复分析积分计算求助：求解∫_{|z|=1} |z-1| |dz|

Hey there! You’ve already nailed the parameterization step with $z=e^{i\theta}$ ($0\leq\theta\leq2\pi$)—great start. Let’s pick up right where you left off and work through the rest of the integral step by step.

Step 1: Calculate $|dz|$ #

You found $\frac{dz}{d\theta} = ie^{i\theta}$. Remember that $|dz|$ is defined as $\left|\frac{dz}{d\theta}\right|d\theta$, so let’s break down the modulus calculation:

• The modulus of $i$ is $|i|=1$
• The modulus of $e^{i\theta}$ is always $1$ for any real $\theta$ (it’s a point on the unit circle, after all)
• The modulus of a product equals the product of moduli, so $|ie^{i\theta}| = |i| \cdot |e^{i\theta}| = 1 \cdot 1 = 1$

That simplifies things nicely: $|dz| = 1 \cdot d\theta = d\theta$.

Step 2: Simplify $|z-1|$ #

Substitute $z=e^{i\theta} = \cos\theta + i\sin\theta$ into $|z-1|$. For a complex number $a+bi$, its modulus is $\sqrt{a^2 + b^2}$, so:
$$|z-1| = |(\cos\theta - 1) + i\sin\theta| = \sqrt{(\cos\theta - 1)^2 + \sin^2\theta}$$

Now let’s expand and simplify using trig identities:

1. Expand $(\cos\theta - 1)^2$: $\cos^2\theta - 2\cos\theta + 1$
2. Add $\sin^2\theta$: $\cos^2\theta - 2\cos\theta + 1 + \sin^2\theta$
3. Use the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$: $1 - 2\cos\theta + 1 = 2(1 - \cos\theta)$
4. Apply the double-angle identity $1 - \cos\theta = 2\sin^2(\theta/2)$: $2 \cdot 2\sin^2(\theta/2) = 4\sin^2(\theta/2)$

Taking the square root, we get:
$$|z-1| = \sqrt{4\sin^2(\theta/2)} = 2\left|\sin\left(\frac{\theta}{2}\right)\right|$$

Since $\theta$ ranges from $0$ to $2\pi$, $\theta/2$ sits between $0$ and $\pi$, where $\sin(\theta/2)$ is non-negative. So we can drop the absolute value:
$$|z-1| = 2\sin\left(\frac{\theta}{2}\right)$$

Step 3: Evaluate the integral #

Now substitute both simplifications into the original integral:
$$\int_{|z|=1} |z-1| |dz| = \int_{0}^{2\pi} 2\sin\left(\frac{\theta}{2}\right) d\theta$$

Let’s use substitution to make this easier: let $u = \frac{\theta}{2}$, so $d\theta = 2du$. When $\theta=0$, $u=0$; when $\theta=2\pi$, $u=\pi$. The integral becomes:
$$\int_{0}^{\pi} 2\sin(u) \cdot 2du = 4\int_{0}^{\pi} \sin(u) du$$

Compute the definite integral:
$$4\left[ -\cos(u) \right]_{0}^{\pi} = 4\left( -\cos(\pi) + \cos(0) \right)$$

We know $\cos(\pi) = -1$ and $\cos(0)=1$, so:
$$4\left( -(-1) + 1 \right) = 4(1+1) = 8$$

That’s the final result! The value of the integral is $8$.

内容的提问来源于stack exchange，提问作者JacobsonRadical