复积分∫_C f(z)d|z|中d|z|含义及∫_C d|z|/overline{z}求解咨询

Hey there! Let's unpack your questions about this arc length-based complex integral—they're great questions, and it's totally normal to get confused about $d|z|$ at first.

First, recall that for any complex number $z = x + iy$, $|z|$ is its modulus (absolute value), which equals $\sqrt{x^2 + y^2}$. The term $d|z|$ here refers to the arc length element of the curve $C$.

For a smooth curve $C$ parameterized by $z(t) = x(t) + iy(t)$ where $t \in [a, b]$, the arc length element is defined as $|z'(t)|dt$—where $z'(t)$ is the derivative of the parameterization with respect to $t$. So $d|z|$ is just shorthand for this real-valued arc length differential.

A key distinction from the standard complex integral $\int_C f(z)dz$: in that case, $dz = z'(t)dt$ is a complex differential, but $\int_C f(z)d|z|$ is a line integral where we integrate against the real arc length of the curve, even if $f(z)$ is complex-valued.

Let's work through this using the definition step by step:

• Parameterize the curve: $C$ is the unit circle, so we use the standard parameterization $z(t) = e^{it} = \cos t + i\sin t$ for $t$ from $0$ to $2\pi$.
• Compute $d|z|$: First, find the derivative of the parameterization: $z'(t) = ie^{it}$. The modulus of this derivative is $|z'(t)| = |i| \cdot |e^{it}| = 1 \cdot 1 = 1$. This means $d|z| = |z'(t)|dt = dt$.
• Simplify the integrand: The conjugate of $z(t)$ is $\overline{z(t)} = \overline{e^{it}} = e^{-it}$. So $\frac{1}{\overline{z(t)}} = \frac{1}{e^{-it}} = e^{it}$.
• Evaluate the integral: Substitute these into the original integral:
  $$
  \int_C \frac{d|z|}{\overline{z}} = \int_0^{2\pi} e^{it} dt
  $$
  Calculating this integral:
  $$
  \int_0^{2\pi} e^{it} dt = \left. \frac{1}{i}e^{it} \right|_0^{2\pi} = \frac{1}{i}(e^{i2\pi} - e^{i0}) = \frac{1}{i}(1 - 1) = 0
  $$

So the value of the integral is $0$.

内容的提问来源于stack exchange，提问作者Joleok