| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To compute the integral $\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) dz$, 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. The function $f(z) = z^2 \sin\left(\frac{1}{z}\right)$ has a singularity at $z = 0$, which is a branch point for the sine function. However, for the purposes of computing the integral, we are interested in the behavior of the function as $z$ approaches $0$ from the outside of the unit circle. The function $f(z)$ can be expanded in a Laurent series around $z = 0$. The Laurent series for $\sin\left(\frac{1}{z}\right)$ is given by: $$\sin\left(\frac{1}{z}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!z^{2n+1}}.$$ Multiplying this by $z^2$, we get: $$f(z) = z^2 \sin\left(\frac{1}{z}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!z^{2n-1}}.$$ The residue of $f(z)$ at $z = 0$ is the coefficient of the $z^{-1}$ term in the Laurent series, which is: $$\text{Res}(f(z), z = 0) = \frac{(-1)^0}{(2\cdot0+1)!} = \frac{1}{1!} = 1.$$ By the residue theorem, the integral of $f(z)$ around the unit circle is: $$\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) dz = 2\pi i \cdot \text{Res}(f(z), z = 0) = 2\pi i \cdot 1 = 2\pi i.$$ Therefore, the answer is $A = 2\pi$. |
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| deepseekr1 | To compute the integral \(\int_{|z| = 1} z^2 \sin(1/z) \, dz\), we use the residue theorem. The function \(z^2 \sin(1/z)\) has an essential singularity at \(z = 0\). We need to find the residue of this function at \(z = 0\). First, we expand \(\sin(1/z)\) as a Laurent series around \(z = 0\): \[ \sin(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-(2n+1)} \] Multiplying this series by \(z^2\), we get: \[ z^2 \sin(1/z) = z^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-(2n+1)} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{1 - 2n} \] We need the coefficient of \(z^{-1}\) in this series, which corresponds to the term when \(1 - 2n = -1\). Solving for \(n\), we get \(n = 1\). The coefficient for this term is: \[ \frac{(-1)^1}{(2 \cdot 1 + 1)!} = -\frac{1}{6} \] Thus, the residue of \(z^2 \sin(1/z)\) at \(z = 0\) is \(-\frac{1}{6}\). By the residue theorem, the integral is: \[ \int_{|z| = 1} z^2 \sin(1/z) \, dz = 2\pi i \left( -\frac{1}{6} \right) = -\frac{\pi i}{3} \] Therefore, the value of \(A\) is \(\boxed{-\dfrac{\pi}{3}}\). |
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| gpto1 | To compute the integral \( I = \int_{|z|=1} z^2 \sin\left(\dfrac{1}{z}\right) dz \), where the integration path is the unit circle \(|z| = 1\) traversed once counterclockwise, we'll use complex analysis techniques, specifically the residue theorem. First, observe that the function \( f(z) = z^2 \sin\left(\dfrac{1}{z}\right) \) has a singularity at \( z = 0 \) (since \(\sin\left(\dfrac{1}{z}\right)\) is not analytic at \( z = 0 \)). This singularity lies within the integration contour \(|z| = 1\), so we can apply the residue theorem: \[ I = 2\pi i \times \text{Res}_{z=0} f(z) \] Our task now is to find the residue of \( f(z) \) at \( z = 0 \). To do this, we'll expand \( f(z) \) into its Laurent series about \( z = 0 \) and identify the coefficient of \( z^{-1} \), which is the residue. We start by expanding \( \sin\left(\dfrac{1}{z}\right) \) using its Taylor series: \[ \sin\left(\dfrac{1}{z}\right) = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} \left(\dfrac{1}{z}\right)^{2n+1} \] Multiplying by \( z^2 \): \[ f(z) = z^2 \sin\left(\dfrac{1}{z}\right) = z^2 \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} \left(\dfrac{1}{z}\right)^{2n+1} = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} z^{2 - (2n+1)} \] Simplify the exponent: \[ 2 - (2n+1) = 1 - 2n \] So the series becomes: \[ f(z) = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} z^{1 - 2n} \] To find the residue at \( z = 0 \), we need the coefficient of \( z^{-1} \). Set the exponent equal to \(-1\): \[ 1 - 2n = -1 \quad \Rightarrow \quad n = 1 \] At \( n = 1 \), the coefficient is: \[ \text{Coefficient} = \dfrac{(-1)^1}{(2(1)+1)!} = \dfrac{-1}{3!} = \dfrac{-1}{6} \] Therefore, the residue is: \[ \text{Res}_{z=0} f(z) = \dfrac{-1}{6} \] Applying the residue theorem: \[ I = 2\pi i \times \left(\dfrac{-1}{6}\right) = \dfrac{-\pi i}{3} \] The integral computes to \( I = \dfrac{-\pi i}{3} \), which is of the form \( A i \) with \( A = \dfrac{-\pi}{3} \). Answer: \(\boxed{\displaystyle -\dfrac{\pi}{3}}\) |
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| GPT-4o | To solve the given integral \(\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) \, dz\), we will use the residue theorem from complex analysis. First, let's rewrite the integrand: \[ z^2 \sin\left(\frac{1}{z}\right). \] The sine function can be expanded into its Taylor series as follows: \[ \sin\left(\frac{1}{z}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{1}{z}\right)^{2n+1}. \] Multiplying this by \(z^2\), we get: \[ z^2 \sin\left(\frac{1}{z}\right) = z^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{1}{z}\right)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{2-2n-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{1-2n}. \] Now, we need to find the residue of this function at \(z=0\). The residue theorem states that if \(f(z)\) has a Laurent series expansion around a singularity, the residue is the coefficient of the \(z^{-1}\) term. Let's identify the term in the series where \(z^{1-2n} = z^{-1}\): \[ 1 - 2n = -1 \implies 2n = 2 \implies n = 1. \] For \(n=1\), the corresponding term in the series is: \[ \frac{(-1)^1}{(2 \cdot 1 + 1)!} z^{1-2 \cdot 1} = \frac{-1}{3!} z^{-1} = \frac{-1}{6} z^{-1}. \] The coefficient of \(z^{-1}\) is \(\frac{-1}{6}\), so the residue at \(z=0\) is \(\frac{-1}{6}\). By the residue theorem, for a function \(f(z)\) analytic inside and on a simple closed contour \(\gamma\) except for isolated singularities: \[ \int_{\gamma} f(z) \, dz = 2\pi i \times \text{(sum of residues inside \(\gamma\))}. \] Here, the only singularity inside \(|z| = 1\) is at \(z=0\), and its residue is \(\frac{-1}{6}\). Therefore: \[ \int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) \, dz = 2\pi i \times \left( \frac{-1}{6} \right) = -\frac{2\pi i}{6} = -\frac{\pi i}{3}. \] Thus, the answer is \(Ai\) where \(A = -\frac{\pi}{3}\). |
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| llama3.1-70B-Instruct | To solve the given problem, we can use the residue theorem from complex analysis. The residue theorem states that for a simple closed curve $C$ and a function $f(z)$ that is analytic inside and on $C$, except for a finite number of singularities $z_1, z_2, ..., z_n$ inside $C$, the following holds: $$\int_C f(z) dz = 2\pi i \sum_{k=1}^{n} \text{Res}(f, z_k)$$ where $\text{Res}(f, z_k)$ is the residue of $f(z)$ at the singularity $z_k$. In this case, we have $f(z) = z^2 \sin(1/z)$ and the curve $C$ is the unit circle $|z| = 1$. The function $f(z)$ has a singularity at $z = 0$, which is inside the unit circle. To find the residue of $f(z)$ at $z = 0$, we can expand the function in a Laurent series around $z = 0$. We have: $$\sin(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n+1}}$$ Therefore, $$z^2 \sin(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n-1}}$$ The residue of $f(z)$ at $z = 0$ is the coefficient of the $1/z$ term in the Laurent series, which is: $$\text{Res}(f, 0) = \frac{(-1)^1}{(2\cdot 1+1)!} = -\frac{1}{3!} = -\frac{1}{6}$$ Therefore, by the residue theorem, we have: $$\int_{|z| = 1} z^2 \sin(1/z) dz = 2\pi i \cdot \left(-\frac{1}{6}\right) = -\frac{\pi}{3} i$$ Comparing this with the given answer $Ai$, we see that $A = -\frac{\pi}{3}$. |
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| llama3.1-405B-Instruct | To compute the integral $\int_{|z| = 1} z^2 \sin(1/z) dz$, we can use the Cauchy Integral Formula. First, we can rewrite the integral as: $$\int_{|z| = 1} z^2 \sin(1/z) dz = \int_{|z| = 1} \frac{z^2}{z} \sin(1/z) dz$$ Now, we can apply the Cauchy Integral Formula, which states that: $$\int_{|z| = 1} \frac{f(z)}{z} dz = 2\pi i f(0)$$ In this case, we have $f(z) = z^2 \sin(1/z)$, so we need to evaluate $f(0)$. Using the Taylor series expansion of $\sin(1/z)$, we have: $$\sin(1/z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{1}{z}\right)^{2n+1}$$ Evaluating this at $z=0$, we get: $$\sin(1/0) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{1}{0}\right)^{2n+1} = 0$$ Therefore, $f(0) = 0^2 \sin(1/0) = 0$. Plugging this into the Cauchy Integral Formula, we get: $$\int_{|z| = 1} z^2 \sin(1/z) dz = 2\pi i f(0) = 2\pi i (0) = 0$$ So, the value of the integral is $0$, which means that $A=0$. The final answer is: $\boxed{0}$ |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to compute the integral \(\int_{|z| = 1} z^2 \sin(1/z) \, dz\). We will use the Cauchy integral formula to find the value of this integral. First, let's identify the function and the contour. The function is \(f(z) = z^2 \sin(1/z)\) and the contour is the unit circle \(|z| = 1\). Notice that the function \(f(z)\) has a singularity at \(z = 0\) because \(\sin(1/z)\) is not defined at \(z = 0\). However, we can use the Cauchy integral theorem to handle this singularity. The Cauchy integral formula states that for any analytic function \(f(z)\) inside and on a simple closed contour \(C\), and any point \(a\) inside \(C\), \[ \int_C \frac{f(z)}{z - a} \, dz = 2\pi i f(a). \] In our case, we can rewrite the integral as: \[ \int_{|z| = 1} z^2 \sin(1/z) \, dz = \int_{|z| = 1} \frac{z^2 \sin(1/z)}{1} \, dz. \] Here, the function \(f(z) = z^2 \sin(1/z)\) is analytic everywhere except at \(z = 0\). Since the singularity is inside the contour, we can use the residue theorem. The residue of \(f(z)\) at \(z = 0\) is the coefficient of the \(1/z\) term in the Laurent series expansion of \(f(z)\) around \(z = 0\). To find the Laurent series expansion of \(z^2 \sin(1/z)\), we start with the Taylor series expansion of \(\sin(w)\): \[ \sin(w) = w - \frac{w^3}{3!} + \frac{w^5}{5!} - \cdots. \] Substituting \(w = 1/z\), we get: \[ \sin(1/z) = \frac{1}{z} - \frac{1}{3! z^3} + \frac{1}{5! z^5} - \cdots. \] Multiplying by \(z^2\), we obtain: \[ z^2 \sin(1/z) = z - \frac{1}{3! z} + \frac{1}{5! z^3} - \cdots. \] The coefficient of \(1/z\) in this expansion is \(-1/6\) (since \(1/3! = 1/6\)). According to the residue theorem, the integral of \(f(z)\) around the contour is \(2\pi i\) times the residue at \(z = 0\). Therefore, \[ \int_{|z| = 1} z^2 \sin(1/z) \, dz = 2\pi i \left(-\frac{1}{6}\right) = -\frac{2\pi i}{6} = -\frac{\pi i}{3}. \] Thus, the value of the integral is \(-\frac{\pi i}{3}\). Comparing this with the given form \(Ai\), we find that \(A = -\frac{\pi}{3}\). Therefore, the value of \(A\) is: \[ \boxed{-\frac{\pi}{3}} \] |
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| Qwen2-72B-Instruct | To compute the integral $\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) dz$, 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. The function $f(z) = z^2 \sin\left(\frac{1}{z}\right)$ has a singularity at $z = 0$, which is a branch point for the sine function. However, for the purposes of computing the integral, we are interested in the behavior of the function as $z$ approaches $0$ from the outside of the unit circle. The function $f(z)$ can be expanded in a Laurent series around $z = 0$. The Laurent series for $\sin\left(\frac{1}{z}\right)$ is given by: $$\sin\left(\frac{1}{z}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!z^{2n+1}}.$$ Multiplying this by $z^2$, we get: $$f(z) = z^2 \sin\left(\frac{1}{z}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!z^{2n-1}}.$$ The residue of $f(z)$ at $z = 0$ is the coefficient of the $z^{-1}$ term in the Laurent series, which is: $$\text{Res}(f(z), z = 0) = \frac{(-1)^0}{(2\cdot0+1)!} = \frac{1}{1!} = 1.$$ By the residue theorem, the integral of $f(z)$ around the unit circle is: $$\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) dz = 2\pi i \cdot \text{Res}(f(z), z = 0) = 2\pi i \cdot 1 = 2\pi i.$$ Therefore, the answer is $A = 2\pi$. |
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| baichuan4 | To solve the integral $\int_{|z| = 1} z^2 \sin(1/z) dz$, we can use the residue theorem. The function $f(z) = z^2 \sin(1/z)$ has a simple pole at $z = 0$, since $\sin(1/z)$ has a singularity there. First, let's find the residue of $f(z)$ at $z = 0$. The residue is the coefficient of $1/z$ in the Laurent series expansion of $f(z)$ around $z = 0$. We have: $$f(z) = z^2 \left( \frac{1}{z} - \frac{1}{3!} \left( \frac{1}{z} \right)^3 + \frac{1}{5!} \left( \frac{1}{z} \right)^5 - \cdots \right)$$ The residue is the coefficient of $1/z$, which is $1$. Now, by the residue theorem, the integral around the unit circle is $2\pi i$ times the residue: $$\int_{|z| = 1} z^2 \sin(1/z) dz = 2\pi i \cdot 1 = 2\pi i$$ So, the value of the integral is $2\pi i$. Therefore, $A = 2\pi$. |
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| doubao | Let $f(z)=z^2\sin(1/z)$. We expand $\sin(1/z)$ into its Laurent series: $\sin(1/z)=\frac{1}{z}-\frac{1}{3!z^3}+\frac{1}{5!z^5}-\cdots$. So $f(z)=z^2(\frac{1}{z}-\frac{1}{3!z^3}+\frac{1}{5!z^5}-\cdots)=z-\frac{1}{3!z}+\frac{1}{5!z^3}-\cdots$. The only term that contributes to the integral is the $z$ term. The residue at $z=0$ is $1$. By the residue theorem, the integral is $2\pi i$ times the residue, so the integral is $2\pi i$. So $A=2\pi$. |
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| minimax | To solve the integral $\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) dz$, we can use the residue theorem from complex analysis. The residue theorem states that if you have a function $f(z)$ that is analytic inside and on a simple closed contour $C$, except for a finite number of singular points $z_1, z_2, ..., z_n$ 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 singular points. The function $f(z) = z^2 \sin\left(\frac{1}{z}\right)$ is analytic everywhere except at $z = 0$. Since the contour $|z| = 1$ does not include the singularity at $z = 0$, the function is analytic on and inside the contour. Therefore, the integral of $f(z)$ around the contour $|z| = 1$ is zero because there are no singularities inside the contour. So, we have: $$\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) dz = 0$$ This means that the answer to the integral is $A = 0$. |
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| kimi | To solve the integral $\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) \, dz$, we can use the residue theorem from complex analysis. The residue theorem states that for a function with isolated singularities inside a simple closed contour, the integral of the function around the contour is equal to $2\pi i$ times the sum of the residues of the function at its singularities inside the contour. The function $f(z) = z^2 \sin\left(\frac{1}{z}\right)$ has a singularity at $z = 0$. To find the residue at this point, we can expand $\sin\left(\frac{1}{z}\right)$ as a Laurent series around $z = 0$. The Laurent series expansion of $\sin\left(\frac{1}{z}\right)$ is given by: \[ \sin\left(\frac{1}{z}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-2n-1} \] Multiplying by $z^2$, we get: \[ z^2 \sin\left(\frac{1}{z}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-2n-1} \cdot z^2 = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{1-2n} \] The residue at $z = 0$ is the coefficient of the $z^{-1}$ term in this series. The only term that contributes to the $z^{-1}$ coefficient is the $n = 0$ term: \[ \frac{(-1)^0}{(2\cdot0+1)!} z^{1-2\cdot0} = \frac{1}{1!} z^{-1} = z^{-1} \] So the residue at $z = 0$ is $1$. Now, applying the residue theorem, we have: \[ \int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) \, dz = 2\pi i \cdot \text{Residue at } z = 0 = 2\pi i \] Therefore, the answer is $A = 2\pi$. |
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| wenxinyiyan | 首先，我们注意到函数$f(z) = z^2 \sin(\frac{1}{z})$在$z=0$处有一个奇点。为了计算这个积分，我们可以使用留数定理。 留数定理告诉我们，如果一个函数在闭合路径内部有奇点，那么沿该路径的积分等于函数在所有这些奇点处的留数之和乘以$2\pi i$。 首先，我们找到函数$f(z)$在$z=0$处的留数。为了找到留数，我们可以将$f(z)$展开为劳伦级数。 我们知道$\sin(z)$的泰勒级数展开是： $\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$因此，我们可以将$\sin(\frac{1}{z})$展开为： $\sin(\frac{1}{z}) = \frac{1}{z} - \frac{1}{3! z^3} + \frac{1}{5! z^5} - \cdots$然后，我们将这个级数乘以$z^2$得到$f(z)$的劳伦级数展开： $f(z) = z^2 \sin(\frac{1}{z}) = z - \frac{1}{3! z} + \frac{1}{5! z^3} - \cdots$我们可以看到，在$z=0$处，$f(z)$的留数是$-\frac{1}{3!}$。 现在我们可以使用留数定理来计算积分： $\int_{|z| = 1} z^2 \sin(\frac{1}{z}) dz = 2\pi i \times (-\frac{1}{3!}) = -\frac{\pi i}{3}$因此，答案是$A = -\frac{\pi}{3}$。 |
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| zhipu | To solve the integral \(\int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) \, dz\), we can use the residue theorem from complex analysis. The contour \(|z| = 1\) is a circle of radius 1 centered at the origin. First, let's rewrite the integrand in a more convenient form: \[ z^2 \sin\left(\frac{1}{z}\right). \] We need to find the singularities of the integrand inside the contour \(|z| = 1\). The function \(\sin\left(\frac{1}{z}\right)\) has singularities at \(z = 0\) because \(\frac{1}{z}\) becomes infinite there. Since \(z = 0\) is inside the contour \(|z| = 1\), we need to consider this singularity. To find the residue at \(z = 0\), we expand \(\sin\left(\frac{1}{z}\right)\) in a Laurent series around \(z = 0\): \[ \sin\left(\frac{1}{z}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{1}{z}\right)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n+1}}. \] Now, multiply this series by \(z^2\): \[ z^2 \sin\left(\frac{1}{z}\right) = z^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n+1}} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{z^2}{z^{2n+1}} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{2n-1}}. \] We are interested in the coefficient of \(\frac{1}{z}\) in this series, which corresponds to the term where \(2n - 1 = -1\). Solving for \(n\), we get: \[ 2n - 1 = -1 \implies 2n = 0 \implies n = 0. \] For \(n = 0\), the term in the series is: \[ \frac{(-1)^0}{(2 \cdot 0 + 1)!} \frac{1}{z^{-1}} = \frac{1}{1!} \cdot z = z. \] However, we need the coefficient of \(\frac{1}{z}\), not \(z\). Let's re-evaluate the series expansion more carefully: \[ z^2 \sin\left(\frac{1}{z}\right) = z^2 \left(\frac{1}{z} - \frac{1}{3!z^3} + \frac{1}{5!z^5} - \cdots \right) = z \left(1 - \frac{1}{3!z^2} + \frac{1}{5!z^4} - \cdots \right). \] The coefficient of \(\frac{1}{z}\) is indeed zero, but we need to consider the residue of the function at \(z = 0\). The residue is the coefficient of \(\frac{1}{z}\) in the Laurent series expansion of the integrand around \(z = 0\). Let's directly compute the residue using the limit definition: \[ \text{Res}\left(z^2 \sin\left(\frac{1}{z}\right), 0\right) = \lim_{z \to 0} z \cdot z^2 \sin\left(\frac{1}{z}\right) = \lim_{z \to 0} z^3 \sin\left(\frac{1}{z}\right). \] Using the fact that \(\sin\left(\frac{1}{z}\right)\) oscillates between \(-1\) and \(1\) as \(z \to 0\), the limit does not converge to a single value. However, we can use the fact that the integral of an odd function over a symmetric interval is zero. Here, \(z^2 \sin\left(\frac{1}{z}\right)\) behaves like an odd function around the origin. To resolve this, we use the fact that the integral of \(z^2 \sin\left(\frac{1}{z}\right)\) around the unit circle can be evaluated using the residue theorem: \[ \int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) \, dz = 2\pi i \cdot \text{Res}\left(z^2 \sin\left(\frac{1}{z}\right), 0\right). \] From the series expansion, we see that the residue at \(z = 0\) is indeed \(i\), as the coefficient of \(\frac{1}{z}\) in the Laurent series is \(i\). Thus, the integral evaluates to: \[ \int_{|z| = 1} z^2 \sin\left(\frac{1}{z}\right) \, dz = 2\pi i \cdot i = 2\pi (-1) = -2\pi. \] Therefore, the value of \(A\) in the given answer \(Ai\) is: \[ A = -2\pi. \] So, the final answer is: \[ \boxed{-2\pi}. \] |
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| gpto1 | To compute the integral \( I = \int_{|z|=1} z^2 \sin\left(\dfrac{1}{z}\right) dz \), where the integration path is the unit circle \(|z| = 1\) traversed once counterclockwise, we'll use complex analysis techniques, specifically the residue theorem. First, observe that the function \( f(z) = z^2 \sin\left(\dfrac{1}{z}\right) \) has a singularity at \( z = 0 \) (since \(\sin\left(\dfrac{1}{z}\right)\) is not analytic at \( z = 0 \)). This singularity lies within the integration contour \(|z| = 1\), so we can apply the residue theorem: \[ I = 2\pi i \times \text{Res}_{z=0} f(z) \] Our task now is to find the residue of \( f(z) \) at \( z = 0 \). To do this, we'll expand \( f(z) \) into its Laurent series about \( z = 0 \) and identify the coefficient of \( z^{-1} \), which is the residue. We start by expanding \( \sin\left(\dfrac{1}{z}\right) \) using its Taylor series: \[ \sin\left(\dfrac{1}{z}\right) = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} \left(\dfrac{1}{z}\right)^{2n+1} \] Multiplying by \( z^2 \): \[ f(z) = z^2 \sin\left(\dfrac{1}{z}\right) = z^2 \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} \left(\dfrac{1}{z}\right)^{2n+1} = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} z^{2 - (2n+1)} \] Simplify the exponent: \[ 2 - (2n+1) = 1 - 2n \] So the series becomes: \[ f(z) = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} z^{1 - 2n} \] To find the residue at \( z = 0 \), we need the coefficient of \( z^{-1} \). Set the exponent equal to \(-1\): \[ 1 - 2n = -1 \quad \Rightarrow \quad n = 1 \] At \( n = 1 \), the coefficient is: \[ \text{Coefficient} = \dfrac{(-1)^1}{(2(1)+1)!} = \dfrac{-1}{3!} = \dfrac{-1}{6} \] Therefore, the residue is: \[ \text{Res}_{z=0} f(z) = \dfrac{-1}{6} \] Applying the residue theorem: \[ I = 2\pi i \times \left(\dfrac{-1}{6}\right) = \dfrac{-\pi i}{3} \] The integral computes to \( I = \dfrac{-\pi i}{3} \), which is of the form \( A i \) with \( A = \dfrac{-\pi}{3} \). Answer: \(\boxed{\displaystyle -\dfrac{\pi}{3}}\) |
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