| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the value of the integral $\int_C \frac{dz}{z(z-2)^2}$, where $C$ is the circle $|z-2| = 1$, we can use the residue theorem from complex analysis. The residue theorem states that the integral of a function around a closed contour is equal to $2\pi i$ times the sum of the residues of the function at its poles inside the contour. First, we identify the poles of the function $f(z) = \frac{1}{z(z-2)^2}$. There are two poles: $z=0$ and $z=2$. However, only $z=2$ is inside the contour $C$, which is the circle centered at $z=2$ with radius $1$. To find the residue at $z=2$, we need to find the coefficient of the $\frac{1}{z-2}$ term in the Laurent series expansion of $f(z)$ around $z=2$. Since $f(z)$ has a pole of order $2$ at $z=2$, we can find the residue by taking the derivative of the function after multiplying it by $(z-2)^2$ and evaluating it at $z=2$. Let $g(z) = (z-2)^2 f(z) = \frac{1}{z}$. Then, we have: \[ g'(z) = -\frac{1}{z^2} \] The residue of $f(z)$ at $z=2$ is then given by $g'(2) = -\frac{1}{4}$. Therefore, by the residue theorem, the integral is: \[ \int_C \frac{dz}{z(z-2)^2} = 2\pi i \cdot \left(-\frac{1}{4}\right) = -\frac{\pi i}{2} \] So, $A = -\frac{\pi}{2}$. |
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| deepseekr1 | To compute the integral \(\int_C \frac{1}{z(z-2)^2} \, dz\) where \(C\) is the contour \(|z - 2| = 1\), we use the residue theorem. 1. **Identify Singularities**: The integrand \(\frac{1}{z(z-2)^2}\) has singularities at \(z = 0\) and \(z = 2\). The contour \(C\) is a circle centered at \(z = 2\) with radius 1. The point \(z = 2\) is inside the contour, while \(z = 0\) is outside. 2. **Residue at \(z = 2\)**: The singularity at \(z = 2\) is a double pole. To find the residue, we use the formula for a pole of order \(m\): \[ \text{Res}(f, a) = \frac{1}{(m-1)!} \lim_{z \to a} \frac{d^{m-1}}{dz^{m-1}} \left[(z - a)^m f(z)\right] \] Here, \(m = 2\). Let \(f(z) = \frac{1}{z(z-2)^2}\). Then, \[ (z - 2)^2 f(z) = \frac{1}{z} \] The first derivative of \(\frac{1}{z}\) is \(-\frac{1}{z^2}\). Evaluating this at \(z = 2\) gives \(-\frac{1}{4}\). Thus, the residue is: \[ \text{Res}(f, 2) = \frac{1}{1!} \left(-\frac{1}{4}\right) = -\frac{1}{4} \] 3. **Apply Residue Theorem**: The integral is \(2\pi i\) times the sum of the residues inside the contour. Since only the residue at \(z = 2\) contributes: \[ \int_C \frac{1}{z(z-2)^2} \, dz = 2\pi i \left(-\frac{1}{4}\right) = -\frac{\pi i}{2} \] 4. **Conclusion**: The integral evaluates to \(-\frac{\pi i}{2}\), so the coefficient \(A\) is \(-\frac{\pi}{2}\). \[ \boxed{-\dfrac{\pi}{2}} \] |
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| gpto1 | To solve the integral: \[ \int_C \frac{dz}{z(z-2)^2}, \] where \(C\) is the circle \(|z - 2| = 1\), we can use the Residue Theorem from complex analysis. **Step 1:** **Identify the singularities of the integrand and determine which ones are inside the contour \(C\).** The integrand \(f(z) = \dfrac{1}{z(z-2)^2}\) has singularities (poles) at: - \(z = 0\) (simple pole) - \(z = 2\) (double pole) Since \(C\) is the circle centered at \(z = 2\) with radius 1, any point \(z\) satisfying \(|z - 2| < 1\) lies inside \(C\). Calculating \(|0 - 2| = 2\), we see that \(z = 0\) is outside \(C\). At \(z = 2\), \(|2 - 2| = 0 < 1\), so the pole at \(z = 2\) is inside \(C\). **Step 2:** **Compute the residue of \(f(z)\) at \(z = 2\).** For a pole of order \(n\) at \(z = z_0\), the residue is: \[ \text{Res}[f, z_0] = \frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}} \left[(z - z_0)^n f(z)\right]. \] Since \(z = 2\) is a double pole (\(n = 2\)), we compute: \[ \text{Res}[f, z = 2] = \lim_{z \to 2} \frac{d}{dz}\left[(z - 2)^2 f(z)\right]. \] First, simplify \((z - 2)^2 f(z)\): \[ (z - 2)^2 f(z) = (z - 2)^2 \cdot \frac{1}{z(z - 2)^2} = \frac{1}{z}. \] Then take the derivative: \[ \frac{d}{dz} \left(\frac{1}{z}\right) = -\frac{1}{z^2}. \] Evaluate at \(z = 2\): \[ \text{Res}[f, z = 2] = -\frac{1}{(2)^2} = -\frac{1}{4}. \] **Step 3:** **Apply the Residue Theorem to compute the integral.** The Residue Theorem states: \[ \int_C f(z) \, dz = 2\pi i \sum \text{Res}[f, \text{inside } C]. \] So: \[ \int_C \frac{dz}{z(z-2)^2} = 2\pi i \left(-\frac{1}{4}\right) = -\frac{\pi i}{2}. \] **Conclusion:** Given that the integral equals \(A i\) (with \(i\) being the imaginary unit), we have: \[ -\frac{\pi i}{2} = A i \implies A = -\frac{\pi}{2}. \] **Answer:** \(-\dfrac{\pi}{2}\) |
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| GPT-4o | To determine the value of \(A\) in the given integral \(\int_C \frac{dz}{z (z-2)^2}\) where \(C\) is the circle \(|z - 2| = 1\), we can utilize the residue theorem. First, let's analyze the integral and the contour: - The integrand is \(\frac{1}{z (z-2)^2}\). - The contour \(C\) is the circle \(|z - 2| = 1\), centered at \(z = 2\) with radius 1. The integrand has singularities at \(z = 0\) and \(z = 2\). We need to determine which singularities lie inside the contour \(C\). - The singularity at \(z = 2\) is on the contour. - The singularity at \(z = 0\) is not inside the contour since \(|0 - 2| = 2\), which is outside the circle \(|z - 2| = 1\). According to the residue theorem, the integral around the closed contour \(C\) is \(2\pi i\) times the sum of the residues of the integrand inside the contour. However, since \(z = 2\) lies on the contour, it technically complicates direct application. But for simplicity in many practical applications, we can often still consider the residue at \(z = 2\) if the contour passes through it. Residue calculation at \(z = 2\): The integrand \(\frac{1}{z (z-2)^2}\) has a pole of order 2 at \(z = 2\). To find the residue at \(z = 2\), we use the formula for the residue at a pole of order 2: \[ \text{Res}\left( \frac{1}{z (z-2)^2}, z = 2 \right) = \lim_{z \to 2} \frac{d}{dz} \left( (z-2)^2 \cdot \frac{1}{z (z-2)^2} \right) \] Simplifying inside the derivative: \[ \lim_{z \to 2} \frac{d}{dz} \left( \frac{1}{z} \right) = \lim_{z \to 2} \left( -\frac{1}{z^2} \right) \] Evaluating the limit: \[ - \frac{1}{(2)^2} = - \frac{1}{4} \] Thus, the residue at \(z = 2\) is \(-\frac{1}{4}\). Using the residue theorem: \[ \int_C \frac{dz}{z (z-2)^2} = 2\pi i \cdot \left( -\frac{1}{4} \right) = -\frac{2\pi i}{4} = -\frac{\pi i}{2} \] The answer is given in the form \(Ai\) where \(i\) is the imaginary unit. Comparing: \[ -\frac{\pi i}{2} = Ai \] Thus, \(A = -\frac{\pi}{2}\). |
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| llama3.1-70B-Instruct | To solve the given problem, we can use the Cauchy's integral formula and the concept of residues. The given integral is: $$\int_C \frac{dz}{z(z-2)^2}$$ where $C$ is the circle $|z-2|=1$. First, we can rewrite the integral as: $$\int_C \frac{dz}{z(z-2)^2} = \int_C \frac{f(z)}{(z-2)^2}dz$$ where $f(z) = \frac{1}{z}$. Since $f(z)$ is analytic inside and on the circle $C$, we can use the Cauchy's integral formula: $$\int_C \frac{f(z)}{(z-2)^2}dz = 2\pi i f'(2)$$ where $f'(2)$ is the derivative of $f(z)$ evaluated at $z=2$. Now, we can compute the derivative of $f(z)$: $$f'(z) = -\frac{1}{z^2}$$ Evaluating this at $z=2$, we get: $$f'(2) = -\frac{1}{4}$$ Therefore, the integral becomes: $$\int_C \frac{dz}{z(z-2)^2} = 2\pi i \left(-\frac{1}{4}\right) = -\frac{\pi i}{2}$$ Comparing this with the given answer $Ai$, we can see that: $$A = -\frac{1}{2}$$ So, the value of $A$ is $-\frac{1}{2}$. |
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| llama3.1-405B-Instruct | To compute the integral $\int_C \frac{dz}{z(z-2)^2}$ where $C: |z - 2| = 1$, we can use the Cauchy Integral Formula. First, we can rewrite the integral as: $$\int_C \frac{dz}{z(z-2)^2} = \int_C \frac{1}{z-2} \cdot \frac{1}{z} dz$$ Now, we can apply the Cauchy Integral Formula, which states that: $$\int_C \frac{f(z)}{z-a} dz = 2\pi i f(a)$$ where $f(z)$ is analytic inside and on the contour $C$, and $a$ is a point inside the contour. In this case, we can choose $f(z) = \frac{1}{z}$ and $a = 2$. Then, we have: $$\int_C \frac{1}{z-2} \cdot \frac{1}{z} dz = 2\pi i \frac{1}{2} = \pi i$$ Therefore, the value of the integral is $\pi i$, so $A = \pi$. The final answer is $\boxed{\pi}$. |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to compute the integral \(\int_C \frac{dz}{z(z-2)^2}\) where \(C\) is the contour defined by \(|z - 2| = 1\). This contour is a circle centered at \(z = 2\) with radius 1. We will use the residue theorem to solve this integral. The residue theorem states that for a contour integral of a function \(f(z)\) around a closed contour \(C\), if \(f(z)\) has singularities inside \(C\), then: \[ \int_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k) \] where the sum is over all singularities \(z_k\) inside \(C\). First, we identify the singularities of the function \(f(z) = \frac{1}{z(z-2)^2}\): - \(z = 0\) (a simple pole) - \(z = 2\) (a pole of order 2) Since the contour \(C\) is \(|z - 2| = 1\), it encloses the singularity at \(z = 2\) but not the singularity at \(z = 0\). Therefore, we only need to compute the residue at \(z = 2\). To find the residue at \(z = 2\), we use the formula for a pole of order \(m\): \[ \text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left( (z - z_0)^m f(z) \right) \] Here, \(z_0 = 2\) and \(m = 2\). Thus, we need to compute: \[ \text{Res}(f, 2) = \lim_{z \to 2} \frac{d}{dz} \left( (z - 2)^2 \cdot \frac{1}{z(z-2)^2} \right) \] Simplifying the expression inside the limit: \[ (z - 2)^2 \cdot \frac{1}{z(z-2)^2} = \frac{1}{z} \] Now, we differentiate: \[ \frac{d}{dz} \left( \frac{1}{z} \right) = -\frac{1}{z^2} \] Evaluating this at \(z = 2\): \[ \text{Res}(f, 2) = -\frac{1}{2^2} = -\frac{1}{4} \] Using the residue theorem, the integral is: \[ \int_C \frac{dz}{z(z-2)^2} = 2\pi i \cdot \text{Res}(f, 2) = 2\pi i \cdot \left( -\frac{1}{4} \right) = -\frac{\pi i}{2} \] Thus, the value of \(A\) in the given problem is: \[ A = -\frac{1}{2} \] So, the final answer is: \[ \boxed{-\frac{1}{2}} \] |
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| Qwen2-72B-Instruct | To find the value of the integral $\int_C \frac{dz}{z(z-2)^2}$, where $C$ is the circle $|z-2| = 1$, we can use the residue theorem from complex analysis. The residue theorem states that the integral of a function around a closed contour is equal to $2\pi i$ times the sum of the residues of the function at its poles inside the contour. First, we identify the poles of the function $f(z) = \frac{1}{z(z-2)^2}$. There are two poles: $z=0$ and $z=2$. However, only $z=2$ is inside the contour $C$, which is the circle centered at $z=2$ with radius $1$. To find the residue at $z=2$, we need to find the coefficient of the $\frac{1}{z-2}$ term in the Laurent series expansion of $f(z)$ around $z=2$. Since $f(z)$ has a pole of order $2$ at $z=2$, we can find the residue by taking the derivative of the function after multiplying it by $(z-2)^2$ and evaluating it at $z=2$. Let $g(z) = (z-2)^2 f(z) = \frac{1}{z}$. Then, we have: \[ g'(z) = -\frac{1}{z^2} \] The residue of $f(z)$ at $z=2$ is then given by $g'(2) = -\frac{1}{4}$. Therefore, by the residue theorem, the integral is: \[ \int_C \frac{dz}{z(z-2)^2} = 2\pi i \cdot \left(-\frac{1}{4}\right) = -\frac{\pi i}{2} \] So, $A = -\frac{\pi}{2}$. |
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| baichuan4 | To solve this problem, we'll use the residue theorem. The function $f(z) = \frac{1}{z(z-2)^2}$ has a simple pole at $z=0$ and a double pole at $z=2$. We need to find the residues at these poles. 1. Residue at $z=0$: The residue of a simple pole can be found by taking the limit of $f(z)(z-0)$ as $z$ approaches $0$: $$\text{Res}(f, 0) = \lim_{z \to 0} (z \cdot \frac{1}{z(z-2)^2}) = \lim_{z \to 0} \frac{1}{(z-2)^2} = \frac{1}{4}$$ 2. Residue at $z=2$: The residue of a double pole can be found by taking the derivative of the coefficient of the $\frac{1}{(z-2)^2}$ term in the Laurent series of $f(z)$ around $z=2$: $$\text{Res}(f, 2) = \frac{d}{dz} \left[ \frac{1}{z(z-2)^2} \right]_{z=2} = \frac{d}{dz} \left[ \frac{1}{2(z-2)^2} \right]_{z=2} = -\frac{1}{2} \cdot \frac{2}{(z-2)^3}_{z=2} = -\frac{1}{4}$$ Now, we can apply the residue theorem: $$\int_C f(z) dz = 2\pi i (\text{Res}(f, 0) + \text{Res}(f, 2)) = 2\pi i (\frac{1}{4} - \frac{1}{4}) = 0$$ So, the integral is $0$. Therefore, $A = 0$. |
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| doubao | Let's use the residue theorem. The function has poles at z = 0 and z = 2. For the pole at z = 0, it is a simple pole and the residue is given by: Res(f, 0) = lim(z→0) z * f(z) = lim(z→0) 1 / (z - 2)^2 = 1 / 4. For the pole at z = 2, it is a pole of order 2. The residue is given by: Res(f, 2) = lim(z→2) d/dz [(z - 2)^2 * f(z)] = lim(z→2) d/dz [1 / z] = -1 / 4. Now, since the contour |z - 2| = 1 encloses only the pole at z = 2, the integral is 2πi * Res(f, 2) = 2πi * (-1 / 4) = -πi / 2. So A = -1 / 2. |
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| minimax | To solve the integral $\int_C \frac{dz}{z(z-2)^2}$, where $C$ is the circle $|z - 2| = 1$, we can use the residue theorem from complex analysis. The residue theorem states that if $f(z)$ is a function that is analytic inside and on a simple closed contour $C$, except for a finite number of singularities inside $C$, then the integral of $f(z)$ around $C$ is $2\pi i$ times the sum of the residues of $f(z)$ at those singularities. The function $f(z) = \frac{1}{z(z-2)^2}$ has singularities at $z = 0$ and $z = 2$. However, since our contour $C$ is the circle $|z - 2| = 1$, only the singularity at $z = 2$ is inside the contour. The singularity at $z = 0$ is outside the contour and does not contribute to the integral. To find the residue at $z = 2$, we need to compute the coefficient of the $\frac{1}{z-2}$ term in the Laurent series expansion of $f(z)$ around $z = 2$. Since we have a pole of order 2 at $z = 2$, we can find the residue by taking the limit as $z$ approaches 2 of the derivative of the numerator of the function obtained by multiplying $f(z)$ by $(z-2)^2$: Residue at $z = 2$ is: $$ \lim_{z \to 2} \frac{d}{dz} \left[ (z-2)^2 \cdot \frac{1}{z(z-2)^2} \right] = \lim_{z \to 2} \frac{d}{dz} \left[ \frac{1}{z} \right] = \lim_{z \to 2} \left[ -\frac{1}{z^2} \right] = -\frac{1}{4} $$ Now, we apply the residue theorem: $$ \int_C \frac{dz}{z(z-2)^2} = 2\pi i \cdot (-\frac{1}{4}) = -\frac{\pi i}{2} $$ The answer is $A = -\frac{\pi i}{2}$. |
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| kimi | To solve the integral $\int_C \frac{dz}{z(z-2)^2}$, where $C: |z - 2| = 1$, we recognize that the contour $C$ is a circle centered at $z = 2$ with a radius of 1. The function $\frac{1}{z(z-2)^2}$ has a simple pole at $z = 0$ and a double pole at $z = 2$. Since the contour $C$ encloses the pole at $z = 2$ but not the pole at $z = 0$, we can use the residue theorem to evaluate the integral. The residue theorem states that for a function with isolated singularities inside a simple closed contour $C$, the integral of the function around $C$ is equal to $2\pi i$ times the sum of the residues of the function at its singularities inside $C$. The residue of $\frac{1}{z(z-2)^2}$ at $z = 2$ can be found by taking the derivative of the function with respect to $z$ and then evaluating it at $z = 2$. However, since the function has a double pole at $z = 2$, we need to take the second derivative. Let $f(z) = \frac{1}{(z-2)^2}$, then the residue at $z = 2$ is given by: $$\text{Res}_{z=2} \left(\frac{1}{z(z-2)^2}\right) = \lim_{z \to 2} (z-2)^2 \frac{1}{z(z-2)^2} = \lim_{z \to 2} \frac{1}{z} = \frac{1}{2}.$$ Now, we can apply the residue theorem: $$\int_C \frac{dz}{z(z-2)^2} = 2\pi i \times \text{Res}_{z=2} \left(\frac{1}{z(z-2)^2}\right) = 2\pi i \times \frac{1}{2} = \pi i.$$ So, in the given problem, $A = \pi$. |
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| wenxinyiyan | 首先，我们识别出这是一个复分析中的围道积分问题，涉及到柯西积分公式和留数定理。 给定的积分是 $\int_C \frac{1}{z(z-2)^2} dz$，其中C是以2为圆心、1为半径的圆，即 $|z - 2| = 1$。 首先，我们找出被积函数 $\frac{1}{z(z-2)^2}$ 的奇点。奇点发生在分母为零的地方，即 $z = 0$ 和 $z = 2$。然而，只有 $z = 2$ 在围道C内。 接下来，我们计算 $z = 2$ 这一点的留数。由于 $z = 2$ 是二阶极点，我们需要使用留数的公式来计算。留数 $Res(f, 2)$ 可以通过以下公式计算： $$Res(f, 2) = \lim_{z \to 2} \frac{d}{dz} [(z-2)^2 f(z)]$$ 其中 $f(z) = \frac{1}{z(z-2)^2}$。 计算留数： $$Res(f, 2) = \lim_{z \to 2} \frac{d}{dz} \left[ \frac{(z-2)^2}{z(z-2)^2} \right] = \lim_{z \to 2} \frac{d}{dz} \left[ \frac{1}{z} \right] = \lim_{z \to 2} -\frac{1}{z^2} = -\frac{1}{4}$$ 现在，根据留数定理，我们有： $$\int_C \frac{1}{z(z-2)^2} dz = 2\pi i \cdot Res(f, 2) = 2\pi i \cdot \left( -\frac{1}{4} \right) = -\frac{\pi i}{2}$$ 因此，积分的结果是 $-\frac{\pi i}{2}$。由于题目中给出的答案是 $Ai$ 形式，我们可以推断出 $A = -\frac{\pi}{2}$。 |
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| zhipu | To solve the integral \(\int_C \frac{dz}{z (z-2)^2}\), where \(C\) is the circle \(|z - 2| = 1\), we can use the residue theorem from complex analysis. ### Step-by-Step Solution: 1. **Identify the Singularities:** The integrand \(\frac{1}{z (z-2)^2}\) has singularities at \(z = 0\) and \(z = 2\). 2. **Determine the Location of Singularities Relative to the Contour:** The contour \(C\) is the circle \(|z - 2| = 1\), which is centered at \(z = 2\) with a radius of 1. Therefore, \(z = 2\) is inside the contour, and \(z = 0\) is outside the contour. 3. **Residue Calculation:** Since \(z = 0\) is outside the contour, we only need to consider the singularity at \(z = 2\). The singularity at \(z = 2\) is a pole of order 2. The residue at \(z = 2\) for a pole of order 2 is given by: \[ \text{Res}\left(\frac{1}{z (z-2)^2}, z = 2\right) = \lim_{z \to 2} \frac{d}{dz} \left( (z-2)^2 \frac{1}{z (z-2)^2} \right) \] Simplifying the expression inside the limit: \[ \frac{d}{dz} \left( \frac{1}{z} \right) = -\frac{1}{z^2} \] Evaluating this at \(z = 2\): \[ -\frac{1}{2^2} = -\frac{1}{4} \] 4. **Apply the Residue Theorem:** The residue theorem states that for a closed contour \(C\) enclosing a pole \(z_0\) with residue \(Res(f, z_0)\): \[ \int_C f(z) \, dz = 2\pi i \cdot \text{Res}(f, z_0) \] Here, \(f(z) = \frac{1}{z (z-2)^2}\) and the residue at \(z = 2\) is \(-\frac{1}{4}\). Therefore: \[ \int_C \frac{dz}{z (z-2)^2} = 2\pi i \left(-\frac{1}{4}\right) = -\frac{\pi i}{2} \] ### Conclusion: The value of the integral is \(-\frac{\pi i}{2}\). Comparing this with the given form \(Ai\), we identify that \(A = -\frac{\pi}{2}\). Thus, the value of \(A\) is: \[ \boxed{-\frac{\pi}{2}} \] |
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| gpto1 | To solve the integral: \[ \int_C \frac{dz}{z(z-2)^2}, \] where \(C\) is the circle \(|z - 2| = 1\), we can use the Residue Theorem from complex analysis. **Step 1:** **Identify the singularities of the integrand and determine which ones are inside the contour \(C\).** The integrand \(f(z) = \dfrac{1}{z(z-2)^2}\) has singularities (poles) at: - \(z = 0\) (simple pole) - \(z = 2\) (double pole) Since \(C\) is the circle centered at \(z = 2\) with radius 1, any point \(z\) satisfying \(|z - 2| < 1\) lies inside \(C\). Calculating \(|0 - 2| = 2\), we see that \(z = 0\) is outside \(C\). At \(z = 2\), \(|2 - 2| = 0 < 1\), so the pole at \(z = 2\) is inside \(C\). **Step 2:** **Compute the residue of \(f(z)\) at \(z = 2\).** For a pole of order \(n\) at \(z = z_0\), the residue is: \[ \text{Res}[f, z_0] = \frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}} \left[(z - z_0)^n f(z)\right]. \] Since \(z = 2\) is a double pole (\(n = 2\)), we compute: \[ \text{Res}[f, z = 2] = \lim_{z \to 2} \frac{d}{dz}\left[(z - 2)^2 f(z)\right]. \] First, simplify \((z - 2)^2 f(z)\): \[ (z - 2)^2 f(z) = (z - 2)^2 \cdot \frac{1}{z(z - 2)^2} = \frac{1}{z}. \] Then take the derivative: \[ \frac{d}{dz} \left(\frac{1}{z}\right) = -\frac{1}{z^2}. \] Evaluate at \(z = 2\): \[ \text{Res}[f, z = 2] = -\frac{1}{(2)^2} = -\frac{1}{4}. \] **Step 3:** **Apply the Residue Theorem to compute the integral.** The Residue Theorem states: \[ \int_C f(z) \, dz = 2\pi i \sum \text{Res}[f, \text{inside } C]. \] So: \[ \int_C \frac{dz}{z(z-2)^2} = 2\pi i \left(-\frac{1}{4}\right) = -\frac{\pi i}{2}. \] **Conclusion:** Given that the integral equals \(A i\) (with \(i\) being the imaginary unit), we have: \[ -\frac{\pi i}{2} = A i \implies A = -\frac{\pi}{2}. \] **Answer:** \(-\dfrac{\pi}{2}\) |
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| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-a94479ef-8cca-90d1-b017-ca9292f95e2e', 'request_id': 'a94479ef-8cca-90d1-b017-ca9292f95e2e'} | -0.3926 |