The automorphism group of the complex plane is $$\text{Aut}(\Bbb C)=\{\text{analytic bijections}\;f:\Bbb C\to \Bbb C\}.$$Such automorphisms have the form $$f(z)=az+b,\quad a,b\in \Bbb C, a\neq 0.$$ (To be proved later) If $f(z)=az+b,g(z)=a'z+b'$, then $$f\circ g(z)=aa'z+(ab'+b)\\f^{-1}(z)=a^{-1}z-a^{-1}b.$$ Expressed in matrix form, we have $$\begin{pmatrix}a&b\\0&1\end{pmatrix}\begin{pmatrix}a'&b'\\0&1\end{pmatrix}=\begin{pmatrix}aa'&ab'+b\\0&1\end{pmatrix}\\ \begin{pmatrix}a&b\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}a^{-1}&-a^{-1}b\\0&1\end{pmatrix},$$ which encode the formulas for composing and inverting automophisms of the plane.
The correspondence $$f(z)=az+b \longleftrightarrow \begin{pmatrix}a&b\\0&1\end{pmatrix}$$ is a natural group isomorphism. We now define the parabolic group as $$P=\bigg\{\begin{pmatrix}a&b\\0&1\end{pmatrix}\bigg|\; a,b\in \Bbb C, a\neq 0\bigg\}.$$ Thus $$\text{Aut}(C)\cong P.$$ Two subgroups of $P$ are its Levi component $$M=\bigg\{\begin{pmatrix}a&0\\0&1\end{pmatrix}\bigg|\; a\in \Bbb C, a\neq 0\bigg\},$$ describing the dilations $f(z)=az$, and its unipotent radical $$N=\bigg\{\begin{pmatrix}1&b\\0&1\end{pmatrix}\bigg|\; b\in \Bbb C\bigg\},$$ describing the translations $f(z)=z+b$. The parabolic group takes the form $$P=MN=NM$$ since $$\begin{pmatrix}a&b\\0&1\end{pmatrix}=\begin{pmatrix}a&0\\0&1\end{pmatrix}\begin{pmatrix}1&\color{blue}{a^{-1}b}\\0&1\end{pmatrix}=\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}.$$ Geometrically, any affine map is the composition of a translation and a dilation, in which the order of translation and dilation does not matter. Also, $M$ normalizes $N$, meaning that $$m^{-1}nm\in N\quad \forall m\in M, n\in N$$ since $$\begin{pmatrix}a^{-1}&0\\0&1\end{pmatrix}\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}=\begin{pmatrix}1&a^{-1}b\\0&1\end{pmatrix}.$$ Geometrically, it says that a dilation followed by a translation followed by the reciprocal dilation is again a translation. On the other hand, $N$ does not normalize $M$.
Saturday, 5 December 2015
Friday, 4 December 2015
Motivation for the definition of a group
We know that $x+1=2$ implies $x=1$. When solving such an equation, we are actually working with the properties of a group. $$\begin{align}x+1&=2\quad &\text{integers under +}\\x+1+(-1)&=2+(-1)\quad &\text{inverses}\\x+1+(-1)&=1\quad &\text{closure under +}\\x+[1+(-1)]&=1\quad &\text{associativity}\\x+0&=1\quad &\text{identity}\\x&=1\end{align}$$ As it turns out, the definition of a group is the simplest definition that will let you solve a basic equation. How amazing is that!
Thursday, 12 November 2015
A typical integral in complex analysis
Show that $$\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2\theta+b^{2}\sin^{2}\theta}\; d\theta=\frac{2\pi}{ab}$$
Method I: Cauchy Integral Formula
Let $C$ be the ellipse $\{z=x+iy\mid x=a\cos \theta,\quad y=b\sin \theta,\quad 0\leq \theta \leq 2\pi\}$.
We consider the integral $$\begin{align}\int_C \frac{1}{z}\; dz&=\int_0^{2\pi} \frac{1}{a\cos\theta+ib\sin\theta}(-a\sin\theta+ib\cos\theta)\; d\theta\\&=\int_0^{2\pi} \frac{a\cos\theta-ib\sin\theta}{a^2\cos^2\theta+b^2\sin^2\theta}(-a\sin\theta+ib\cos\theta)\; d\theta\\&=\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta + iab}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta.\end{align}$$ Since we know that $$\int_{C} \dfrac{1}{z}dz=2\pi i,$$ we have $$\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=0\\ \int_0^{2\pi} \frac{iab}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=2\pi i.$$ We therefore have $$\int_{0}^{2\pi}\frac{\sin\theta\cos\theta}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=0\\ \int_{0}^{2\pi}\frac{1}{a^{2}\cos^2\theta+b^{2}\sin^{2}\theta}\; d\theta=\frac{2\pi}{ab}.$$
Method II: Residue theorem
Put $z=e^{i\theta}$. Then $dz=ie^{i\theta}\; d\theta$ and $d\theta=1/(iz)\; dz$. $$\begin{align}\int_0^{2\pi}\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta&=\int_{|z|=1} \frac{4}{a^2(z+1/z)^2 - b^2(z-1/z)^2} \frac{1}{iz}\; dz\\&=\frac{1}{i}\int_{|z|=1} \frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2} \; dz.\end{align}$$ This has the following four simple poles: $$z= \pm i\sqrt{\frac{a-b}{a+b}}, \pm i\sqrt{\frac{a+b}{a-b}}.$$ Now suppose that $0<b<a$ (other cases are treated similarly). This leaves only $\pm i\sqrt{\dfrac{a-b}{a+b}}$ inside the unit circle.
We now find the residue at $i\sqrt{\dfrac{a-b}{a+b}}$, which happens to be the same as that at $- i\sqrt{\dfrac{a-b}{a+b}}$: $$\begin{align}\mathrm{Res}\left(\frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2}; z_0=i\sqrt{\frac{a-b}{a+b}} \right)&=\left.\frac{4z}{2a^2(z^2+1)2z - 2b^2(z^2-1)2z}\right|_{z=z_0}\\&= \left.\frac{1}{(a^2-b^2)z^2 + (a^2+b^2)}\right|_{z=z_0}\\ &=\frac{1}{-(a^2-b^2)(a-b)/(a+b)+(a^2+b^2)}\\&=\frac{1}{-(a-b)^2 + (a^2+b^2)}\\&=\frac{1}{2ab}.\end{align}$$ Therefore $$\int_{0}^{2\pi}\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=\frac{1}{i} 2\pi i \times \left(\frac{1}{2ab} + \frac{1}{2ab}\right)= \frac{2\pi}{ab}.$$
Method I: Cauchy Integral Formula
Let $C$ be the ellipse $\{z=x+iy\mid x=a\cos \theta,\quad y=b\sin \theta,\quad 0\leq \theta \leq 2\pi\}$.
We consider the integral $$\begin{align}\int_C \frac{1}{z}\; dz&=\int_0^{2\pi} \frac{1}{a\cos\theta+ib\sin\theta}(-a\sin\theta+ib\cos\theta)\; d\theta\\&=\int_0^{2\pi} \frac{a\cos\theta-ib\sin\theta}{a^2\cos^2\theta+b^2\sin^2\theta}(-a\sin\theta+ib\cos\theta)\; d\theta\\&=\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta + iab}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta.\end{align}$$ Since we know that $$\int_{C} \dfrac{1}{z}dz=2\pi i,$$ we have $$\int_0^{2\pi} \frac{(b^2-a^2)\sin\theta\cos\theta}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=0\\ \int_0^{2\pi} \frac{iab}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=2\pi i.$$ We therefore have $$\int_{0}^{2\pi}\frac{\sin\theta\cos\theta}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=0\\ \int_{0}^{2\pi}\frac{1}{a^{2}\cos^2\theta+b^{2}\sin^{2}\theta}\; d\theta=\frac{2\pi}{ab}.$$
Method II: Residue theorem
Put $z=e^{i\theta}$. Then $dz=ie^{i\theta}\; d\theta$ and $d\theta=1/(iz)\; dz$. $$\begin{align}\int_0^{2\pi}\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta&=\int_{|z|=1} \frac{4}{a^2(z+1/z)^2 - b^2(z-1/z)^2} \frac{1}{iz}\; dz\\&=\frac{1}{i}\int_{|z|=1} \frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2} \; dz.\end{align}$$ This has the following four simple poles: $$z= \pm i\sqrt{\frac{a-b}{a+b}}, \pm i\sqrt{\frac{a+b}{a-b}}.$$ Now suppose that $0<b<a$ (other cases are treated similarly). This leaves only $\pm i\sqrt{\dfrac{a-b}{a+b}}$ inside the unit circle.
We now find the residue at $i\sqrt{\dfrac{a-b}{a+b}}$, which happens to be the same as that at $- i\sqrt{\dfrac{a-b}{a+b}}$: $$\begin{align}\mathrm{Res}\left(\frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2}; z_0=i\sqrt{\frac{a-b}{a+b}} \right)&=\left.\frac{4z}{2a^2(z^2+1)2z - 2b^2(z^2-1)2z}\right|_{z=z_0}\\&= \left.\frac{1}{(a^2-b^2)z^2 + (a^2+b^2)}\right|_{z=z_0}\\ &=\frac{1}{-(a^2-b^2)(a-b)/(a+b)+(a^2+b^2)}\\&=\frac{1}{-(a-b)^2 + (a^2+b^2)}\\&=\frac{1}{2ab}.\end{align}$$ Therefore $$\int_{0}^{2\pi}\frac{1}{a^2\cos^2\theta + b^2\sin^2\theta}\; d\theta=\frac{1}{i} 2\pi i \times \left(\frac{1}{2ab} + \frac{1}{2ab}\right)= \frac{2\pi}{ab}.$$
Wednesday, 11 November 2015
An interesting integral
This is a problem related to A problem in complex analysis. $$\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5}$$
Method I: Real analysis
Letting $a=\cos \dfrac{2\pi}{5},b=\cos \dfrac{4\pi}{5}$, we have $$\begin{align}\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}&=\int_{-\infty}^\infty \dfrac{1}{(x^2-2x\cos \dfrac{2\pi}{5}+1)(x^2-2x\cos \dfrac{4\pi}{5}+1)}dx\\&=\dfrac{1}{2(a-b)}\int_{-\infty}^\infty \bigg(\dfrac{-x+2a}{x^2+1-2xa}-\dfrac{-x+2b}{x^2+1-2xb}\bigg)dx\\&=\dfrac{1}{2(a-b)}\bigg(\dfrac{1}{2}\ln \bigg|\dfrac{x^2+1-2xb}{x^2+1-2xa}\bigg| \bigg]_{-\infty}^\infty\\& +\int_{-\infty}^\infty \dfrac{a}{x^2+1-2xa}dx-\int_{-\infty}^\infty \dfrac{b}{x^2+1-2xb}dx\bigg)\\&=\dfrac{1}{2(a-b)}\bigg(0+\int_{-\infty}^\infty \dfrac{a}{x^2+1-2xa}dx-\int_{-\infty}^\infty \dfrac{b}{x^2+1-2xb}dx\bigg)\\&=\dfrac{1}{2(a-b)}\bigg(\int_{-\infty}^\infty \dfrac{a}{(x-a)^2+1-a^2}dx-\int_{-\infty}^\infty \dfrac{b}{(x-b)^2+1-b^2}dx\bigg)\\&=\dfrac{\pi}{2(a-b)} \bigg( \dfrac{a}{\sqrt{1-a^2}}-\dfrac{b}{\sqrt{1-b^2}}\bigg)\\&=\dfrac{\pi}{\sqrt 5}\bigg(\dfrac{\cos \dfrac{2\pi}{5}}{\sin \dfrac{2\pi}{5}}-\dfrac{\cos \dfrac{4\pi}{5}}{\sin \dfrac{4\pi}{5}}\bigg)\\&=\dfrac{\pi}{\sqrt 5} \dfrac{\sin \dfrac{2\pi}{5}}{\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}}\end{align}$$ It remains to prove that $\sqrt{5}\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}=\dfrac{5}{4}$. We have $$\begin{align}\sqrt{5}\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}&=2\sqrt{5}\sin^2 \dfrac{2\pi}{5} \cos \dfrac{2\pi}{5}\\&=2\sqrt{5}(\cos \dfrac{2\pi}{5}-\cos^3 \dfrac{2\pi}{5})\\&=2\sqrt{5}\dfrac{\sqrt{5}-1}{4}\bigg(1-(\dfrac{\sqrt{5}-1}{4})^2\bigg)\\&=\dfrac{\sqrt{5}(\sqrt{5}-1)(10+2\sqrt{5})}{32}\\&=\dfrac{5}{4}.\end{align}$$ Finally, $$\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5}.$$ Sorry for the clumsy solution. I venture there should be a much faster way... But it's nice to review so many techniques in one problem!
Remark: We have used nth roots of unity, partial fraction decomposition (algebraic identity), concept of limit, tangent substitution, trigonometry identity, value of $\cos \dfrac{2\pi}{5}$ and $\cos \dfrac{4\pi}{5}$.
Method II: Complex analysis residue theorem
$$\int_{-\infty}^\infty \dfrac{P(z)}{Q(z)}\;dz=2\pi i\sum_{i=1}^m \dfrac{P(a_i)}{Q'(a_i)},$$ where $a_1,\cdots,a_m$ are the poles of $P/Q$ in the upper half plane. $$\begin{align}\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}&=\int_{-\infty}^\infty \dfrac{x-1}{x^5-1}\; dx\\&=2\pi i\bigg(\dfrac{a-1}{5a^4}+\dfrac{a^2-1}{5a^8}\bigg)\\&=\dfrac{2\pi i}{5}(a^2-a+a^{-1}-a^2)\\&=\dfrac{2\pi i}{5}(-2i\sin \dfrac{2\pi}{5})\\&=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5},\end{align}$$ where $a=e^{2\pi i/5}$ and $a^5=1$.
Method I: Real analysis
Letting $a=\cos \dfrac{2\pi}{5},b=\cos \dfrac{4\pi}{5}$, we have $$\begin{align}\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}&=\int_{-\infty}^\infty \dfrac{1}{(x^2-2x\cos \dfrac{2\pi}{5}+1)(x^2-2x\cos \dfrac{4\pi}{5}+1)}dx\\&=\dfrac{1}{2(a-b)}\int_{-\infty}^\infty \bigg(\dfrac{-x+2a}{x^2+1-2xa}-\dfrac{-x+2b}{x^2+1-2xb}\bigg)dx\\&=\dfrac{1}{2(a-b)}\bigg(\dfrac{1}{2}\ln \bigg|\dfrac{x^2+1-2xb}{x^2+1-2xa}\bigg| \bigg]_{-\infty}^\infty\\& +\int_{-\infty}^\infty \dfrac{a}{x^2+1-2xa}dx-\int_{-\infty}^\infty \dfrac{b}{x^2+1-2xb}dx\bigg)\\&=\dfrac{1}{2(a-b)}\bigg(0+\int_{-\infty}^\infty \dfrac{a}{x^2+1-2xa}dx-\int_{-\infty}^\infty \dfrac{b}{x^2+1-2xb}dx\bigg)\\&=\dfrac{1}{2(a-b)}\bigg(\int_{-\infty}^\infty \dfrac{a}{(x-a)^2+1-a^2}dx-\int_{-\infty}^\infty \dfrac{b}{(x-b)^2+1-b^2}dx\bigg)\\&=\dfrac{\pi}{2(a-b)} \bigg( \dfrac{a}{\sqrt{1-a^2}}-\dfrac{b}{\sqrt{1-b^2}}\bigg)\\&=\dfrac{\pi}{\sqrt 5}\bigg(\dfrac{\cos \dfrac{2\pi}{5}}{\sin \dfrac{2\pi}{5}}-\dfrac{\cos \dfrac{4\pi}{5}}{\sin \dfrac{4\pi}{5}}\bigg)\\&=\dfrac{\pi}{\sqrt 5} \dfrac{\sin \dfrac{2\pi}{5}}{\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}}\end{align}$$ It remains to prove that $\sqrt{5}\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}=\dfrac{5}{4}$. We have $$\begin{align}\sqrt{5}\sin \dfrac{2\pi}{5}\sin \dfrac{4\pi}{5}&=2\sqrt{5}\sin^2 \dfrac{2\pi}{5} \cos \dfrac{2\pi}{5}\\&=2\sqrt{5}(\cos \dfrac{2\pi}{5}-\cos^3 \dfrac{2\pi}{5})\\&=2\sqrt{5}\dfrac{\sqrt{5}-1}{4}\bigg(1-(\dfrac{\sqrt{5}-1}{4})^2\bigg)\\&=\dfrac{\sqrt{5}(\sqrt{5}-1)(10+2\sqrt{5})}{32}\\&=\dfrac{5}{4}.\end{align}$$ Finally, $$\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5}.$$ Sorry for the clumsy solution. I venture there should be a much faster way... But it's nice to review so many techniques in one problem!
Remark: We have used nth roots of unity, partial fraction decomposition (algebraic identity), concept of limit, tangent substitution, trigonometry identity, value of $\cos \dfrac{2\pi}{5}$ and $\cos \dfrac{4\pi}{5}$.
Method II: Complex analysis residue theorem
$$\int_{-\infty}^\infty \dfrac{P(z)}{Q(z)}\;dz=2\pi i\sum_{i=1}^m \dfrac{P(a_i)}{Q'(a_i)},$$ where $a_1,\cdots,a_m$ are the poles of $P/Q$ in the upper half plane. $$\begin{align}\int_{-\infty}^\infty \dfrac{dx}{x^4+x^3+x^2+x+1}&=\int_{-\infty}^\infty \dfrac{x-1}{x^5-1}\; dx\\&=2\pi i\bigg(\dfrac{a-1}{5a^4}+\dfrac{a^2-1}{5a^8}\bigg)\\&=\dfrac{2\pi i}{5}(a^2-a+a^{-1}-a^2)\\&=\dfrac{2\pi i}{5}(-2i\sin \dfrac{2\pi}{5})\\&=\dfrac{4\pi}{5}\sin \dfrac{2\pi}{5},\end{align}$$ where $a=e^{2\pi i/5}$ and $a^5=1$.
Tuesday, 10 November 2015
A problem in complex analysis
Determine the roots of $z^5=1$, and deduce that $$z^5-1=(z-1)(z^2-2z\cos \dfrac{2\pi}{5}+1)(z^2-2z\cos \dfrac{4\pi}{5}+1).$$ Deduce that $$\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}=-\dfrac{1}{2},\quad \cos \dfrac{2\pi}{5}\cos \dfrac{4\pi}{5}=-\dfrac{1}{4},$$ and hence show that $$\cos \dfrac{\pi}{5}=\dfrac{\sqrt{5}+1}{4},\quad \cos \dfrac{2\pi}{5}=\dfrac{\sqrt{5}-1}{4}.$$
Note that $$z^5=1 \Rightarrow (e^{2\pi ki/5})^5=1,\quad k=0,1,2,3,4.$$ We know the roots to $z^5-1=0$ are $1,e^{2\pi i/5},e^{4\pi i/5},e^{-2\pi i/5},e^{-4\pi i/5}$. We thus have $$(z-1)(z-e^{2\pi i/5})(z-e^{-2\pi i/5})(z-e^{4\pi i/5})(z-e^{-4\pi i/5})=0,$$ or $$(z-1)(z^2-z(e^{2\pi i/5}+e^{-2\pi i/5})+1)(z^2-z(e^{4\pi i/5}+e^{-4\pi i/5})+1)=0,$$ which is $$(z-1)(z^2-2z\cos{\dfrac{2\pi i}{5}}+1)(z^2-2z\cos{\dfrac{4\pi i}{5}}+1)=0.$$ Note also that $$z^5-1=(z-1)(z^4+z^3+z^2+z+1).$$ Now expand the expression $$(z^2-2z\cos{\dfrac{2\pi i}{5}}+1)(z^2-2z\cos{\dfrac{4\pi i}{5}}+1)\\=z^4-2z^3(\cos \dfrac{4\pi}{5}+\cos \dfrac{2\pi}{5})+z^2(4\cos \dfrac{2\pi}{5}\cos \dfrac{4\pi}{5}+2)-2z(\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5})+1.$$ Comparing the two equations, we have $$\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}=-\dfrac{1}{2},\quad \cos \dfrac{2\pi}{5} \cos \dfrac{4\pi}{5}=-\dfrac{1}{4}.$$ Treating $\cos \dfrac{2\pi}{5}$ as $a$ and $\cos \dfrac{4\pi}{5}$ as $b$, it remains to solve the linear system: $$\begin{cases}a+b=-\dfrac{1}{2}\\ ab=-\dfrac{1}{4}.\end{cases}$$ Using quadratic formula, we have $$a=\cos \dfrac{2\pi}{5}=\dfrac{\sqrt{5}-1}{4},\quad b=\cos \dfrac{4\pi}{5}=\dfrac{-\sqrt{5}-1}{4}.$$ Therefore $$\cos \dfrac{\pi}{5}=\cos (\pi-\dfrac{4\pi}{5})=-\dfrac{-\sqrt{5}-1}{4}=\dfrac{\sqrt{5}+1}{4}.$$
Note that $$z^5=1 \Rightarrow (e^{2\pi ki/5})^5=1,\quad k=0,1,2,3,4.$$ We know the roots to $z^5-1=0$ are $1,e^{2\pi i/5},e^{4\pi i/5},e^{-2\pi i/5},e^{-4\pi i/5}$. We thus have $$(z-1)(z-e^{2\pi i/5})(z-e^{-2\pi i/5})(z-e^{4\pi i/5})(z-e^{-4\pi i/5})=0,$$ or $$(z-1)(z^2-z(e^{2\pi i/5}+e^{-2\pi i/5})+1)(z^2-z(e^{4\pi i/5}+e^{-4\pi i/5})+1)=0,$$ which is $$(z-1)(z^2-2z\cos{\dfrac{2\pi i}{5}}+1)(z^2-2z\cos{\dfrac{4\pi i}{5}}+1)=0.$$ Note also that $$z^5-1=(z-1)(z^4+z^3+z^2+z+1).$$ Now expand the expression $$(z^2-2z\cos{\dfrac{2\pi i}{5}}+1)(z^2-2z\cos{\dfrac{4\pi i}{5}}+1)\\=z^4-2z^3(\cos \dfrac{4\pi}{5}+\cos \dfrac{2\pi}{5})+z^2(4\cos \dfrac{2\pi}{5}\cos \dfrac{4\pi}{5}+2)-2z(\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5})+1.$$ Comparing the two equations, we have $$\cos \dfrac{2\pi}{5}+\cos \dfrac{4\pi}{5}=-\dfrac{1}{2},\quad \cos \dfrac{2\pi}{5} \cos \dfrac{4\pi}{5}=-\dfrac{1}{4}.$$ Treating $\cos \dfrac{2\pi}{5}$ as $a$ and $\cos \dfrac{4\pi}{5}$ as $b$, it remains to solve the linear system: $$\begin{cases}a+b=-\dfrac{1}{2}\\ ab=-\dfrac{1}{4}.\end{cases}$$ Using quadratic formula, we have $$a=\cos \dfrac{2\pi}{5}=\dfrac{\sqrt{5}-1}{4},\quad b=\cos \dfrac{4\pi}{5}=\dfrac{-\sqrt{5}-1}{4}.$$ Therefore $$\cos \dfrac{\pi}{5}=\cos (\pi-\dfrac{4\pi}{5})=-\dfrac{-\sqrt{5}-1}{4}=\dfrac{\sqrt{5}+1}{4}.$$
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