| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | D |
D |
| deepseekr1 | To solve \((1+i)^{10}\), we convert \(1+i\) to polar form. The modulus is \(\sqrt{2}\) and the argument is \(\pi/4\). Using De Moivre's theorem: \[ (1+i)^{10} = \left(\sqrt{2}\right)^{10} \left[\cos\left(10 \cdot \frac{\pi}{4}\right) + i \sin\left(10 \cdot \frac{\pi}{4}\right)\right] \] Calculating the modulus part: \(\left(\sqrt{2}\right)^{10} = 2^{5} = 32\). For the angle: \(10 \cdot \frac{\pi}{4} = \frac{5\pi}{2}\), which reduces to \(\frac{\pi}{2}\) modulo \(2\pi\). Thus: \[ \cos\left(\frac{\pi}{2}\right) = 0, \quad \sin\left(\frac{\pi}{2}\right) = 1 \] So, \((1+i)^{10} = 32(0 + i \cdot 1) = 32i\). Alternatively, using \((1+i)^2 = 2i\), then: \[ (1+i)^{10} = \left((1+i)^2\right)^5 = (2i)^5 = 32i^5 = 32i \] **Answer:** D |
D |
| gpto1 | D | D |
| GPT-4o | To solve \((1+i)^{10}\), we can use polar form and De Moivre's Theorem. First, we write \(1+i\) in polar form: \[ 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \] Now, using De Moivre's Theorem: \[ (1+i)^{10} = \left( \sqrt{2} \right)^{10} \left( \cos \frac{10\pi}{4} + i \sin \frac{10\pi}{4} \right) \] \[ = 2^5 \left( \cos \frac{5\pi}{2} + i \sin \frac{5\pi}{2} \right) \] \[ = 32 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \] \[ = 32 \left( 0 + i \cdot 1 \right) \] \[ = 32i \] Thus, the answer is: \[ D. 32i \] |
D |
| llama3.1-70B-Instruct | D |
D |
| llama3.1-405B-Instruct | D |
D |
| DeepSeek-Coder-V2 | C. 32 | D |
| Qwen2-72B-Instruct | C. 32 Explanation: (1+i)^10 = ((1+i)^2)^5 = (2i)^5 = 32i^5 = 32(i^4 * i) = 32(1 * i) = 32i. But since i^5 = i, the answer simplifies to 32. |
D |
| baichuan4 | B. i | D |
| doubao | D. 32i. \[ \begin{align*} (1+i)^{10}&=[(1+i)^2]^5\\ &=(1+2i+i^2)^5\\ &=(1+2i-1)^5\\ &=(2i)^5\\ &=32i^5\\ &=32i \end{align*} \] |
D |
| minimax | To solve the expression \((1+i)^{10}\), we can use the binomial theorem or recognize that \(1+i\) is a complex number that can be expressed in polar form. However, since this is a multiple-choice question, we can also use a shortcut by recognizing that \(1+i\) is a complex number whose magnitude is \(\sqrt{1^2 + 1^2} = \sqrt{2}\), and its argument (angle with the positive real axis) is \(45^\circ\) or \(\pi/4\) radians. When raised to the power of 10, the magnitude \(\sqrt{2}\) will be raised to the power of 10, and the argument will be multiplied by 10. So we have: Magnitude: \((\sqrt{2})^{10} = 2^{5} = 32\) Argument: \(10 \cdot \frac{\pi}{4} = \frac{10\pi}{4} = \frac{5\pi}{2}\) Since \(5\pi/2\) is an odd multiple of \(\pi/2\), the resulting complex number will have an argument that is an odd multiple of \(90^\circ\) or \(\pi/2\) radians, which corresponds to the imaginary axis in the complex plane. Therefore, the result will be purely imaginary. The magnitude of the result is 32, and since it's purely imaginary, the correct answer is: D. 32i |
D |
| kimi | To solve this problem, we can use the properties of complex numbers and De Moivre's theorem. The complex number \(1 + i\) can be expressed in polar form as \(\sqrt{2} \text{cis}(45^\circ)\), where \(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\). Using De Moivre's theorem, which states that \((r \text{cis}(\theta))^n = r^n \text{cis}(n\theta)\), where \(r\) is the magnitude and \(\theta\) is the argument of the complex number, we can find the 10th power of \(1 + i\). So, \((1 + i)^{10} = (\sqrt{2} \text{cis}(45^\circ))^{10} = (\sqrt{2})^{10} \text{cis}(450^\circ)\). Since \((\sqrt{2})^{10} = 2^5 = 32\), and \(450^\circ\) is equivalent to \(90^\circ\) (because \(450^\circ - 360^\circ = 90^\circ\)), we have \(\text{cis}(450^\circ) = \text{cis}(90^\circ) = i\). Therefore, \((1 + i)^{10} = 32i\). Answer: D. 32i |
D |
| wenxinyiyan | D Explanation: To solve this problem, we can use the polar form of complex numbers and De Moivre's theorem. First, convert $1+i$ to polar form: $1+i = \sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))$Now, we can use De Moivre's theorem to find $(1+i)^{10}$: $(1+i)^{10} = (\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})))^{10}$$= (\sqrt{2})^{10}(\cos(10 \times \frac{\pi}{4}) + i\sin(10 \times \frac{\pi}{4}))$$= 32(\cos(\frac{5\pi}{2}) + i\sin(\frac{5\pi}{2}))$$= 32i$Therefore, the answer is D. $32i$. |
D |
| zhipu | To solve \((1+i)^{10}\), we can use De Moivre's Theorem, which states that for any complex number \(z = r(\cos \theta + i \sin \theta)\) and any positive integer \(n\), \[ z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \] First, express \(1+i\) in polar form. The magnitude \(r\) of \(1+i\) is: \[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \] The argument \(\theta\) of \(1+i\) is: \[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \] So, \(1+i\) in polar form is: \[ 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \] Using De Moivre's Theorem: \[ (1+i)^{10} = \left( \sqrt{2} \right)^{10} \left( \cos \left( 10 \cdot \frac{\pi}{4} \right) + i \sin \left( 10 \cdot \frac{\pi}{4} \right) \right) \] Calculate \(\left( \sqrt{2} \right)^{10}\): \[ \left( \sqrt{2} \right)^{10} = (2^{1/2})^{10} = 2^5 = 32 \] Calculate the argument: \[ 10 \cdot \frac{\pi}{4} = \frac{10\pi}{4} = \frac{5\pi}{2} \] Since \(\cos\) and \(\sin\) are periodic with period \(2\pi\): \[ \cos \left( \frac{5\pi}{2} \right) = \cos \left( 2\pi + \frac{\pi}{2} \right) = \cos \left( \frac{\pi}{2} \right) = 0 \] \[ \sin \left( \frac{5\pi}{2} \right) = \sin \left( 2\pi + \frac{\pi}{2} \right) = \sin \left( \frac{\pi}{2} \right) = 1 \] So, \[ (1+i)^{10} = 32 \left( \cos \left( \frac{5\pi}{2} \right) + i \sin \left( \frac{5\pi}{2} \right) \right) = 32 (0 + i \cdot 1) = 32i \] Therefore, the answer is: D. 32i |
D |
| gpto1 | D | D |
| deepseekv3 | To solve \((1 + i)^{10}\), we can use the polar form of complex numbers. 1. **Express \(1 + i\) in polar form:** - The modulus of \(1 + i\) is: \[ |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \] - The argument (angle) of \(1 + i\) is: \[ \theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4} \] - Therefore, \(1 + i\) in polar form is: \[ \sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right) \] 2. **Raise to the 10th power using De Moivre's Theorem:** - De Moivre's Theorem states: \[ \left[ r \left( \cos\theta + i \sin\theta \right) \right]^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right) \] - Applying this: \[ (1 + i)^{10} = \left(\sqrt{2}\right)^{10} \left( \cos\left(10 \cdot \frac{\pi}{4}\right) + i \sin\left(10 \cdot \frac{\pi}{4}\right) \right) \] - Simplify: \[ \left(\sqrt{2}\right)^{10} = 2^{5} = 32 \] \[ 10 \cdot \frac{\pi}{4} = \frac{5\pi}{2} \] - We can reduce \(\frac{5\pi}{2}\) modulo \(2\pi\): \[ \frac{5\pi}{2} - 2\pi = \frac{\pi}{2} \] - Therefore: \[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \sin\left(\frac{\pi}{2}\right) = 1 \] - So: \[ (1 + i)^{10} = 32 \left(0 + i \cdot 1\right) = 32i \] **Answer:** D. 32i |
D |