}\left(2-2 \cos \left(\frac{k \pi}{n}\right)\right) $$ Recall that $$ 1-\cos \left(\frac{k \pi}{n}\right)=2 \sin ^{2}\left(\frac{k \pi}{2 n}\right) $$ Hence $$ \begin{aligned} n &=\prod_{k=1}^{n-1}\left(2-2 \cos \left(\frac{k \pi}{n}\right)\right) \\ &=\prod_{k=1}^{n-1} 4 \sin ^{2}\left(\frac{k \pi}{2 n}\right) \\ &=4^{n-1} \prod_{k=1}^{n-1} \sin ^{2}\left(\frac{k \pi}{2 n}\right) \end{aligned} $$ and hence the result by taking the square root of both sides since the numbers \(\sin \left(\frac{k \pi}{2 n}\right)>0\) for \(k=1, \ldots, n-1 .\) In view of the proof of the previous formula we note $$ \frac{n}{2^{n-1}}=\prod_{k=1}^{n-1}\left(1-\cos \left(\frac{k \pi}{n}\right)\right) $$ which follows from the previous arguments. Finally, we prove the last formula, We set \(z=e^{i t}\) in the previous one to obtain $$ e^{2 i n t}-1=\left(e^{i t}+1\right)\left(e^{i t}-1\right) \prod_{k=1}^{n-1}\left(e^{2 i t}-2 e^{i t} \cos \left(\frac{k \pi}{n}\right)+1\right) $$ Thus, $$ \begin{aligned} e^{i n t}\left(e^{i n t}-e^{-i n t}\right)=& e^{i t / 2}\left(e^{i t / 2}+e^{-i t / 2}\right) e^{i t / 2}\left(e^{i t / 2}-e^{-i t / 2}\right) \\ & \times \prod_{k=1}^{n-1} e^{i t}\left(e^{i t}+e^{-i t}-2 \cos \left(\frac{k \pi}{n}\right)\right) \end{aligned} $$ Dividing both sides by \(2 i e^{i n t}\) we obtain $$ \sin (n t)=2 \cos (t / 2) \sin (t / 2) \prod_{k=1}^{n-1}\left(2 \cos t-2 \cos \left(\frac{k \pi}{n}\right)\right) $$ and hence the result.

证明 . Solution of Exercise \(1.5 .7 .\) Let \(c=\max _{j=1, \ldots, n}\left|c_{j}\right|^{\frac{1}{j}} .\) By hypothesis \(c>0 .\) Let \(z\) be a root of the polynomial equation $$ z^{n}+c_{1} z^{n-1}+\cdots+c_{n}=0 $$ and let \(u=\frac{z}{c} .\) Dividing both sides of the previous formula by \(c^{n}\) we obtain $$ u^{n}+\frac{c_{1}}{c} u^{n-1}+\cdots+\frac{c_{n}}{c^{n}}=0 $$ By definition of \(c\) we have \(\left|c_{j}\right| \leq c^{j} .\) Therefore, \((1.6 .26)\) leads to $$ |u|^{n} \leq|u|^{n-1}+\cdots+1 $$ Assume that \(|u| \geq 2\). Then \(1 /|u| \leq 1 / 2\). Dividing both sides of \((1.6 .27)\) by \(|u|^{n}\) leads to $$ \begin{aligned} 1 & \leq \frac{1}{|u|}+\cdots+\frac{1}{|u|^{n}} \\ & \leq \frac{1}{2}+\cdots+\frac{1}{2^{n}} \\ &<1 \end{aligned} $$ which is a contradiction. Thus \(|u|<2,\) that is \(|z|<2 \max _{j=1, \ldots, n}\left|c_{j}\right|^{\frac{1}{j}}\).