Problems

Solution:

1. Note that the vector \(\overrightarrow{AB}\) is \(b-a\), and if we rotate this by \(\frac13\pi\) we get \(e^{-i\pi/3}(b-a)\) after rotating it. Therefore the point \(K\) is represented by \(a + e^{-i\pi/3}(b-a)\) and so \(G_{AB}\) is \begin{align*} && g_{ab} &= \tfrac13(a + b + a + e^{-i\pi/3}(b-a)) \\ &&&= \tfrac13((1+ e^{-i\pi/3})b+(2-e^{-i\pi/3})a)\\ &&&= \tfrac13((1+\tfrac12 - \tfrac{\sqrt3}{2}i)b + ((2-\tfrac12+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac13((\tfrac32 - \tfrac{\sqrt3}{2}i)b + ((\tfrac32+\tfrac{\sqrt3}{2}i)a) \\ &&&= \tfrac1{\sqrt3}((\tfrac{\sqrt3}2 - \tfrac{1}{2}i)b + ((\tfrac{\sqrt3}2+\tfrac{1}{2}i)a) \\ &&&= \frac{1}{\sqrt3}(\omega^* b + \omega a) \end{align*}
2. First note that \(Q_1\) is a parallelogram iff \(c - a = (b-a) + (d-a)\) ie \(a + c = b+d\) (indeed this is true for all quadrilaterals), so. \begin{align*} && Q_1 &\text{ is a parallelogram} \\ \Longleftrightarrow && a + b &= c + d \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega - \omega^*)(a + c) &= \frac{1}{\sqrt{3}}(\omega -\omega^*)(b + d) \\ \Longleftrightarrow && \frac{1}{\sqrt{3}}(\omega a + \omega^*b)+\frac{1}{\sqrt{3}}(\omega c + \omega^*d) &=\frac{1}{\sqrt{3}}(\omega b + \omega^*c)+\frac{1}{\sqrt{3}}(\omega d + \omega^*a) \\ \Longleftrightarrow && g_{ab}+g_{cd} &=g_{bc}+g_{da} \\ \Longleftrightarrow && Q_2 &\text{ is a parallelogram} \\ \end{align*}
3. We consider \(\frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}}\) so \begin{align*} && \frac{g_{ab}-g_{bc}}{g_{ca}-g_{bc}} &= \frac{(\omega a + \omega^*b)-(\omega b + \omega^* c)}{(\omega c + \omega^*a)-(\omega b + \omega^* c)} \\ &&&= \frac{\omega a- \omega^* c -(\omega- \omega^*)b }{\omega^*a-\omega b -(\omega^* -\omega )c} \\ &&&= \frac{\omega^2 a- c -(\omega^2- 1)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- \omega^4 c -(1- \omega^4)b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&=\omega^2\frac{ a- (1-\omega^2) c -\omega^2b }{a-\omega^2 b -(1 -\omega^2 )c} \\ &&&= \omega^2 \end{align*} Therefore the triangle is equilateral.

Solution: Notice that \(\omega^n = e^{2\pi i} = 1\), so \(\omega\) is a root of \(z^n - 1\), notice also that \((\omega^k)^n =1\) so therefore the \(n\) roots are \(1, \omega, \omega^2, \cdots, \omega^{n-1}\) and so \((z-1)(z-\omega) \cdots (z-\omega^{n-1}) = C(z^n-1)\). By considering the coefficient of \(z^n\) we can see that \(C = 1\).

1. \(P\) lies on the perpendicular bisect of \(1\) and \(\omega\), so \(p = re^{\pi i/n}\), where \(r\) can be positive or negative, but \(|r| = |OP|\). \begin{align*} && |PX_0| \times |PX_1| \times \cdots \times |PX_{n-1}| &= |(p-1)(p-\omega) \cdots (p-\omega^{n-1})| \\ &&&= |p^n - 1| \\ &&&= |r^ne^{\pi i} - 1| \\ &&&= |-|OP|^n - 1| \tag{since \(n\) even} \\ &&&= |OP|^n+1 \end{align*} If \(n\) is odd, depending on the sign of \(r\) we get \(|OP|^n+1\) or \(||OP|^n-1|\).
2. \(\,\) \begin{align*} && (z-\omega) \cdots(z-\omega^{n-1}) &= \frac{z^n-1}{z-1} \\ &&&= 1 + z +\cdots + z^{n-1} \\ && |X_0X_1| \times |X_0X_2| \times \cdots \times |X_0X_{n-1}| &= |(1 - \omega)\cdots(1-\omega^{n-1})| \\ &&&= 1+1+1^2+\cdots + 1^{n-1} \\ &&&= n \end{align*}

Solution:

1. \(G(x) = \frac{1}{6} (1 + x + x^2 + x^3 + x^4 + x^5)\) The pgf for \(R_2\) is: \begin{align*} \frac1{36}x^2 + \frac{2}{36}x^3 + \frac{3}{36}x^4 + \frac{4}{36}x^5 + \frac{5}{36} +\\ \quad \quad + \frac{6}{36}x^1 + \frac{5}{36}x^2 + \frac4{36}x^3 + \frac3{36}x^4 + \frac{2}{36}x^5 + \frac{1}{36} \\ = \frac{1}{6}(1 + x + x^2 + x^3 + x^4 + x^5) = G(x) \end{align*} Since rolling the dice twice is the same as rolling the dice once, rolling the dice \(n\) times will be the same as rolling it once, ie the pgf for \(R_n\) will be \(G(x)\) and the probability \(S_n\) is divisible by \(6\) is \(\frac16\)
2. \(G_1(x) = \frac{1}{6} + \frac{1}{3}x^1 + \frac{1}{6}x^2 + \frac16x^3+ \frac16x^4 = \frac16(1 + 2x+x^2+x^3+x^4)\). If \(G_n\) is the probability generating function for \(T_n\) then we can obtain \(G_n\) by multiplying \(G_{n-1}\) by \(G(x)\) and replacing any terms of order higher than \(5\) with their remainder on division by \(5\). (Or equivalently, working over \(\mathbb{R}[x]/(x^5-1)\). If \(y = 1 + x + x^2 + x^3 + x^4\) then: \begin{align*} xy &= x + x^2 + x^3 + x^4 +x^5 \\ &= x + x^2 + x^3 + x^4 + 1 \\ &= y \\ \\ y^2 &= (1 + x+x^2 + x^3+x^4)^2 \\ &= 1 + 2x + 3x^2 + 4x^3+5x^4+4x^5+3x^6 + 2x^7 + x^8 \\ &= (1+4) + (2+3)x+(3+2)x^2 + (4+1)x^3 + 5x^4 \\ &= 5y \end{align*} \begin{align*} \frac{1}{36}(y+x)(y+x) &= \frac1{36}(y^2 + 2xy + x^2) \\ &= \frac1{36}(5y + 2y + x^2 ) \\ &= \frac1{36}(7y + x^2) \end{align*} Similarly, \begin{align*} G_n(x) &= \l\frac{1}{6}(x+y) \r^n \\ &= \frac1{6^n} \l \sum_{i=0}^n \binom{n}{i} y^ix^{n-i} \r \\ &= \frac1{6^n} \l \sum_{i=1}^n \binom{n}{i} y^i + x^n \r \\ &= \frac1{6^n} \l \sum_{i=1}^n \binom{n}{i} 5^{i-1}y + x^n \r \\ &= \frac1{6^n} \l \frac{1}{5}y((5+1)^n-1) + x^n \r \\ &= \frac1{6^n} \l \frac{1}{5}y(6^n-1) + x^n \r \\ \end{align*} Therefore if \(n \not \equiv 0 \pmod{5}\), we can find the probability of \(T_n = 0\) by looking at the constant coefficient, ie plugging in \(x = 0\), which is: \[\frac1{6^n} \l \frac{1}{5}(6^n-1) \r = \frac{1}{5} \l 1- \frac{1}{6^n} \r \] When \(n \equiv 0 \pmod{5}\) we can also find the constant coefficient by plugging in \(x = 0\), which is: \[\frac1{6^n} \l \frac{1}{5}(6^n-1) + 1 \r = \frac{1}{5} \l 1+ \frac{4}{6^n} \r \]

Solution: \begin{align*} && (z-e^{i \theta})(z-e^{-i\theta}) &= z^2 - (e^{i\theta}+e^{-i\theta})z + 1 \\ &&&= z^2-2\cos \theta z + 1 \end{align*} The \(2n\)th roots of \(-1\) are \(e^{\frac{i (2k+1)\pi}{2n}}, k \in \{-n, \cdots, n-1 \}\) or \(e^{\frac{i k \pi}{2n}}, k \in \{-2n+1, -2n+3, \cdots, 2n-3, 2n-1 \}\) \begin{align*} && z^{2n}+1 &= (z-e^{-i(2n-1)/2n})\cdot (z-e^{-i(2n-3)/2n})\cdots (z-e^{i(2n-3)/2n})\cdot (z-e^{i(2n-1)/2n}) \\ &&&= \prod_{k=1}^n \left (z - e^{i \frac{2k-1}{2n}\pi} \right)\left (z - e^{-i \frac{2k-1}{2n}\pi} \right)\\ &&&= \prod_{k=1}^n \left (z^2 - 2z \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \end{align*}

1. \begin{align*} && i^{2n} + 1 &= \prod_{k=1}^n \left (i^2 - 2i \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \\ \Rightarrow && (-1)^n + 1 &= (-1)^n2^ni^n\prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) \\ \Rightarrow && \prod_{k=1}^n \cos \left ( \frac{(2k-1)\pi}{2n}\right) &= 2^{1-n}(-1)^{n/2} \tag{if \(n\equiv 0\pmod{2}\)} \end{align*}
2. When \(n\) is odd, we notice that two of the roots are \(i\) and \(-i\), if we exclude those, (ie by factoring out \(z^2+1\), we see that \begin{align*} && 1-z^2+z^4-\cdots + z^{2n-2} &= \prod_{k=1, 2k-1\neq n}^n \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=(n+1)/2}^{n} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ &&&= \prod_{k=1}^{(n-1)/2} \left (z^2-2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right) \prod_{k=1}^{(n-1)/2} \left (z^2+2z \cos \left ( \frac{(2k-1)\pi}{2n} \right)+1 \right)\\ \Rightarrow && 1-i^2 + i^4 + \cdots + i^{2n-2} &= \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right) \prod_{k=1}^{(n-1)/2} \left (2 \cos \left ( \frac{(2k-1)\pi}{2n} \right) \right)\\ \Rightarrow && n &= 2^{n-1} \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) \\ \Rightarrow && \prod_{k=1}^{(n-1)/2}\cos^2 \left ( \frac{(2k-1)\pi}{2n} \right) &= n2^{1-n} \end{align*}

Solution: The primitive 4th roots of unity are \(i\) and \(-i\). (Since the other two roots of \(x^4-1\) are also roots of \(x^2-1\) \({\rm C}_4(x) = (x-i)(x+i) = x^2+1\) as required.

1. \(\,\) \begin{align*} && {\rm C}_1 (x) &= x-1 \\ && {\rm C}_2 (x) &= x+1 \\ && {\rm C}_3 (x) &= x^2+x+1 \\ && {\rm C}_5 (x) &= x^4+x^3+x^2+x+1 \\ && {\rm C}_6 (x) &= x^2-x+1 \\ \end{align*}
2. Since \((x^4+1)(x^4-1) = x^8-1\) we must have \(n \mid 8\). But \(n \neq 1,2,4\) so \(n = 8\).
3. \({\rm C}_p(x) = x^{p-1} +x^{p-2}+\cdots+x+1\)
4. Suppose \({\rm C_q}(x) \equiv {\rm C}_r(x){\rm C}_s(x)\), then if \(\omega\) is a primitive \(q\)th root of unity we must \({\rm C}_q(\omega) = 0\), but that means that one of \({\rm C}_r(\omega)\), \({\rm C}_s(\omega)\) is \(0\). But that's only possible if \(r\) or \(s\) \(=q\). If this were the case, then what would the other value be? There are no possible values, hence it's not possible.

Solution: \begin{align*} && 0 &= z^4+z^3+z^2+z+1 \\ \Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{\(z \neq 0\)} \\ &&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\ &&&= u^2+u-1 \\ \Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && 0 &= z^2-\left ( \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\ \Rightarrow && z &= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left ( \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} \end{align*} Since \(z^4+z^3+z^2+z+1 = 0\) we can multiply both sides by \(z-1\) to obtain \(z^5-1 = 0\). Therefore if \(z = r(\cos \theta + i \sin \theta)\) we see that \(z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi\) ie \(z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}\). We are looking for a solution in the first quadrant, therefore \(\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4\) and \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)

Solution: \begin{align*} 0 &= z^2+az+b \\ &= (z-u)(z-v) \\ &= z^2-(u+v)z+uv \end{align*} Therefore by comparing coefficients, \(a = -u-v\) and \(b = uv\). Suppose \(\alpha = \cos(2\pi/7) + i \sin (2\pi/7)\), then by De Moivre, \(\alpha^7 = \cos(2\pi) + i \sin (2\pi) = 1\), ie \(\alpha^7-1 = 0\). Notice that \((\alpha+\alpha^2 + \alpha^4) + (\alpha^3+\alpha^5+\alpha^6) = -1\) and \begin{align*} P &= (\alpha+\alpha^2 + \alpha^4)(\alpha^3+\alpha^5+\alpha^6) \\ &= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^{7}+\alpha^{9}+\alpha^{10} \\ &= 3 + \alpha+ \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 \\ &= 2 \end{align*} Therefore it is a root of \(x^2+x+2 = 0 \Rightarrow x = \frac{-1 \pm i\sqrt{7}}{2}\) Therefore $\cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} = \textrm{Re}(\alpha + \alpha^2 + \alpha^4) = -\frac12$ And \(\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7} = \textrm{Im}(\alpha + \alpha^2 + \alpha^4) = \pm\frac{\sqrt{7}}2\) since it is positive it is \(\frac{\sqrt{7}}{2}\)

Solution: \(S_3 \) $\begin{array}{c | c |c |c |c |c |c |} \text{elements} & e & (12) & (13) & (23) & (123) & (132) \\ \text{order} & 1 & 2 & 2 & 2 & 3 & 3 \\ \end{array}$ \(\mathbb{Z}_6\) $\begin{array}{c | c |c |c |c |c |c |} \text{elements} & 0 & 1 & 2 & 3 & 4 & 5 \\ \text{order} & 1 & 6 & 3 & 2 & 3 & 6 \\ \end{array}$ \(S_3\) is not isomorphic to \(\mathbb{Z}_6\) since \(\mathbb{Z}_6\) has two elements of order \(6\) but \(S_3\) has none. Consider the map \(f : \mathbb{Z}_6 \to C_6\) with \(i \mapsto \alpha^i\). This is an isomorphism, since \(i + j \mapsto \alpha^{i+j} = \alpha^i\alpha^j\) \(S_3\) is non-abelian, since \((12)(123) = (23) \neq (13) = (123)(12)\) but multiplication and addition of complex numbers is commutative.

Solution: Note that \(\cos 4 \phi = \cos \phi, \cos 3 \phi = \cos 2 \phi\) \begin{align*} && LHS & = 2y^2 \\ &&&= 2 \left ( \cos \phi + \cos 2 \phi \right)^2 \\ &&&= 2 \cos ^2 \phi + 2 \cos^2 2 \phi + 4 \cos \phi \cos 2 \phi \\ &&&= \cos 2 \phi+1+ \cos4 \phi+1+2 \left ( \cos \phi + \cos 3 \phi \right) \\ &&&= \cos 2 \phi + 2 + \cos \phi + 2 \cos \phi + 2 \cos 2 \phi \\ &&&= 3(\cos \phi + \cos 2 \phi) + 2 \\ &&&= 3 y + 2 \\ &&&= RHS \end{align*} Therefore \(y\) satisfies \(2y^2 = 3y+2\), which we can solve: \begin{align*} && 0 &= 2y^2-3y-2 \\ &&&= (2y+1)(y-2) \\ \Rightarrow && y &= -\frac12,2 \end{align*} Since \(\cos \phi \neq 1\), \(y \neq 2\), therefore \(y = -\frac12\). \begin{align*} && \sum_{k=0}^{16} \cos k \theta &= \sum_{k=0}^{17} \textrm{Re} \left ( e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \sum_{k=0}^{16}e^{ k \theta i} \right ) \\ &&&= \textrm{Re} \left ( \frac{1-e^{17 \theta i}}{1-e^{i \theta}} \right ) \\ &&&= 0 \end{align*} Suppose \(z = \cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta\) \begin{align*} z^2 &= \left (\cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta \right)^2 \\ &= \cos^2 \theta + \cos^2 2 \theta + \cos^2 4 \theta + \cos^2 8 \theta \\ & \quad \quad 2( \cos \theta \cos 2 \theta + \cos \theta \cos 4 \theta + \cos \theta \cos 8 \theta + \\ & \quad \quad \quad \cos 2 \theta \cos 4 \theta + \cos 2 \theta \cos 8 \theta + \cos 4 \theta \cos 8 \theta) \\ &= \frac12 \left (\cos 2 \theta + 1+ \cos 4 \theta + 1 + \cos 8 \theta + 1 + \cos 16 \theta + 1 \right ) + \\ &\quad \quad ( \cos \theta + \cos 3 \theta + \cos 3 \theta + \cos 5 \theta + \cos 7 \theta + \cos 9 \theta + \\ & \quad \quad \quad \cos 2 \theta + \cos 6 \theta + \cos 6 \theta + \cos 10 \theta +\cos 4 \theta + \cos 12 \theta ) \\ &= \frac12 z + 2 + \\ & \quad \quad ( \cos 3 \theta + \cos 6 \theta - \cos 8 \theta - \cos 11 \theta \\ & \quad \quad \quad - \cos 13 \theta - \cos 14 \theta - \cos 15 \theta - \cos 16 \theta - 1) \\ &= \frac12 z + 1 - z \\ &= -\frac12 z +1 \end{align*} Therefore \(z\) satisfies \(z^2=-\frac12 z+1 \Rightarrow z = \frac{-\frac12 \pm \sqrt{\frac14+4}}{2} = \frac{-1 \pm \sqrt{17}}{4}\) Therefore \(z = \frac{\sqrt{17}-1}{4}\) since \(z > 0\)

Solution: Since \(\omega^3 = 1\) and \(\omega \neq 1\) we must have that \((\omega-1)(1 + \omega + \omega^2) = 0\) but by dividing by \(\omega - 1\) we obtain the desired result. \(1+\omega^2 = -\omega\) so \(|1 + \omega^2| = |-\omega| = 1\) and \(\arg ( 1 + \omega^2) = \arg(-\omega) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}\) \begin{align*} && (1 + 1)^n &= \sum_{k=0}^n \binom{n}{k}\\ && (1+ \omega)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{k} \\ && (1+ \omega^2)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{2k} \\ \Rightarrow && 2^n+(-\omega^2)^n + (-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n (1+1+1)\binom{n}{k} + \sum_{k=0, k \equiv 1 \pmod{3}}^n (1 + \omega + \omega^2) \binom{n}{k} + \sum_{k=0, k \equiv 2 \pmod{3}}^n (1 + \omega^2 + \omega) \binom{n}{k} \\ \Rightarrow && 2^n +((-\omega)^n)^{-1}+(-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n \binom{n}{k} \end{align*} \(2^n +((-\omega)^n)^{-1}+(-\omega)^n = 2^n + 2 \textrm{Re}(-\omega^n) = 2^n + 2 \cos \frac{n\pi}{3}\) Therefore our answer follows. \begin{align*} \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} &= \frac13 \l 2^{25} + 2\cos (\frac{25 \pi}{3}) \r \\ &= \frac13 \l 2^{25} + 2 \cos \frac{\pi}{3} \r \\ &= \frac13 \l 2^{25} + 1 \r \\ &= \frac13 \l (4096 \cdot 4096 \cdot 2) + 1 \r \\ &= 11\,184\,811 \end{align*} Notice that \(S_2 = \binom{24}{2} + \cdots +\binom{24}{23} = \binom{24}{1} + \cdots + \binom{24}{22} = S_1\) and \(S_0 = \binom{24}0 + \cdots + \binom{24}{21} = \frac13 \l 2^{24} + 2 \r\) Therefore since \(S_0 + 2 \cdot S_2 = 2^{24}\) we must have \begin{align*} S_2 &= \frac12 \l 2^{24} - \frac13 \l 2^{24} + 2 \r \r \\ &= \frac13 \l 2^{24} - 1 \r \\ &= \frac13 \l 16777216- 1 \r \\ &= \frac13 \cdot 16777215 \\ &= 5\,592\,405 \end{align*}

Solution: \(R(0) = 0\), \(R(1) = \Omega^2 1 = \Omega^2\), \(R(\Omega^2) = \Omega^4 = -\Omega\), \(R(-\Omega) = -\Omega^3 = 1\) \(T(0) = \frac{\Omega^2}1 = \Omega^2\), \(T(1) = \frac{\Omega + \Omega^2}{2\Omega^2+1} = \frac{2\Omega - 1}{2\Omega-1} = 1\) \(T(\Omega^2) = \frac{\Omega^3 + \Omega^2}{2\Omega^4+1} = \Omega \frac{\Omega^2+\Omega}{-2\Omega+1} = \Omega \frac{2\Omega-1}{-2\Omega+1} = - \Omega\) \(T(-\Omega) = \frac{-\Omega^2 + \Omega^2}{-2\Omega^3+1} = \frac{0}{3} = 0\) Thinking of \(R\) and \(S\) as elements of \(S_4\), we have that \(R = (234), S = (134)\), we can also construct \(RS = (14)(23), R^2S = (12)(34), RSR^2S = (13)(24)\). Therefore we have the subgroups \(\{e, (234), (243)\}\) of order \(3\) and the subgroup \(\{e, (12)(34), (13)(24), (14)(23) \}\) of order \(4\). By Lagrange's theorem this means that both \(3\) and \(4\) divide the order of the group, therefore the group has order divisible by \(12\) (and therefore is at least \(12\)). Yes, we cannot produce any odd permutation, for example \((12)\) cannot be produced. (Since all our generators are even permutations).

Solution: Suppose \(z^7 = 1\), then we can write \(z = \cos \theta + i \sin \theta\) and we must have that: \begin{align*} 0 &= \textrm{Im}((\cos \theta + i \sin \theta)^7) \\ &= \binom{7}{6}\cos^6 \theta \sin \theta - \binom{7}{4} \cos^4 \theta \sin^3 \theta + \binom{7}{2} \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= 7 \cos^6 \sin \theta - 35 \cos^4 \theta \sin ^3 \theta + 21 \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= -\cos^7 \theta \l \tan^7 \theta - 21 \tan^5 \theta + 35 \tan^3 \theta - 7 \tan \theta\r \\ &= \cos^7 \theta \cdot t (t^7-21t^4+35t^2-7) \end{align*} Where \(t = \tan \theta\). So if \(z\) is a root of \(z^7 = 1\) and \(\cos \theta \neq 0, \tan \theta \neq 0\) then \(t\) is a root of the equation. Thererefore the roots are: \(\tan \frac{2\pi k}{7}\) where \(k = 1, 2, \ldots 6\). Noting that \(\tan \frac{\pi}7 = -\tan \frac{6\pi}{7}, \tan \frac{3\pi}{7} = -\tan \frac{4 \pi}{7}, \tan \frac{5\pi}{7} = -\tan \frac{2 \pi}{7}\) we can conclude that: \begin{align*} && 7 &= \prod_{k=1}^k \tan \frac{k \pi}{6} \\ &&&= \l \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \r^2 \\ \Rightarrow&& \pm \sqrt{7} &= \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \end{align*} However, we know that \(\tan \frac{2\pi}{7}\) is positive, \(\tan \frac{4\pi}{7},\tan \frac{6\pi}{7}\) are negative, therefore the result must be positive, ie \(+\sqrt{7}\) Using a similar method, we notice that: \begin{align*} 0 &= \textrm{Im} \l (\cos \theta + i \sin \theta)^n \r \\ &= \cos^n \theta \cdot t (t^{n-1} + \cdots - \binom{n}{n-1}) \end{align*} Therefore \(\prod_{k=0}^{n-1} \tan \frac{k \pi}{n} = n\) and since \(\tan \frac{(2k+1) \pi}{n} = \tan \frac{(n-2k-1)\pi}{n}\) is a map of all the odd numbers to the even numbers (and vice versa) when \(n\) is odd. We also know that the terms less where \(\tan \theta\) has \(\theta < \frac{\pi}{2}\) are positive, and the others even, we can determine the signs: \begin{align*} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} \tan \frac{6 \pi}{9} \tan \frac{8 \pi}{9} & = 3 \\ \tan \frac{2 \pi}{11} \tan \frac{4 \pi}{11} \tan \frac{6 \pi}{11} \tan \frac{8 \pi}{11} \tan \frac{10 \pi}{11} &= -\sqrt{11} \end{align*}