| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the integral \(\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz\), we need to use the residue theorem. However, before we can apply the residue theorem, we need to identify the singularities of the function \(f(z) = \frac{5z - 2}{z(z - 1)}\) and determine which of them lie inside the contour \(|z| = 2\). The singularities are at \(z = 0\) and \(z = 1\), both of which are inside the contour \(|z| = 2\). Next, we find the residues of \(f(z)\) at these singularities. The residue at a simple pole \(z = c\) of a function \(f(z)\) is given by \(\lim_{z \to c} (z - c) f(z)\). 1. For \(z = 0\), the residue is: \[ \lim_{z \to 0} z \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} \frac{5z - 2}{z - 1} = -2 \] 2. For \(z = 1\), the residue is: \[ \lim_{z \to 1} (z - 1) \frac{5z - 2}{z(z - 1)} = \lim_{z \to 1} \frac{5z - 2}{z} = 3 \] 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 singularities inside the contour. Therefore, \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz = 2\pi i \left( -2 + 3 \right) = 2\pi i \] Thus, \(A = 2\pi\). Solution: \(A = 2\pi\). |
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| deepseekr1 | To compute the integral \(\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz\), we use the residue theorem. The integrand has singularities at \(z = 0\) and \(z = 1\), both of which lie inside the contour \(|z| = 2\). First, we find the residues at these singularities: - The residue at \(z = 0\) is calculated as: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 0\right) = \lim_{z \to 0} z \cdot \frac{5z - 2}{z(z - 1)} = \frac{-2}{-1} = 2. \] - The residue at \(z = 1\) is calculated as: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 1\right) = \lim_{z \to 1} (z - 1) \cdot \frac{5z - 2}{z(z - 1)} = \frac{5 - 2}{1} = 3. \] Summing these residues gives \(2 + 3 = 5\). By the residue theorem, the integral is \(2\pi i\) times the sum of the residues: \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz = 2\pi i \cdot 5 = 10\pi i. \] To verify, we decompose the integrand using partial fractions: \[ \frac{5z - 2}{z(z - 1)} = \frac{2}{z} + \frac{3}{z - 1}. \] Integrating each term separately around \(|z| = 2\): - The integral of \(\frac{2}{z}\) is \(2 \cdot 2\pi i = 4\pi i\). - The integral of \(\frac{3}{z - 1}\) is \(3 \cdot 2\pi i = 6\pi i\). Adding these results gives \(4\pi i + 6\pi i = 10\pi i\). Thus, the value of \(A\) is \(\boxed{10\pi}\). |
31.4 |
| gpto1 | To compute the integral \( I = \int_{|z|=2} \frac{5z - 2}{z(z - 1)}\,dz \), we can use the residue theorem from complex analysis. The residue theorem states that for a function \( f(z) \) with isolated singularities inside a closed contour \( C \), the integral over \( C \) is: \[ I = \int_C f(z)\,dz = 2\pi i \sum \text{Residues of } f(z) \text{ inside } C \] **Step 1: Identify the Singularities** The function \( f(z) = \frac{5z - 2}{z(z - 1)} \) has singularities (poles) where the denominator is zero, i.e., at \( z = 0 \) and \( z = 1 \). Both singularities are inside the contour \( |z| = 2 \) since the circle of radius 2 centered at the origin encompasses these points. **Step 2: Compute the Residues** We need to compute the residues at \( z = 0 \) and \( z = 1 \). **Residue at \( z = 0 \):** Since \( z = 0 \) is a simple pole, the residue is: \[ \text{Res}_{z=0} f(z) = \lim_{z \to 0} z \cdot f(z) = \lim_{z \to 0} \frac{5z - 2}{z - 1} \] Evaluating the limit: \[ \text{Res}_{z=0} f(z) = \frac{5 \cdot 0 - 2}{0 - 1} = \frac{-2}{-1} = 2 \] **Residue at \( z = 1 \):** Similarly, for \( z = 1 \): \[ \text{Res}_{z=1} f(z) = \lim_{z \to 1} (z - 1) \cdot f(z) = \lim_{z \to 1} \frac{5z - 2}{z} \] Evaluating the limit: \[ \text{Res}_{z=1} f(z) = \frac{5 \cdot 1 - 2}{1} = 3 \] **Step 3: Sum the Residues** Total residue inside the contour: \[ \sum \text{Residues} = \text{Res}_{z=0} f(z) + \text{Res}_{z=1} f(z) = 2 + 3 = 5 \] **Step 4: Apply the Residue Theorem** Using the residue theorem: \[ I = 2\pi i \cdot \left( \text{Sum of residues inside } C \right) = 2\pi i \cdot 5 = 10\pi i \] So, the integral evaluates to \( I = 10\pi i \). Therefore, \( A = 10\pi \). **Answer:** \(10\pi\) |
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| GPT-4o | To solve the integral \( \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz \), we will use the residue theorem from complex analysis. The residue theorem states that for a function \( f(z) \) which is analytic inside and on a simple closed contour \( C \), except for isolated singularities, the contour integral is given by \[ \int_C f(z) \, dz = 2\pi i \sum \text{Res}(f, z_k) \] where the sum is taken over all the residues of \( f \) inside \( C \). First, let's identify the singularities of the integrand \( \frac{5z - 2}{z(z - 1)} \): - \( z = 0 \) (simple pole) - \( z = 1 \) (simple pole) Next, we need to check if these singularities are inside the contour \( |z| = 2 \): - \( z = 0 \) is inside \( |z| = 2 \). - \( z = 1 \) is also inside \( |z| = 2 \). Both singularities are inside the contour. Now, we calculate the residues at these points. ### Residue at \( z = 0 \) For \( f(z) = \frac{5z - 2}{z(z - 1)} \), the residue at \( z = 0 \) is: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 0\right) = \lim_{z \to 0} z \cdot \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} \frac{5z - 2}{z - 1} = \frac{5(0) - 2}{0 - 1} = \frac{-2}{-1} = 2 \] ### Residue at \( z = 1 \) For \( f(z) = \frac{5z - 2}{z(z - 1)} \), the residue at \( z = 1 \) is: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 1\right) = \lim_{z \to 1} (z - 1) \cdot \frac{5z - 2}{z(z - 1)} = \lim_{z \to 1} \frac{5z - 2}{z} = \frac{5(1) - 2}{1} = \frac{5 - 2}{1} = 3 \] Now, we add the residues: \[ \sum \text{Res}(f, z_k) = 2 + 3 = 5 \] Therefore, applying the residue theorem, we get: \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz = 2\pi i \cdot 5 = 10\pi i \] Thus, the answer is \( A = 10 \). |
31.4 |
| llama3.1-70B-Instruct | To solve the given integral, we can use the Cauchy Integral Formula, which states that for a function f(z) that is analytic inside and on a simple closed curve C, and a point a inside C, the integral of f(z) / (z - a) around C is equal to 2πi times the value of f(a). In this case, we have the function f(z) = (5z - 2) / z, which is analytic inside and on the circle |z| = 2, except at the point z = 0, where it has a pole. We can rewrite the integral as: ∫_{|z| = 2} (5z - 2) / (z * (z - 1)) dz = ∫_{|z| = 2} (5z - 2) / z^2 * (1 - 1/z) dz Now, we can use the Cauchy Integral Formula with a = 0 to evaluate the integral: ∫_{|z| = 2} (5z - 2) / z^2 * (1 - 1/z) dz = 2πi * (5 * 0 - 2) / 0^2 * (1 - 1/0) However, this is not well-defined, since 0^2 is zero. To fix this, we can use the fact that the function f(z) = (5z - 2) / z is analytic at z = 0, and evaluate the integral using the Cauchy Integral Formula with a = 1 instead: ∫_{|z| = 2} (5z - 2) / (z * (z - 1)) dz = ∫_{|z| = 2} (5z - 2) / z * (1 - 1/z) dz Now, we can use the Cauchy Integral Formula with a = 1 to evaluate the integral: ∫_{|z| = 2} (5z - 2) / z * (1 - 1/z) dz = 2πi * (5 * 1 - 2) / 1 * (1 - 1/1) Simplifying, we get: ∫_{|z| = 2} (5z - 2) / (z * (z - 1)) dz = 2πi * 3 Therefore, A = 3. |
31.4 |
| llama3.1-405B-Instruct | To compute the integral $\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz$, we can use the method of partial fractions to decompose the integrand into simpler terms. First, we can write the integrand as: $$\frac{5z - 2}{z(z - 1)} = \frac{A}{z} + \frac{B}{z - 1}$$ where $A$ and $B$ are constants to be determined. To find $A$ and $B$, we can multiply both sides of the equation by $z(z - 1)$ to get: $$5z - 2 = A(z - 1) + Bz$$ Expanding the right-hand side, we get: $$5z - 2 = Az - A + Bz$$ Comparing coefficients of $z$ and constant terms, we get: $$A + B = 5$$ $$-A = -2$$ Solving for $A$ and $B$, we get: $$A = 2$$ $$B = 3$$ Therefore, we can write the integrand as: $$\frac{5z - 2}{z(z - 1)} = \frac{2}{z} + \frac{3}{z - 1}$$ Now, we can integrate each term separately: $$\int_{|z| = 2} \frac{2}{z} dz = 2\int_{|z| = 2} \frac{1}{z} dz = 2(2\pi i) = 4\pi i$$ $$\int_{|z| = 2} \frac{3}{z - 1} dz = 3\int_{|z| = 2} \frac{1}{z - 1} dz = 3(2\pi i) = 6\pi i$$ Adding the two integrals, we get: $$\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz = 4\pi i + 6\pi i = 10\pi i$$ Therefore, the answer is $A = 10\pi$. |
31.4 |
| DeepSeek-Coder-V2 | To solve the given problem, we need to compute the integral \(\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz\). We will use the residue theorem to find the value of this integral. First, let's identify the singularities of the integrand \(\frac{5z - 2}{z(z - 1)}\). The singularities occur at \(z = 0\) and \(z = 1\), both of which lie inside the contour \(|z| = 2\). Next, we need to find the residues of the integrand at these singularities. 1. **Residue at \(z = 0\)**: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 0\right) = \lim_{z \to 0} z \cdot \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} \frac{5z - 2}{z - 1} = -2 \] 2. **Residue at \(z = 1\)**: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 1\right) = \lim_{z \to 1} (z - 1) \cdot \frac{5z - 2}{z(z - 1)} = \lim_{z \to 1} \frac{5z - 2}{z} = 3 \] According to the residue theorem, the integral of a function around a closed contour is \(2\pi i\) times the sum of the residues of the function inside the contour. Therefore, \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz = 2\pi i \left( \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 0\right) + \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 1\right) \right) \] Substituting the residues we found: \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz = 2\pi i (-2 + 3) = 2\pi i (1) = 2\pi i \] Thus, the value of the integral is \(2\pi i\), and the coefficient \(A\) in the given answer \(Ai\) is \(2\pi\). Therefore, the final answer is: \[ \boxed{2\pi} \] |
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| Qwen2-72B-Instruct | To solve the integral $\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz$, we need to use the residue theorem. However, before we can apply the residue theorem, we need to identify the singularities of the function $f(z) = \frac{5z - 2}{z(z - 1)}$ and determine which of them lie inside the contour $|z| = 2$. The singularities are at $z = 0$ and $z = 1$, both of which are inside the contour $|z| = 2$. Next, we find the residues of $f(z)$ at these singularities. The residue at a simple pole $z = c$ of a function $f(z)$ is given by $\lim_{z \to c} (z - c) f(z)$. 1. For $z = 0$, the residue is: \[ \lim_{z \to 0} z \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} \frac{5z - 2}{z - 1} = -2 \] 2. For $z = 1$, the residue is: \[ \lim_{z \to 1} (z - 1) \frac{5z - 2}{z(z - 1)} = \lim_{z \to 1} \frac{5z - 2}{z} = 3 \] 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 singularities inside the contour. Therefore, \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz = 2\pi i \left( -2 + 3 \right) = 2\pi i \] Thus, $A = 2\pi$. Solution: $A = 2\pi$. |
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| baichuan4 | To solve this problem, we will use the residue theorem. The residue theorem states that for a simply connected domain $D$ in the complex plane containing a finite number of isolated singularities, and a function $f(z)$ that is analytic on $D$ except at the isolated singularities, the integral of $f(z)$ around the boundary of $D$ is equal to $2\pi i$ times the sum of the residues of $f(z)$ at the singularities within $D$. The function $f(z) = \frac{5z - 2}{z(z - 1)}$ has two singularities: a simple pole at $z = 0$ and a simple pole at $z = 1$. We need to find the residues at these poles. Residue at $z = 0$: $$ \text{Res}(f, 0) = \lim_{z \to 0} (z - 0) \cdot f(z) = \lim_{z \to 0} \frac{5z - 2}{z - 1} = -2 $$ Residue at $z = 1$: $$ \text{Res}(f, 1) = \lim_{z \to 1} (z - 1) \cdot f(z) = \lim_{z \to 1} \frac{5z - 2}{z} = 3 $$ Now, we can apply the residue theorem: $$ \int_{|z| = 2} f(z) dz = 2\pi i (\text{Res}(f, 0) + \text{Res}(f, 1)) = 2\pi i (-2 + 3) = 2\pi i $$ So, the value of the integral is $2\pi i$. Therefore, $A = 2\pi$. |
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| doubao | Let's decompose the integrand using partial fraction decomposition: $\frac{5z-2}{z(z-1)}=\frac{A}{z}+\frac{B}{z-1}$. Multiplying through by $z(z-1)$ gives $5z-2=A(z-1)+Bz$. Setting $z=0$ gives $-2=-A$, so $A=2$. Setting $z=1$ gives $3=B$. So the integrand can be written as $\frac{2}{z}+\frac{3}{z-1}$. Now we can evaluate the integral: $\int_{|z|=2}\frac{5z-2}{z(z-1)}dz=\int_{|z|=2}\frac{2}{z}dz+\int_{|z|=2}\frac{3}{z-1}dz$. The integral of $\frac{2}{z}$ around a circle centered at the origin of radius 2 is $2\times2\pi i=4\pi i$. For the integral of $\frac{3}{z-1}$, it has a pole at $z=1$ which is inside the circle $|z|=2$. The residue at $z=1$ is 3. So the integral of $\frac{3}{z-1}$ around the circle is $3\times2\pi i=6\pi i$. Adding these two integrals gives $4\pi i+6\pi i=10\pi i$. So $A=10$. |
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| minimax | To solve the integral $\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz$, 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 $f(z) = \frac{5z - 2}{z(z - 1)}$ has singularities at $z = 0$ and $z = 1$. However, since the contour $|z| = 2$ is a circle of radius 2 centered at the origin, only the singularity at $z = 0$ is inside the contour. The singularity at $z = 1$ is outside the contour and does not contribute to the integral. We only need to find the residue of $f(z)$ at $z = 0$. The residue at a simple pole (which is the case here since $z = 0$ is a simple zero of the denominator) can be found by taking the limit as $z$ approaches the pole of $z$ times the function: $$ \text{Res}(f, 0) = \lim_{z \to 0} z \cdot \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} \frac{5z - 2}{z - 1} $$ Now we can compute the limit: $$ \text{Res}(f, 0) = \lim_{z \to 0} \frac{5z - 2}{z - 1} = \frac{-2}{-1} = 2 $$ According to the residue theorem, the integral is then: $$ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz = 2\pi i \cdot \text{Res}(f, 0) = 2\pi i \cdot 2 = 4\pi i $$ Comparing this result to the given answer $A i$, we find that $A = 4\pi$. Therefore, the value of $A$ is $4\pi$. |
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| kimi | To solve the integral $\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz$, we can use the method of residues. The integral is a contour integral over a circle of radius 2 centered at the origin. The function has singularities at $z = 0$ and $z = 1$. Since the contour $|z| = 2$ encloses both singularities, we need to find the residues at these points. First, let's find the residue at $z = 0$. The function has a simple pole at $z = 0$, so we can find the residue by evaluating the limit as $z$ approaches 0: $$\text{Res}_{z=0} \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} (z - 0) \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} \frac{5z - 2}{z - 1} = \frac{-2}{-1} = 2.$$ Next, let's find the residue at $z = 1$. The function has a simple pole at $z = 1$, so we can find the residue by evaluating the limit as $z$ approaches 1: $$\text{Res}_{z=1} \frac{5z - 2}{z(z - 1)} = \lim_{z \to 1} (z - 1) \frac{5z - 2}{z(z - 1)} = \lim_{z \to 1} \frac{5z - 2}{z} = 5.$$ Now, we can use the residue theorem to evaluate the contour integral: $$\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz = 2\pi i \left(\text{Res}_{z=0} + \text{Res}_{z=1}\right) = 2\pi i (2 + 5) = 14\pi i.$$ So, the answer is $A = 14$. |
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| wenxinyiyan | 首先，我们需要找到函数$f(z) = \frac{5z - 2}{z(z - 1)}$在$|z| = 2$内的所有奇点。 奇点是$z = 0$和$z = 1$。 接着，我们计算每个奇点的留数： 对于$z = 0$，我们有： $Res(f, 0) = \lim_{z \to 0} z \cdot \frac{5z - 2}{z(z - 1)} = 2$ 对于$z = 1$，我们有： $Res(f, 1) = \lim_{z \to 1} (z - 1) \cdot \frac{5z - 2}{z(z - 1)} = 3$ 根据留数定理，我们有： $\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz = 2\pi i (Res(f, 0) + Res(f, 1)) = 2\pi i (2 + 3) = 10\pi i$ 因此，$A = 10\pi$。 |
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| zhipu | To solve the integral \(\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz\), we can use the Residue Theorem from complex analysis. The Residue Theorem states that for a function \(f(z)\) that is analytic inside and on a simple closed contour \(C\) except for a finite number of singularities inside \(C\), \[ \int_{C} f(z) \, dz = 2\pi i \sum \text{Res}(f, a_k), \] where the sum is over all singularities \(a_k\) inside \(C\). ### Step-by-Step Solution: 1. **Identify the Singularities:** The function \(f(z) = \frac{5z - 2}{z(z - 1)}\) has singularities where the denominator is zero. These are: - \(z = 0\) - \(z = 1\) 2. **Determine the Singularities Inside the Contour:** The contour \(|z| = 2\) is a circle of radius 2 centered at the origin. Both singularities \(z = 0\) and \(z = 1\) lie inside this contour. 3. **Compute the Residues:** - **Residue at \(z = 0\):** The function \(f(z)\) has a simple pole at \(z = 0\). The residue at a simple pole \(z = a\) is given by: \[ \text{Res}(f, a) = \lim_{z \to a} (z - a) f(z). \] For \(z = 0\): \[ \text{Res}(f, 0) = \lim_{z \to 0} z \cdot \frac{5z - 2}{z(z - 1)} = \lim_{z \to 0} \frac{5z - 2}{z - 1} = \frac{-2}{-1} = 2. \] - **Residue at \(z = 1\):** The function \(f(z)\) also has a simple pole at \(z = 1\). For \(z = 1\): \[ \text{Res}(f, 1) = \lim_{z \to 1} (z - 1) \cdot \frac{5z - 2}{z(z - 1)} = \lim_{z \to 1} \frac{5z - 2}{z} = \frac{5 \cdot 1 - 2}{1} = 3. \] 4. **Apply the Residue Theorem:** Summing the residues inside the contour \(|z| = 2\): \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} \, dz = 2\pi i \left( \text{Res}(f, 0) + \text{Res}(f, 1) \right) = 2\pi i (2 + 3) = 2\pi i \cdot 5 = 10\pi i. \] Therefore, the value of the integral is \(10\pi i\), and comparing this with the given form \(Ai\), we find that \(A = 10\pi\). ### Final Answer: \[ A = 10\pi \] |
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| gpto1 | To compute the integral \( I = \int_{|z|=2} \frac{5z - 2}{z(z - 1)}\,dz \), we can use the residue theorem from complex analysis. The residue theorem states that for a function \( f(z) \) with isolated singularities inside a closed contour \( C \), the integral over \( C \) is: \[ I = \int_C f(z)\,dz = 2\pi i \sum \text{Residues of } f(z) \text{ inside } C \] **Step 1: Identify the Singularities** The function \( f(z) = \frac{5z - 2}{z(z - 1)} \) has singularities (poles) where the denominator is zero, i.e., at \( z = 0 \) and \( z = 1 \). Both singularities are inside the contour \( |z| = 2 \) since the circle of radius 2 centered at the origin encompasses these points. **Step 2: Compute the Residues** We need to compute the residues at \( z = 0 \) and \( z = 1 \). **Residue at \( z = 0 \):** Since \( z = 0 \) is a simple pole, the residue is: \[ \text{Res}_{z=0} f(z) = \lim_{z \to 0} z \cdot f(z) = \lim_{z \to 0} \frac{5z - 2}{z - 1} \] Evaluating the limit: \[ \text{Res}_{z=0} f(z) = \frac{5 \cdot 0 - 2}{0 - 1} = \frac{-2}{-1} = 2 \] **Residue at \( z = 1 \):** Similarly, for \( z = 1 \): \[ \text{Res}_{z=1} f(z) = \lim_{z \to 1} (z - 1) \cdot f(z) = \lim_{z \to 1} \frac{5z - 2}{z} \] Evaluating the limit: \[ \text{Res}_{z=1} f(z) = \frac{5 \cdot 1 - 2}{1} = 3 \] **Step 3: Sum the Residues** Total residue inside the contour: \[ \sum \text{Residues} = \text{Res}_{z=0} f(z) + \text{Res}_{z=1} f(z) = 2 + 3 = 5 \] **Step 4: Apply the Residue Theorem** Using the residue theorem: \[ I = 2\pi i \cdot \left( \text{Sum of residues inside } C \right) = 2\pi i \cdot 5 = 10\pi i \] So, the integral evaluates to \( I = 10\pi i \). Therefore, \( A = 10\pi \). **Answer:** \(10\pi\) |
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| deepseekv3 | To compute the integral \(\int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz\), we can use the **Residue Theorem**, which is a powerful tool in complex analysis for evaluating contour integrals. ### Step 1: Identify the Singularities The integrand \(\frac{5z - 2}{z(z - 1)}\) has singularities at \(z = 0\) and \(z = 1\). Both of these points lie inside the contour \(|z| = 2\) (since \(|0| = 0 < 2\) and \(|1| = 1 < 2\)). ### Step 2: Compute the Residues The Residue Theorem states that: \[ \oint_C f(z) dz = 2\pi i \sum \text{Res}(f, a_k) \] where \(a_k\) are the singularities inside the contour \(C\). #### Residue at \(z = 0\): The function has a simple pole at \(z = 0\). The residue at \(z = 0\) is: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 0\right) = \lim_{z \to 0} z \cdot \frac{5z - 2}{z(z - 1)} = \frac{5 \cdot 0 - 2}{0 - 1} = \frac{-2}{-1} = 2 \] #### Residue at \(z = 1\): The function also has a simple pole at \(z = 1\). The residue at \(z = 1\) is: \[ \text{Res}\left(\frac{5z - 2}{z(z - 1)}, 1\right) = \lim_{z \to 1} (z - 1) \cdot \frac{5z - 2}{z(z - 1)} = \frac{5 \cdot 1 - 2}{1} = 3 \] ### Step 3: Apply the Residue Theorem The total residue is the sum of the residues at \(z = 0\) and \(z = 1\): \[ \text{Total Residue} = 2 + 3 = 5 \] Thus, the integral is: \[ \int_{|z| = 2} \frac{5z - 2}{z(z - 1)} dz = 2\pi i \cdot 5 = 10\pi i \] ### Final Answer The integral evaluates to \(10\pi i\), so \(A = 10\). |
31.4 |