$|e^{2x\pi i}-1|\leq 2\pi |x|$

Show $|e^{2x\pi i}-1|\leq 2\pi |x|$ for real $x$.

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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 $$|e^{2x\pi i}-1|\leq 2\pi |x|.$$

A problem in complex analysis

Show that if $f$ is holomorphic, then $$\Delta|f(z)+z|^2=4|f'(z)+1|^2.$$

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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 $|f'(z)+1|=(u_x+1)^2+u_y^2$.

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 $z=x+iy$.

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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!