关于单位圆内外解析函数积分计算中复共轭函数取值的疑问

Hey there, let's break down this confusion together! First, let's recap the key details to ensure we're aligned:

We're calculating the integral:
$$I = {1 \over {2\pi i}}\int_r {{{\omega (t) \cdot \overline {\varphi(t)} } \over {t - {z_0}}}} dt$$
where:

• $r$ is a curve on the positive unit circle,
• $z_0$ lies outside the unit circle,
• $\omega(z)$ has a first-order pole $z_k$ inside the unit circle and is analytic on the unit circle,
• $\varphi(z)$ is analytic outside the unit circle ($|z| \geq 1$) with the form $\varphi(z) = \frac{a_1}{z} + \frac{a_2}{z^2} + \dots + \frac{a_n}{z^n}$.

Your confusion comes from: why does the paper use $\overline{\varphi\left(\frac{1}{\bar{z}_k}\right)}$ instead of $\overline{\varphi(z_k)}$ when evaluating the residue?

The core issue here is the domain of $\varphi(z)$:

• $\varphi(z)$ is only defined and analytic for $|z| \geq 1$ (outside or on the unit circle). The point $z_k$ is inside the unit circle ($|z_k| < 1$), so $\varphi(z_k)$ is not defined at all! We can't substitute $z = z_k$ directly into $\varphi(z)$ because it's outside the function's valid domain.

So how do we connect the unit-circle-inner analytic function $\overline{\varphi(z)}$ to the outer-defined $\varphi(z)$? We use the unit circle inversion transformation:
For any point $z$ inside the unit circle, define $w = \frac{1}{\bar{z}}$. Since $|z| < 1$, $|w| = \frac{1}{|\bar{z}|} = \frac{1}{|z|} > 1$, which means $w$ lies in the domain of $\varphi(z)$ (outside the unit circle).

Now let's compute $\varphi(w)$ and its conjugate:
$$\varphi(w) = \varphi\left(\frac{1}{\bar{z}}\right) = \sum_{k=1}^n \frac{a_k}{\left(\frac{1}{\bar{z}}\right)^k} = \sum_{k=1}^n a_k \bar{z}^k$$
Taking the complex conjugate of this expression gives:
$$\overline{\varphi\left(\frac{1}{\bar{z}}\right)} = \sum_{k=1}^n \bar{a}_k z^k$$

Notice that this result is exactly the analytic extension of $\overline{\varphi(t)}$ (from the unit circle $t$) into the interior of the unit circle. This is the function we actually need for the residue calculation at $z = z_k$, since it's the valid analytic function inside the unit circle corresponding to $\overline{\varphi(t)}$ on the circle.

To summarize:

• $\varphi(z_k)$ is undefined (since $z_k$ is inside the unit circle, outside $\varphi$'s domain)
• $\frac{1}{\bar{z}_k}$ is outside the unit circle (in $\varphi$'s domain), so $\varphi\left(\frac{1}{\bar{z}_k}\right)$ is valid, and its conjugate gives us the correct value of the inner-analytic function at $z_k$.

备注：内容来源于stack exchange，提问作者Elliot