复数 第五部分 - 棣莫弗公式

复数 - 第五部分

棣莫弗公式、乘法、除法、幂与根

✖️ 极坐标形式的复数乘法

若 \(z_1 = r_1(\cos\theta_1 + i\sin\theta_1)\) 与 \(z_2 = r_2(\cos\theta_2 + i\sin\theta_2)\)

\(z_1 \cdot z_2 = r_1 r_2 \left(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right)\)

💡 用文字表达:

  • 模相乘:\(|z_1 \cdot z_2| = |z_1| \cdot |z_2|\)
  • 辐角相加:\(\arg(z_1 \cdot z_2) = \arg(z_1) + \arg(z_2)\)
z₁ θ₁ z₂ θ₂ z₁·z₂ θ₁+θ₂ Re Im

➗ 极坐标形式的复数除法

\(\frac{z_1}{z_2} = \frac{r_1}{r_2} \left(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right)\)

💡 用文字表达:

  • 模相除:\(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}\)
  • 辐角相减:\(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)\)

✏️ 例 1:极坐标乘法

计算:

\(z_1 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\)

\(z_2 = 3\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\)

求 \(z_1 \cdot z_2\)

解答:

模:\(r_1 \cdot r_2 = 2 \cdot 3 = 6\)

辐角:\(\theta_1 + \theta_2 = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\)

答案:\(z_1 \cdot z_2 = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 6i\)

⭐ 棣莫弗公式(De Moivre)

\(\left[r(\cos\theta + i\sin\theta)\right]^n = r^n(\cos n\theta + i\sin n\theta)\)

💡 用文字表达:

  • 模的 n 次幂:\(|z^n| = |z|^n\)
  • 辐角乘以 n:\(\arg(z^n) = n \cdot \arg(z)\)

⚡ 这对负数和分数 n 也成立!

✏️ 例 2:极坐标幂运算

计算:\((1 + i)^8\)

解答:

步骤 1:转换为极坐标形式

\(r = \sqrt{1^2 + 1^2} = \sqrt{2}\)

\(\theta = \frac{\pi}{4}\)(第一象限)

\(1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\)

步骤 2:应用棣莫弗公式

\((1+i)^8 = (\sqrt{2})^8 \left(\cos\frac{8\pi}{4} + i\sin\frac{8\pi}{4}\right)\)

\(= (\sqrt{2})^8 (\cos 2\pi + i\sin 2\pi)\)

步骤 3:计算

\((\sqrt{2})^8 = (2^{1/2})^8 = 2^4 = 16\)

\(\cos 2\pi = 1, \quad \sin 2\pi = 0\)

答案:\((1+i)^8 = 16 \cdot (1 + 0 \cdot i) = 16\)

✏️ 例 3:再算一个幂

计算:\(\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^{12}\)

解答:

步骤 1:注意到这已经在单位圆上

\(r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1\)

\(\theta = \frac{\pi}{3}\)(因为 \(\cos\frac{\pi}{3} = \frac{1}{2}, \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\))

步骤 2:棣莫弗公式

\(z^{12} = 1^{12} \left(\cos\frac{12\pi}{3} + i\sin\frac{12\pi}{3}\right)\)

\(= \cos 4\pi + i\sin 4\pi = 1\)

答案:\(1\)

√ 复数的根

\(z = r(\cos\theta + i\sin\theta)\) 的 n 次根为:

\(z_k = \sqrt[n]{r}\left(\cos\frac{\theta + 2\pi k}{n} + i\sin\frac{\theta + 2\pi k}{n}\right)\)

其中 \(k = 0, 1, 2, ..., n-1\)

💡 根的性质:

  • 恰好有 n 个不同的根
  • 所有根都在圆上,半径为 \(\sqrt[n]{r}\)
  • 这些根将圆等分成 n 份
  • 相邻根之间的角度:\(\frac{2\pi}{n}\)

✏️ 例 4:8 的三次根

求:\(\sqrt[3]{8}\)(所有复数根)

解答:

步骤 1:把 8 写成极坐标形式

\(8 = 8(\cos 0 + i\sin 0)\)

\(r = 8, \quad \theta = 0\)

步骤 2:计算三个根(k = 0, 1, 2)

\(\sqrt[3]{r} = \sqrt[3]{8} = 2\)

\(z_k = 2\left(\cos\frac{0 + 2\pi k}{3} + i\sin\frac{0 + 2\pi k}{3}\right)\)

k = 0:

\(z_0 = 2(\cos 0 + i\sin 0) = 2\)

k = 1:

\(z_1 = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -1 + \sqrt{3}i\)

k = 2:

\(z_2 = 2\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right) = 2\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -1 - \sqrt{3}i\)

答案:\(z = 2, \, -1+\sqrt{3}i, \, -1-\sqrt{3}i\)

2 -1+√3i -1-√3i r=2

⭕ 单位根(1 的根)

n 次单位根(\(z^n = 1\) 的解):

\(\omega_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}\)

其中 \(k = 0, 1, 2, ..., n-1\)

💡 性质:

  • 所有根都在单位圆上(r = 1)
  • 第一个根总是 \(\omega_0 = 1\)
  • 有时记作:\(\omega = e^{2\pi i/n}\)

例:四次单位根

\(z^4 = 1\) 的解:

\(\omega_0 = 1, \quad \omega_1 = i, \quad \omega_2 = -1, \quad \omega_3 = -i\)

✏️ 例 5:-1 的四次根

求:\(\sqrt[4]{-1}\)

解答:

步骤 1:把 \(-1\) 写成极坐标形式

\(-1 = 1 \cdot (\cos\pi + i\sin\pi)\)

步骤 2:根的公式

\(z_k = \cos\frac{\pi + 2\pi k}{4} + i\sin\frac{\pi + 2\pi k}{4}\)

k = 0:\(\theta = \frac{\pi}{4}\) → \(z_0 = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\)

k = 1:\(\theta = \frac{3\pi}{4}\) → \(z_1 = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\)

k = 2:\(\theta = \frac{5\pi}{4}\) → \(z_2 = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\)

k = 3:\(\theta = \frac{7\pi}{4}\) → \(z_3 = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\)

答案:\(\pm\frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2}i\)(4 个根)

📋 总结表 - 棣莫弗公式

运算 模 辐角
乘法 \(z_1 \cdot z_2\) \(r_1 \cdot r_2\) \(\theta_1 + \theta_2\)
除法 \(\frac{z_1}{z_2}\) \(\frac{r_1}{r_2}\) \(\theta_1 - \theta_2\)
幂 \(z^n\) \(r^n\) \(n\theta\)
根 \(\sqrt[n]{z}\) \(\sqrt[n]{r}\) \(\frac{\theta + 2\pi k}{n}\)

💡 考试提示

1️⃣ 幂

先转换为极坐标,再应用棣莫弗公式

2️⃣ 根

有 n 个根!不要忘记 k=0,1,...,n-1

3️⃣ 简化角度

若 \(\theta > 2\pi\),减去 \(2\pi\)

4️⃣ 画图

根都在圆上 - 画出来!

📝 第五部分总结

棣莫弗公式:

\([r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta)\)

这就是本主题的结尾!现在你已经准备好考试了 🎉