7.1. Integrals of Complex-Valued Functions
Remark
Let \(\mathbb{C}\) represent the field of complex numbers. If \(z=x+iy\in \mathbb{C}\), where \(x,y\in \mathbb{R}\), then \(x=\text{Real}(z)\) and \(y=\text{Image}(z)\) are called the real and imaginary parts of \(z\), and \(|z|=\sqrt{x^2+y^2}\) and \(\bar z=x-iy\) are called the modulus and the complex conjucate of \(z\), respectively.
Recall that in its polar representation, \(z\) is written in the form \[ z=|z|e^{i\theta}=|z|(\cos \theta+ i\sin \theta), \] and the angle \(\theta=\arg z\), measured in radians, is called the argument of \(z\). Of course, \(\theta\) is determined by the equations \(\cos \theta=x/|z|\) and \(\sin \theta=y/|z|\).
- Note: \(\theta\)가 angle일 때
\[ \begin{cases}x=\text{Real}(z)&\implies x=|z|\cos \theta,\\ y=\text{Image}(z) &\implies y=|z| \sin \theta. \end{cases}\implies z=x+iy= |z|(\cos\theta+i\sin\theta). \]
여기서 절대값 \(|z|\)는, \(xy\)-plane에서 원점으로부터 좌표 \(z\)에서의 길이를 나타낸다. 때문에 complex axis를 포함한 공간 \(\mathbb{C}\)라 할지라도 \(|z|\in\mathbb{R}\)이다.
공식유도: 테일러전개 at \(\theta=0\) 에 의해 \[\begin{eqnarray*} \sin(\theta)&=& \sin(0)+ \theta\sin'(0)+\frac{\theta^2}{2!}\sin''(0)+\cdots\\ &=& \theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}\cdots=\sum_{k=0}^\infty(-1)^{k}\frac{\theta^{2k+1}}{(2k+1)!},\\ \cos(\theta)&=& \cos(0)+ \theta\cos'(0)+\frac{\theta^2}{2!}\cos''(0)+\cdots\\ &=& 1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}\cdots=\sum_{k=1}^\infty(-1)^{k+1}\frac{\theta^{2k}}{(2k)!} \\ \end{eqnarray*}\] 때문에 \[\begin{eqnarray*} e^{i\theta}&=& \sum_{k=0}^\infty \frac{(i\theta)^k}{k!}=1+i\theta+\frac{i^2\theta^2}{2!}+\frac{i^3\theta^3}{3!}+\cdots\\ &=& \left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right) +i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots\right)\\ &=&\cos\theta+i\sin\theta \end{eqnarray*}\] 그러므로, \[ z=|z|e^{i\theta}=|z|(\cos\theta+i\sin\theta). \]
- 중요 : \(e^{-i\theta}\)는 위 그림에서 반지름이 1인 좌표평면 상의 단위원의 한 점(각도 \(\theta\))이므로, \(|e^{-i\theta}|=1\)이다.
Remark
For a measure space \((\Omega,\mathcal{F},\mu)\), and a function \(f:\Omega\rightarrow \mathbb{C}\), with \(f=g+ih\), where \(g,h:\Omega\rightarrow \mathbb{R}\), the function \(f\) is (Borel) measurable if and only if both its real and imagenary parts \(g\) and \(h\) are measurable, and we define \[ \int f \mbox{ }d\mu= \int g \mbox{ }d\mu+i \int h \mbox{ }d\mu, \] as long as both integrals on the right hand side exists. Similarly we say that \(f\) is integrable if
\[ \int |f|\mbox{ }d\mu<\infty, \] and since \(|f|=(g^2+h^2)^{1/2}\le |g|+|h|\) while both \(|g|\le |f|\) and \(|h|\le |f|\), we see that \(f\) is integrable if and only if both its real and imaginary parts are integrable.
- 즉 \(f\)는 real part와 imaginary part가 모두 integrable할 때 integrable하다.
매우 중요하다.
Remark (Modulus Inequality)
If \(f\) is integrable w.r.t \(\mu\), then for any \(\theta\in \mathbb{R}\), we have \[ \left|\int f\mbox{ }d\mu\right|\le \int|f|\mbox{ }d\mu. \]
증명: Note \[\begin{eqnarray*} \left|\text{Re}\left\{e^{-i\theta}\cdot \int f\mbox{ }d\mu\right\}\right|&=& \left|\text{Re}\left\{\int e^{-i\theta}\cdot f\mbox{ }d\mu\right\}\right|\\ &=& \left|\int \text{Re}\left\{e^{-i\theta}\cdot f\right\}\mbox{ }d\mu\right|\\ &\le&\int \left|\text{Re}\left\{e^{-i\theta}\cdot f\right\}\right|\mbox{ }d\mu\\ &\le&\int \left|e^{-i\theta}\cdot f\right|\mbox{ }d\mu= \int |e^{-i\theta}|\cdot |f|\mbox{ }d\mu=\int|f|\mbox{ }d\mu. \end{eqnarray*}\] With \(\theta=\arg\left\{\int f\mbox{ }d\mu \right\}\), we have \[ e^{-i\theta}\int f\mbox{ }d\mu= e^{-i\theta} \cdot \left|\int f\mbox{ }d\mu\right|e^{i\theta}=\left|\int f\mbox{ }d\mu\right|\in \mathbb{R}. \]
즉, \(e^{-i\theta}\int f\mbox{ }d\mu\)는 원래 real value이다.
Thus, we have \[ \left|\int f\mbox{ }d\mu\right|\le \int|f|\mbox{ }d\mu. \]