| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To evaluate the integral $\int_c \frac{z^2}{z-5} dz$ where $c$ is the circle $|z|=2$, we can use the Cauchy integral theorem or the residue theorem. The integrand $\frac{z^2}{z-5}$ has a simple pole at $z=5$, which lies outside the contour $|z|=2$. According to the Cauchy integral theorem, if a function is analytic inside and on a simple closed contour, then the integral of the function over the contour is zero. Since the pole at $z=5$ is outside the circle $|z|=2$, the integrand is analytic inside and on the contour. Therefore, by the Cauchy integral theorem, we have: $$\int_c \frac{z^2}{z-5} dz = 0$$ So, the integral $\int_c \frac{z^2}{z-5} dz$ where $c$ is the circle $|z|=2$ is equal to $0$. |
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| deepseekr1 | To evaluate the integral \(\int_c \frac{z^2}{z - 5} \, dz\) where \(c\) is the circle \(|z| = 2\), we start by noting that the integrand has a singularity at \(z = 5\). Since the circle \(|z| = 2\) has a radius of 2 and is centered at the origin, the point \(z = 5\) is outside this circle. By Cauchy's integral theorem, if a function is analytic inside and on a closed contour, the integral around that contour is zero. Here, the function \(\frac{z^2}{z - 5}\) is analytic inside and on the contour \(|z| = 2\) because the singularity at \(z = 5\) is outside the contour. Therefore, the integral is zero. To confirm, we can use the residue theorem, which states that the integral is \(2\pi i\) times the sum of the residues of the function inside the contour. Since there are no singularities inside the contour, the sum of the residues is zero, leading to the same result. Additionally, parameterizing the integral using \(z = 2e^{i\theta}\) and expanding the integrand as a power series (which converges uniformly) also shows that all terms integrate to zero, confirming the result. Thus, the final answer is: \[ \boxed{0} \] |
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| gpto1 | To evaluate the integral: \[ \int_c \frac{z^2}{z - 5} \, dz \] where \( c \) is the circle \( |z| = 2 \) traversed once in the positive (counter-clockwise) direction. **Solution:** First, observe that the circle \( |z| = 2 \) is centered at the origin with a radius of 2. **Step 1: Analyze the Singularities** The integrand has a singularity (pole) at \( z = 5 \), since the denominator becomes zero there: \[ z - 5 = 0 \implies z = 5 \] The modulus of this singularity is \( |5| = 5 \), which lies **outside** the circle \( |z| = 2 \). Thus, the function \( \frac{z^2}{z - 5} \) is analytic (holomorphic) everywhere on and inside the contour \( c \). **Step 2: Apply Cauchy's Theorem** Cauchy's Theorem states that if a function is analytic inside and on a closed contour \( C \), then: \[ \int_C f(z) \, dz = 0 \] Since \( \frac{z^2}{z - 5} \) is analytic on and inside \( c \), we can directly apply Cauchy's Theorem: \[ \int_c \frac{z^2}{z - 5} \, dz = 0 \] **Conclusion:** The value of the integral is **zero**. --- **Final Answer:** 0 |
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| GPT-4o | To evaluate the contour integral \(\int_c \frac{z^2}{z - 5} \, dz\), where \(c\) is the circle \(|z| = 2\), we will use the residue theorem. The integrand is \(\frac{z^2}{z - 5}\). Notice that the integrand has a singularity at \(z = 5\). 1. **Check if the singularity is inside the contour**: The contour \(c\) is the circle with radius 2 centered at the origin, i.e., \(|z| = 2\). The singularity at \(z = 5\) lies outside this circle because \(|5| = 5 > 2\). 2. **Analyze the integral using the residue theorem**: According to the residue theorem, if \(f(z)\) is analytic inside and on a simple closed contour \(c\) except for isolated singularities, then \[ \int_c f(z) \, dz = 2\pi i \sum \text{(residues of \(f\) inside \(c\))}. \] Since the singularity at \(z = 5\) is outside the contour \(c\), there are no residues inside the contour. Therefore, the sum of the residues inside \(c\) is zero. 3. **Conclusion**: Since there are no residues inside the contour \(c\), \[ \int_c \frac{z^2}{z - 5} \, dz = 0. \] Thus, the value of the integral \(\int_c \frac{z^2}{z - 5} \, dz\) over the circle \(|z| = 2\) is \(0\). |
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| llama3.1-70B-Instruct | The problem is to evaluate the integral $\int_c \frac{z^2}{z-5} dz$, where $c$ is the circle $|z| = 2$. ## Step 1: Identify the type of integral The integral is a contour integral of a complex function over a closed curve (the circle $|z| = 2$). ## Step 2: Check if the function has any singularities within the contour The function $\frac{z^2}{z-5}$ has a singularity at $z = 5$, but since $|z| = 2$, the singularity is outside the contour. ## Step 3: Apply Cauchy's Integral Theorem Since the function is analytic within and on the contour, and the contour is closed and simple, Cauchy's Integral Theorem states that the integral is equal to zero. The final answer is: $\boxed{0}$ |
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| llama3.1-405B-Instruct | To evaluate the integral $\int_c z^2 / (z - 5) dz$, where $c$ is the circle $|z| = 2$, we can use the Cauchy Integral Formula. The Cauchy Integral Formula states that if $f(z)$ is analytic inside and on a simple closed curve $c$, and $a$ is a point inside $c$, then $$\int_c \frac{f(z)}{z - a} dz = 2\pi i f(a).$$ In this case, we have $f(z) = z^2$ and $a = 5$. Since $f(z)$ is analytic inside and on the circle $|z| = 2$, and $5$ is a point inside the circle, we can apply the Cauchy Integral Formula. Substituting $f(z) = z^2$ and $a = 5$ into the formula, we get $$\int_c \frac{z^2}{z - 5} dz = 2\pi i (5^2) = 2\pi i (25) = 50\pi i.$$ Therefore, the value of the integral is $50\pi i$. |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to evaluate the integral \(\int_c \frac{z^2}{z - 5} \, dz\), where \(c\) is the circle \(|z| = 2\). ### Step-by-Step Solution: 1. **Identify the contour and the function:** - The contour \(c\) is the circle \(|z| = 2\). - The function to be integrated is \(f(z) = \frac{z^2}{z - 5}\). 2. **Analyze the singularity:** - The function \(f(z)\) has a singularity at \(z = 5\). 3. **Check if the singularity is inside the contour:** - The circle \(|z| = 2\) is centered at the origin with radius 2. - The point \(z = 5\) lies outside this circle since \(|5| = 5 > 2\). 4. **Apply Cauchy's Theorem:** - According to Cauchy's Theorem, if a function \(f(z)\) is analytic inside and on a simple closed contour \(c\), then \(\int_c f(z) \, dz = 0\). - Since \(z = 5\) is outside the contour \(c\), the function \(f(z) = \frac{z^2}{z - 5}\) is analytic inside and on the contour \(c\). 5. **Conclusion:** - Therefore, by Cauchy's Theorem, the integral \(\int_c \frac{z^2}{z - 5} \, dz = 0\). Thus, the solution to the given problem is: \[ \boxed{0} \] |
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| Qwen2-72B-Instruct | To evaluate the integral $\int_c \frac{z^2}{z-5} dz$ where $c$ is the circle $|z|=2$, we can use the Cauchy integral formula or the residue theorem, since the integrand is a rational function with a pole inside the contour. The integrand has a simple pole at $z=5$, which is outside the contour $|z|=2$. However, it has a pole at $z=5$ of order $1$. Since this pole is outside the circle $|z|=2$, the integral is equal to $0$ according to the Cauchy's integral theorem, which states that the integral of an analytic function over a closed contour is zero if there are no singularities enclosed by the contour. Alternatively, we can use the residue theorem to confirm this. The residue theorem states that for a function $f(z)$ with isolated singularities inside a simple closed contour $C$, $$\int_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k)$$ where the sum is over all singularities $z_k$ inside $C$. In this case, since there are no singularities inside the circle $|z|=2$, the sum over residues is zero. Therefore, by the residue theorem, we have $$\int_c \frac{z^2}{z-5} dz = 2\pi i \cdot 0 = 0$$ So, the integral $\int_c \frac{z^2}{z-5} dz$ where $c$ is the circle $|z|=2$ is equal to $0$. |
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| baichuan4 | To evaluate the integral $\int_c \frac{z^2}{z - 5} dz$, where $c$ is the circle $|z| = 2$, we can use the Cauchy Integral Formula, which states that for a function $f(z)$ analytic inside a closed contour $C$, and a point $a$ inside $C$, the integral of $f(z)$ around $C$ is given by: $$\oint_C \frac{f(z)}{z - a} dz = 2\pi i f(a)$$ In our case, we have $f(z) = z^2$ and $a = 5$, but $a$ is not inside $C$, since $|5| > 2$. Therefore, we cannot directly apply the Cauchy Integral Formula. However, we can use a different approach. We can rewrite the integral as: $$\int_c \frac{z^2}{z - 5} dz = \int_c \frac{z^2 - 5z + 5z - 25 + 25}{z - 5} dz = \int_c \left(z + 5 + \frac{25}{z - 5}\right) dz$$ Now, the first two terms $z + 5$ are analytic everywhere, including inside the contour $C$, and their integrals around $C$ are zero by Cauchy's Integral Theorem. So, we are left with: $$\int_c \frac{25}{z - 5} dz$$ This is in the form of the Cauchy Integral Formula, but since $5$ is outside the contour $C$, the integral is zero. Therefore, the original integral evaluates to: $$\int_c \frac{z^2}{z - 5} dz = 0$$ This is because the function $z^2$ is analytic within and on the contour $C$, and there are no singularities of the integrand within $C$. Thus, by Cauchy's Integral Theorem, the integral around a closed contour of an analytic function is zero. |
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| doubao | The function $f(z)=\frac{z^2}{z-5}$ has a singularity at $z=5$. But the circle $|z|=2$ is a closed curve centered at the origin with radius 2 and does not enclose the singularity at $z=5$. By Cauchy's theorem, the integral is zero. | 0 |
| minimax | To evaluate the integral $\int_c \frac{z^2}{z - 5} dz$, where $c$ is the circle $|z| = 2$, 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 singular points $z_1, z_2, ..., z_n$ inside $C$, then the integral of $f(z)$ around $C$ is given by: $$\int_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k)$$ where $\text{Res}(f, z_k)$ is the residue of $f$ at the point $z_k$. In our case, the function $f(z) = \frac{z^2}{z - 5}$ has a single pole of order 1 at $z = 5$, which lies outside the contour $c$ since the circle is centered at the origin with radius 2. Therefore, the function is analytic inside and on the contour $c$, and the integral of $f(z)$ around $c$ is zero by Cauchy's integral theorem, which states that the integral of an analytic function over a closed contour is zero. So, the solution to the integral is: $$\int_c \frac{z^2}{z - 5} dz = 0$$ |
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| kimi | To evaluate the integral $\int_c \frac{z^2}{z-5} dz$, we need to consider the contour $C$, which is a circle with radius 2 centered at the origin. The contour encloses the point $z = 5$, which is a singularity of the function $\frac{z^2}{z-5}$. The function has a simple pole at $z = 5$. Since the contour encloses this singularity, 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 each singularity inside $C$. The residue of a function at a simple pole $z = a$ is given by: $$ \text{Res}\left(\frac{f(z)}{g(z)}, a\right) = \lim_{z \to a} (z - a) f(z) $$ For the function $\frac{z^2}{z-5}$, the residue at $z = 5$ is: $$ \text{Res}\left(\frac{z^2}{z-5}, 5\right) = \lim_{z \to 5} (z - 5) \frac{z^2}{z-5} = \lim_{z \to 5} z^2 = 5^2 = 25 $$ Now, applying the residue theorem, we have: $$ \int_c \frac{z^2}{z-5} dz = 2\pi i \times \text{Residue} = 2\pi i \times 25 = 50\pi i $$ So the solution to the integral is $\boxed{50\pi i}$. |
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| wenxinyiyan | 首先，我们可以将$z^2$写为$(z - 5 + 5)^2$，然后展开得到$(z - 5)^2 + 10(z - 5) + 25$。 因此，$\int_c z^2 / (z - 5) dz$可以写为$\int_c [(z - 5)^2 + 10(z - 5) + 25] / (z - 5) dz$。 这个式子可以进一步拆分为$\int_c [(z - 5) + 10 + 25 / (z - 5)] dz$。 现在我们可以逐项积分： $\int_c (z - 5) dz$ = 0，因为$(z - 5)$在圆$c$上是解析的。 $\int_c 10 dz$ = 0，因为常数在圆$c$上也是解析的。 $\int_c 25 / (z - 5) dz$ = $25 * 2\pi i$，根据柯西积分公式。 因此，$\int_c z^2 / (z - 5) dz = 0 + 0 + 25 * 2\pi i = 50\pi i$。 |
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| zhipu | To evaluate the integral \(\int_c \frac{z^2}{z - 5} \, dz\), where \(c\) is the circle \(|z| = 2\), we can use the Residue Theorem from complex analysis. The Residue Theorem is particularly useful for evaluating integrals of complex functions over closed contours. ### Step-by-Step Solution: 1. **Identify the Contour and the Function:** - The contour \(c\) is the circle \(|z| = 2\). - The function to integrate is \(f(z) = \frac{z^2}{z - 5}\). 2. **Find the Singularities of the Function:** - The function \(f(z) = \frac{z^2}{z - 5}\) has a simple pole at \(z = 5\). 3. **Check if the Singularity Lies Inside the Contour:** - The contour \(|z| = 2\) is a circle centered at the origin with radius 2. - The singularity \(z = 5\) lies outside this contour because \(|5| = 5 > 2\). 4. **Apply the Residue Theorem:** - The Residue Theorem states that if \(f(z)\) is analytic inside and on a simple closed contour \(c\), except for a finite number of singularities inside \(c\), then: \[ \int_c f(z) \, dz = 2\pi i \sum \text{Residues of } f(z) \text{ inside } c. \] - In this case, since the singularity \(z = 5\) is outside the contour \(|z| = 2\), there are no singularities inside the contour. 5. **Conclude the Integral Value:** - Because there are no singularities of \(f(z)\) inside the contour \(|z| = 2\), the sum of the residues inside the contour is zero. - Therefore, by the Residue Theorem: \[ \int_c \frac{z^2}{z - 5} \, dz = 2\pi i \cdot 0 = 0. \] ### Final Answer: \[ \int_c \frac{z^2}{z - 5} \, dz = 0. \] This result is consistent with the fact that the integrand is analytic everywhere inside and on the contour \(|z| = 2\), except at \(z = 5\), which lies outside the contour. Thus, the integral over the closed contour is zero. |
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| gpto1 | To evaluate the integral: \[ \int_c \frac{z^2}{z - 5} \, dz \] where \( c \) is the circle \( |z| = 2 \) traversed once in the positive (counter-clockwise) direction. **Solution:** First, observe that the circle \( |z| = 2 \) is centered at the origin with a radius of 2. **Step 1: Analyze the Singularities** The integrand has a singularity (pole) at \( z = 5 \), since the denominator becomes zero there: \[ z - 5 = 0 \implies z = 5 \] The modulus of this singularity is \( |5| = 5 \), which lies **outside** the circle \( |z| = 2 \). Thus, the function \( \frac{z^2}{z - 5} \) is analytic (holomorphic) everywhere on and inside the contour \( c \). **Step 2: Apply Cauchy's Theorem** Cauchy's Theorem states that if a function is analytic inside and on a closed contour \( C \), then: \[ \int_C f(z) \, dz = 0 \] Since \( \frac{z^2}{z - 5} \) is analytic on and inside \( c \), we can directly apply Cauchy's Theorem: \[ \int_c \frac{z^2}{z - 5} \, dz = 0 \] **Conclusion:** The value of the integral is **zero**. --- **Final Answer:** 0 |
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| deepseekv3 | To evaluate the integral \(\int_c \frac{z^2}{z - 5} \, dz\), where \(c\) is the circle \(|z| = 2\), we can use **Cauchy's Integral Formula**. ### Step 1: Identify the singularity The integrand \(\frac{z^2}{z - 5}\) has a singularity at \(z = 5\). However, the circle \(|z| = 2\) is centered at the origin with radius 2, and the point \(z = 5\) lies **outside** this circle (since \(|5| = 5 > 2\)). ### Step 2: Apply Cauchy's Integral Theorem If the singularity lies outside the contour of integration, the integral evaluates to **0** by Cauchy's Integral Theorem. This is because the integrand is analytic (holomorphic) everywhere inside and on the contour \(c\). ### Final Answer: \[ \boxed{0} \] |
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