複素関数論

次の各問に答えよ。

1. 複素関数

$$f(z) = \frac{1}{z(z - 2)^2}$$

を $z = 0$ でローラン展開せよ。 2. 複素関数

$$g(z) = z \sin\left(\frac{1}{z + 2}\right)$$

を $z = -2$ でローラン展開し、級数が収束する領域を示せ。次に、$z = -2$ における留数を求めよ。

GPT-4o

(1)

要在 $z = 0$ 处对 $f(z) = \frac{1}{z(z - 2)^2}$ 进行 Laurent展开，我们需要将 $f(z)$ 分解成 $z$ 的幂级数形式。使用 Laurent 展开的方法，我们不需要直接使用积分公式计算系数 $a_n$，而是通过代数操作和幂级数展开的方法来完成。

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分步求解

1. 分解函数形式
   $f(z) = \frac{1}{z(z - 2)^2}$ 中，$\frac{1}{(z-2)^2}$ 是 $z$ 的幂级数展开的核心部分。将 $z - 2$ 重写为 $-2(1 - \frac{z}{2})$，得到：

   $$f(z) = \frac{1}{z \cdot (-2)^2} \cdot \frac{1}{(1 - \frac{z}{2})^2} = \frac{1}{4z} \cdot \frac{1}{(1 - \frac{z}{2})^2}.$$

2. 展开 $\frac{1}{(1 - \frac{z}{2})^2}$ 的幂级数
   对于 $(1 - x)^{-2}$，可以直接用公式展开：

   $$(1 - x)^{-2} = \sum_{n=0}^\infty (n+1)x^n.$$

   令 $x = \frac{z}{2}$，则：

   $$\frac{1}{(1 - \frac{z}{2})^2} = \sum_{n=0}^\infty (n+1) \left(\frac{z}{2}\right)^n = \sum_{n=0}^\infty \frac{n+1}{2^n} z^n.$$

3. 合并到 $f(z)$ 中
   将展开代入 $f(z) = \frac{1}{4z} \cdot \frac{1}{(1 - \frac{z}{2})^2}$：

   $$f(z) = \frac{1}{4z} \sum_{n=0}^\infty \frac{n+1}{2^n} z^n.$$

   分解为两部分：

   $$f(z) = \sum_{n=0}^\infty \frac{n+1}{4 \cdot 2^n} z^{n-1}.$$

4. 整理系数
   这里 $z^{n-1}$ 的指数意味着：
   • 当 $n=0$ 时，对应 $z^{-1}$ 项；
   • 当 $n > 0$ 时，对应 $z^n$ 项。 整理后得到 Laurent 展开形式：

   $$f(z) = \frac{1}{4z} + \frac{1}{8} + \frac{3z}{32} + \frac{2z^2}{32} + \cdots.$$

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总结

通过幂级数展开和代数操作，无需直接计算积分，我们可以得到 $f(z)$ 的 Laurent 展开形式为：

$$f(z) = \sum_{n=-1}^\infty a_n z^n,$$

其中：

$$a_{-1} = \frac{1}{4}, \quad a_0 = \frac{1}{8}, \quad a_1 = \frac{3}{32}, \quad a_2 = \frac{2}{32}, \ldots.$$

(2)

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DeepSeekV3

(1)

To find the Laurent series expansion of the function

$$f(z) = \frac{1}{z(z - 2)^2}$$

around $z = 0$, we first identify the singularities at $z = 0$ and $z = 2$. The expansion will be valid in the annulus $0 < |z| < 2$. We decompose the function into partial fractions:

$$\frac{1}{z(z - 2)^2} = \frac{A}{z} + \frac{B}{z - 2} + \frac{C}{(z - 2)^2}.$$

Solving for $A$, $B$, and $C$, we find:

$$A = \frac{1}{4}, \quad B = -\frac{1}{4}, \quad C = \frac{1}{2}.$$

Thus, the partial fraction decomposition is:

$$\frac{1}{z(z - 2)^2} = \frac{1}{4z} - \frac{1}{4(z - 2)} + \frac{1}{2(z - 2)^2}.$$

Next, we expand each term in powers of $z$ for $0 < |z| < 2$.

1. The term $\frac{1}{4z}$ is already in the correct form.
2. For $-\frac{1}{4(z - 2)}$, we write:

$$-\frac{1}{4(z - 2)} = \frac{1}{8} \sum_{n=0}^{\infty} \left( \frac{z}{2} \right)^n,$$

which converges for $|z| < 2$. 3. For $\frac{1}{2(z - 2)^2}$, we differentiate the series for $\frac{1}{z - 2}$:

$$\frac{1}{(z - 2)^2} = -\frac{1}{4} \sum_{n=0}^{\infty} (n + 1) \left( \frac{z}{2} \right)^n.$$

Combining these expansions, we get:

$$f(z) = \frac{1}{4z} + \sum_{n=0}^{\infty} \frac{z^n}{2^{n+3}} - \frac{1}{8} \sum_{n=0}^{\infty} (n + 1) \frac{z^n}{2^n}.$$

Simplifying the coefficients:

$$\frac{1}{2^{n+3}} - \frac{n + 1}{8 \cdot 2^n} = -\frac{n}{2^{n+3}},$$

we obtain the final Laurent series:

$$f(z) = \frac{1}{4z} - \sum_{n=0}^{\infty} \frac{n}{2^{n+3}} z^n.$$

This is the Laurent series expansion of $f(z)$ around $z = 0$, valid for $0 < |z| < 2$.

(2)

To solve the problem of finding the Laurent series expansion of $g(z) = z \sin\left(\frac{1}{z + 2}\right)$ around $z = -2$, determining the region of convergence, and finding the residue at $z = -2$, we proceed as follows:

1. Substitution and Series Expansion:
   • Let $\xi = z + 2$, so $z = \xi - 2$.
   • The function becomes $g(z) = (\xi - 2) \sin\left(\frac{1}{\xi}\right)$.
   • Expand $\sin\left(\frac{1}{\xi}\right)$ using its Taylor series:

     $$\sin\left(\frac{1}{\xi}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{1}{\xi}\right)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{-2n-1}$$

2. Laurent Series Expansion:
   • Multiply by $\xi - 2$:

     $$g(z) = (\xi - 2) \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{-2n-1}$$

   • Distribute $(\xi - 2)$:

     $$g(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (\xi^{-2n} - 2 \xi^{-2n-1})$$

   • This can be written as:

     $$g(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{-2n} - 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{-2n-1}$$

3. Region of Convergence:
   • The series converges for all $\xi \neq 0$, which translates to $0 < |z + 2| < \infty$.
4. Residue at $z = -2$:
   • Identify the coefficient of $\xi^{-1}$ in the Laurent series:

     $$g(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{-2n} - 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{-2n-1}$$

   • The term with $\xi^{-1}$ comes from the second sum when $n = 0$:

     $$-2 \cdot \frac{(-1)^0}{1!} \xi^{-1} = -2 \xi^{-1}$$

   • Therefore, the residue at $z = -2$ is $-2$.

Final Answer:

The residue of $g(z) = z \sin\left(\frac{1}{z + 2}\right)$ at $z = -2$ is $\boxed{-2}$.

Explanation

Using the definition of the Laurent series to calculate the coefficients $a_n$ directly via contour integration is indeed a valid approach. However, there are several practical difficulties and complexities associated with this method, especially for functions with higher-order singularities or more complicated forms. Let's discuss these challenges in detail:

1. Complexity of Integration:

• For the function $g(z) = z \sin\left(\frac{1}{z + 2}\right)$, the integrand in the formula for $a_n$ becomes:

  $$\frac{g(z)}{(z + 2)^{n+1}} = \frac{z \sin\left(\frac{1}{z + 2}\right)}{(z + 2)^{n+1}}$$

• This integrand is highly complex due to the presence of the sine function with a reciprocal argument. Evaluating such integrals directly is non-trivial and often requires advanced techniques or special functions.

2. Higher-Order Singularities:

• The function $g(z)$ has an essential singularity at $z = -2$. Essential singularities are more challenging to handle than poles because the behavior of the function near the singularity is much more complicated.
• For essential singularities, the Laurent series has an infinite number of negative powers, making it difficult to compute each coefficient $a_n$ individually.

3. Contour Selection:

• Choosing an appropriate contour $C$ for the integration is crucial. The contour must lie within the annulus of convergence and avoid other singularities.
• For functions with essential singularities, ensuring that the contour does not pass through or too close to the singularity requires careful consideration.

4. Computational Intensity:

• Directly computing the integrals for each coefficient $a_n$ is computationally intensive, especially for higher-order terms. This method is not efficient for obtaining a general expression for the Laurent series.

5. Analytical Challenges:

• The integrals involved may not have closed-form solutions, or they may require special functions that are not elementary. This makes it difficult to obtain explicit expressions for the coefficients.

Example Calculation:

To illustrate the complexity, let's attempt to compute the residue (which corresponds to $a_{-1}$) using the contour integration method:

1. Residue Calculation:
   • The residue at $z = -2$ is given by:

     $$\text{Res}(g, -2) = \frac{1}{2\pi i} \oint_C g(z) \, dz$$

   • Substituting $g(z) = z \sin\left(\frac{1}{z + 2}\right)$:

     $$\text{Res}(g, -2) = \frac{1}{2\pi i} \oint_C z \sin\left(\frac{1}{z + 2}\right) \, dz$$

   • This integral is challenging to evaluate directly due to the oscillatory nature of the sine function with a reciprocal argument.

Conclusion:

While the definition of the Laurent series via contour integration is theoretically sound, it is often impractical for functions with essential singularities or complex forms. The method of series expansion and partial fractions, as demonstrated earlier, is usually more straightforward and efficient for obtaining the Laurent series and residues. Therefore, the standard approach of expanding the function into a series and identifying the coefficients is preferred due to its simplicity and effectiveness in handling such problems.