Problem 1. Write in cartesian and polar forms the complex number $$z=\frac{1}{1+\cos t+i \sin t}$$ where $t$ is real and not an odd multiple of $\pi .$
Proof . 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 $$z=\frac{-1}{\cos (t / 2)}(\cos ((t / 2)+\pi)-i \sin ((t / 2)+\pi))$$
Problem 2. Let $z$ be in the open unit disk $\mathbb{D}$. Show that $$\left|\frac{z^{2 n}}{2+z^{n}+z^{5 n}}\right| \leq \frac{|z|^{2 n}}{2(1-|z|)}$$
Proof . 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.
Problem 3. Let $z, w, v \in \mathbb{C}$ be such that $$1-z \bar{w} \neq 0, \quad 1-z \bar{v} \neq 0 \quad \text { and } \quad 1-v \bar{w} \neq 0$$ Show that $$\frac{1-b_{w}(z) \overline{b_{w}(v)}}{1-z \bar{v}}=\frac{1-|w|^{2}}{(1-z \bar{w})(1-\bar{v} w)}$$
Proof . 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.
Problem 4. Let $z_{1}$ and $z_{2}$ be complex numbers. Show that $$\sin z_{1}=\sin z_{2} \Longleftrightarrow\left\{\begin{array}{ll} z_{1}=z_{2}+2 k \pi, & \text { for some } k \in \mathbb{Z}, \quad \text { or } \\ z_{1}+z_{2}=(2 k+1) \pi, & \text { for some } k \in \mathbb{Z} \end{array}\right.$$
Proof . 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 $$\sin z_{1}=\sin z_{2} \Longleftrightarrow\left\{\begin{array}{ll} \frac{z_{1}+z_{2}}{2} \in \frac{\pi}{2}+\pi \mathbb{Z}, & \text { or } \\ \frac{z_{1}-z_{2}}{2} \in \pi \mathbb{Z} \end{array}\right.$$
Problem 5. Exercise 1.5.5. (See for instance $[98,$ pp. $43-44]$ ) Prove the following classical factorizations: \begin{aligned} z^{2 n}+1 &=\prod_{k=0}^{n-1}\left(z^{2}-2 z \cos \left(\frac{2 k+1}{2 n} \pi\right)+1\right) \\ z^{2 n+1}+1 &=(z+1) \prod_{k=1}^{n}\left(z^{2}-2 z \cos \left(\frac{2 k-1}{2 n+1} \pi\right)+1\right) \\ z^{2 n}-1 &=(z+1)(z-1) \prod_{k=1}^{n-1}\left(z^{2}-2 z \cos \left(\frac{k}{n} \pi\right)+1\right) \\ z^{2 n+1}-1 &=(z-1) \prod_{k=1}^{n}\left(z^{2}-2 z \cos \left(\frac{2 k}{2 n+1} \pi\right)+1\right) \end{aligned} Using the third identity, decompose the polynomial $p(z)=\sum_{k=0}^{n-1} z^{2 k}$ into irreducible factors and prove the identity $$\prod_{k=1}^{n-1} \sin \left(\frac{k \pi}{2 n}\right)=\frac{\sqrt{n}}{2^{n-1}}$$ Still using the third identity, prove (see [98, p. 44]) that $$\frac{\sin n t}{\sin t}=2^{n-1} \prod_{k=1}^{n-1}\left(\cos t-\cos \frac{k \pi}{n}\right) .$$
Proof . 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.
Problem 6. Exercise 1.5.7. Given complex numbers $c_{1}, \ldots, c_{n}$ not all equal to 0 , show that $$z^{n}+c_{1} z^{n-1}+\cdots+c_{n}=0 \quad \Longrightarrow \quad|z|<2 \max _{j=1, \ldots, n}\left|c_{j}\right|^{\frac{1}{j}}$$
Proof . 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 $|z|<2 \max _{j=1, \ldots, n}\left|c_{j}\right|^{\frac{1}{j}}$.

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