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 $a^5=1$.
Wednesday, 11 November 2015
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 $$\cos \dfrac{\pi}{5}=\cos (\pi-\dfrac{4\pi}{5})=-\dfrac{-\sqrt{5}-1}{4}=\dfrac{\sqrt{5}+1}{4}.$$
Saturday, 24 October 2015
$|e^{2x\pi i}-1|\leq 2\pi |x|$
Show $|e^{2x\pi i}-1|\leq 2\pi |x|$ for real $x$.
We first prove another result: $$\bigg| \int_b^a f(t) dt\bigg|\leq \int_b^a |f(t)| dt.$$ $$\bigg| \int_b^a f(t) dt\bigg|\stackrel{(*)}{=}e^{-i\phi}\int_b^a f(t) dt=\int_b^a e^{-i\phi}f(t) dt$$ $(*)\quad F=|F|e^{i\phi}$ and $e^{i\phi}e^{-i\phi}=1$ means $|F|=Fe^{-i\phi}$.
Since $\bigg| \int_b^a f(t) dt\bigg|$ is real and $\text{Re}(z)\leq |z|$, we have $$\begin{align}\bigg| \int_b^a f(t) dt\bigg|&=\text{Re}\bigg(\int_b^a e^{-i\phi}f(t) dt\bigg)\\&=\int_b^a \text{Re}(e^{-i\phi}f(t)) dt\\&\leq \int_b^a |e^{-i\phi}f(t)| dt\\&=\int_b^a |f(t)| dt.\end{align}$$ One can also prove this using Cauchy Schwarz inequality.
Onto the problem, let $f(t)=e^{ixt}$ for $x,t\in \Bbb R$. $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|\leq \int_0^{2\pi}|e^{ixt}|dt=2\pi$$ Also $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|=\bigg[\bigg|\frac{e^{ixt}}{ix}\bigg|\bigg]_0^{2\pi}=\frac{|e^{2x\pi i}-1|}{|x|}.$$ Therefore $$|e^{2x\pi i}-1|\leq 2\pi |x|.$$
We first prove another result: $$\bigg| \int_b^a f(t) dt\bigg|\leq \int_b^a |f(t)| dt.$$ $$\bigg| \int_b^a f(t) dt\bigg|\stackrel{(*)}{=}e^{-i\phi}\int_b^a f(t) dt=\int_b^a e^{-i\phi}f(t) dt$$ $(*)\quad F=|F|e^{i\phi}$ and $e^{i\phi}e^{-i\phi}=1$ means $|F|=Fe^{-i\phi}$.
Since $\bigg| \int_b^a f(t) dt\bigg|$ is real and $\text{Re}(z)\leq |z|$, we have $$\begin{align}\bigg| \int_b^a f(t) dt\bigg|&=\text{Re}\bigg(\int_b^a e^{-i\phi}f(t) dt\bigg)\\&=\int_b^a \text{Re}(e^{-i\phi}f(t)) dt\\&\leq \int_b^a |e^{-i\phi}f(t)| dt\\&=\int_b^a |f(t)| dt.\end{align}$$ One can also prove this using Cauchy Schwarz inequality.
Onto the problem, let $f(t)=e^{ixt}$ for $x,t\in \Bbb R$. $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|\leq \int_0^{2\pi}|e^{ixt}|dt=2\pi$$ Also $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|=\bigg[\bigg|\frac{e^{ixt}}{ix}\bigg|\bigg]_0^{2\pi}=\frac{|e^{2x\pi i}-1|}{|x|}.$$ Therefore $$|e^{2x\pi i}-1|\leq 2\pi |x|.$$
Sunday, 11 October 2015
A problem in complex analysis
Show that if $f$ is holomorphic, then $$\Delta|f(z)+z|^2=4|f'(z)+1|^2.$$
There are two ways to deal with the modulus: either use the formula $|z|^2=z\bar{z}$ or separate $z$ into real and imaginary parts. For ease of calculations, we separate $f(z)+z$ into real and imaginary parts, having $$\Delta|f(z)+z|^2=\Delta\bigg((u+x)^2+(v+y)^2\bigg).$$ [Here $u,v$ are actually $u(x,y)$ and $v(x,y)$ respectively.] $$\partial_x \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_x+1)+2(v+y)v_x\\ \partial_x^2 \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)u_{xx}+2(v+y)v_{xx}+2(u_x+1)^2+2v_x^2$$ Therefore, $$\Delta\bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_{xx}+u_{yy})+2(v+y)(v_{xx}+v_{yy})\\+2\bigg( (u_x+1)^2+v_x^2+u_y^2+(v_y+1)^2\bigg).$$ Since $f$ is holomorphic, we have $$u_{xx}+u_{yy}=0\\ v_{xx}+v_{yy}=0\\ u_x=v_y,\quad u_y=-v_x.$$ Finally, $$\Delta|f(z)+z|^2=4\bigg( (u_x+1)^2+u_y^2\bigg)=4|f'(z)+1|^2$$ since $f'(z)+1=u_x+1+iu_y$ and $|f'(z)+1|=(u_x+1)^2+u_y^2$.
There are two ways to deal with the modulus: either use the formula $|z|^2=z\bar{z}$ or separate $z$ into real and imaginary parts. For ease of calculations, we separate $f(z)+z$ into real and imaginary parts, having $$\Delta|f(z)+z|^2=\Delta\bigg((u+x)^2+(v+y)^2\bigg).$$ [Here $u,v$ are actually $u(x,y)$ and $v(x,y)$ respectively.] $$\partial_x \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_x+1)+2(v+y)v_x\\ \partial_x^2 \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)u_{xx}+2(v+y)v_{xx}+2(u_x+1)^2+2v_x^2$$ Therefore, $$\Delta\bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_{xx}+u_{yy})+2(v+y)(v_{xx}+v_{yy})\\+2\bigg( (u_x+1)^2+v_x^2+u_y^2+(v_y+1)^2\bigg).$$ Since $f$ is holomorphic, we have $$u_{xx}+u_{yy}=0\\ v_{xx}+v_{yy}=0\\ u_x=v_y,\quad u_y=-v_x.$$ Finally, $$\Delta|f(z)+z|^2=4\bigg( (u_x+1)^2+u_y^2\bigg)=4|f'(z)+1|^2$$ since $f'(z)+1=u_x+1+iu_y$ and $|f'(z)+1|=(u_x+1)^2+u_y^2$.
Thursday, 1 October 2015
Harmonic conjugate and analyticity
a) For which values of $a,b,c,d$ is the function $u(x,y)=ax^3+bx^2y+cxy^2+dy^3$ harmonic?
b) Find a harmonic conjugate $v(x,y)$ of $u(x,y)$ in this case.
c) Find an analytic function $f(z)=u(x,y)+iv(x,y)$, where $z=x+iy$.
a) Use the Laplace equation for $u$ to get $$(\partial_x^2+\partial_y^2)u(x,y)=(6a+2c)x+(2b+6d)y=0,$$ thus $c=-3a,b=-3d$.
b) $$u(x,y)=ax^3-3dx^2y-3axy^2+dy^3.$$ Harmonic conjugate satisfies the Cauchy-Riemann condition $$v_y=u_x=3a(x^2-y^2)-6dxy,\\ v_x=-u_y=3d(x^2-y^2)+6axy.$$ Integrating these, we have $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ Alternatively, we can put $$f(z)=2u(\dfrac{z}{2},\dfrac{z}{2i})-u(0,0).$$ Then $$\begin{align}f(z)&=2\bigg(a(\dfrac{z}{2})^3-3d(\dfrac{z}{2})^2(\dfrac{z}{2i})-3a(\dfrac{z}{2})(\dfrac{z}{2i})^2+d(\dfrac{z}{2i})^3\bigg)\\&=(a+3id+3a+id)\dfrac{z^3}{4}\\&=(a+id)z^3\\&=(a+id)(x+iy)^3\end{align}$$ The imaginary part is $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ c) $$\begin{align}f(z)&=u(x,y)+iv(x,y)\\&=ax^3-3dx^2y-3axy^2+dy^3+i(dx^3+3ax^2y-3dxy^2-ay^3)\\&=a(x^3+3ix^2y-3xy^2-iy^3)+d(ix^3-3x^2y-3ixy^2+y^3)\\&=a(x+iy)^3+id(x+iy)^3\\&=(a+id)z^3\end{align}$$ Remark: The coefficients $1,3,3,1$ hint the use of $(?+?)^3$. We can also substitute $y=0,x=z$ to get $(a+id)z^3$ immediately!
b) Find a harmonic conjugate $v(x,y)$ of $u(x,y)$ in this case.
c) Find an analytic function $f(z)=u(x,y)+iv(x,y)$, where $z=x+iy$.
a) Use the Laplace equation for $u$ to get $$(\partial_x^2+\partial_y^2)u(x,y)=(6a+2c)x+(2b+6d)y=0,$$ thus $c=-3a,b=-3d$.
b) $$u(x,y)=ax^3-3dx^2y-3axy^2+dy^3.$$ Harmonic conjugate satisfies the Cauchy-Riemann condition $$v_y=u_x=3a(x^2-y^2)-6dxy,\\ v_x=-u_y=3d(x^2-y^2)+6axy.$$ Integrating these, we have $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ Alternatively, we can put $$f(z)=2u(\dfrac{z}{2},\dfrac{z}{2i})-u(0,0).$$ Then $$\begin{align}f(z)&=2\bigg(a(\dfrac{z}{2})^3-3d(\dfrac{z}{2})^2(\dfrac{z}{2i})-3a(\dfrac{z}{2})(\dfrac{z}{2i})^2+d(\dfrac{z}{2i})^3\bigg)\\&=(a+3id+3a+id)\dfrac{z^3}{4}\\&=(a+id)z^3\\&=(a+id)(x+iy)^3\end{align}$$ The imaginary part is $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ c) $$\begin{align}f(z)&=u(x,y)+iv(x,y)\\&=ax^3-3dx^2y-3axy^2+dy^3+i(dx^3+3ax^2y-3dxy^2-ay^3)\\&=a(x^3+3ix^2y-3xy^2-iy^3)+d(ix^3-3x^2y-3ixy^2+y^3)\\&=a(x+iy)^3+id(x+iy)^3\\&=(a+id)z^3\end{align}$$ Remark: The coefficients $1,3,3,1$ hint the use of $(?+?)^3$. We can also substitute $y=0,x=z$ to get $(a+id)z^3$ immediately!
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