复数幂公式

复数是具有公式 a + ib 的数，其中 a 和 b 是实数，I (iota) 是虚数部分，表示 (-1)，通常以矩形或标准形式表示。例如，10 + 5i 是一个复数，其中 10 代表实部，5i 代表虚部。根据 a 和 b 的值，它们可能是完全真实的或纯虚构的。当a + ib 中a = 0 时，ib 是一个全虚数，而当b = 0 时，我们得到a，它是一个严格实数。

复数幂公式

要根据指定的指数展开复数，必须首先将其转换为其极坐标形式，该形式具有模数和自变量作为分量。之后，应用 DeMoivre 定理，该定理指出：

De Moivre 公式指出，对于一个数字的所有实数值，比如 x，

(cos x + isinx)ⁿ = cos(nx) + isin(nx), where n is any integer.

公式的推导

DeMoivre 定理可以在数学归纳法的帮助下导出/证明如下：

As stated previously, (cos x + isinx)ⁿ = cos(nx) + isin(nx)     …. (1)

For n = 1, we obtain:

           (cos x + i sin x)¹ = cos(1x) + i sin(1x) = cos(x) + i sin(x), which is true.

Assuming that the formula holds true for any integer, say n = k. Then,

          (cos x + i sin x)^k = cos(kx) + i sin(kx)      …. (2)

Now, we just have to prove that the formula holds true for n = k + 1.

          (cos x + i sin x)^k+1 = (cos x + i sin x)k (cos x + i sin x)

          = (cos (kx) + i sin (kx)) (cos x + i sin x) [Using (i)]

          = cos (kx) cos x − sin(kx) sinx + i (sin(kx) cosx + cos(kx) sinx)

          = cos {(k+1)x} + i sin {(k+1)x}

          ⇒ (cos x + i sin x)k+1 = cos {(k+1)x} + i sin {(k+1)x}

Hence the result is proved.

示例问题

问题 1. 展开 (1 + i) ⁵ 。

解决方案：

Here, r = $\sqrt{(1^2+1^2)}$= $\sqrt{2}$, θ = π/4

The polar form of (1 + i) = $[\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]$

According to De Moivre’s Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + i sin(nθ).

Thus, (1 + i)⁵ = $[\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^5$

= $(\sqrt{2})^{5}[cos(\frac{5\pi}{4})+i\ sin(\frac{5\pi}{4})]$

= −4 − 4i

问题 2：展开 (2 + 2i) ⁶ 。

解决方案：

Here, r = $\sqrt{(2^2+2^2)} = 2\sqrt{2}$, θ = π/4

The polar form of  (2 + 2i) = $[2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]$

According to De Moivre’s Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + isin(nθ).

Thus, (2 + 2i)⁶ = $[2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^6$

= $(\sqrt{2})^{6}[cos(\frac{6\pi}{4})+i\ sin(\frac{6\pi}{4})]$

= 512 (-i)

= −512i

问题 3：展开 (1 + i) ¹⁸ 。

解决方案：

Here, r = $\sqrt{(1^2+1^2)} = \sqrt{2}$, θ = π/4

The polar form of (1+i) = $[\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]$

According to De Moivre’s Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + isin(nθ).

Thus, (1 + i)¹⁸ = $[\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{18}$

= $(\sqrt{2})^{18}[cos(\frac{18\pi}{4})+i\ sin(\frac{18\pi}{4})]$

= 512i

问题 4：展开 (-√3 + 3i) ³¹ .

解决方案：

Here, r = $\sqrt{((-\sqrt{3})^2+3^2)} = 2\sqrt{3}$, θ = 2π/3

The polar form of  (-√3 + 3i) = $[2\sqrt{3}cos(\frac{2\pi}{3})+i\ sin(\frac{2\pi}{3})]$

According to De Moivre’s Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + isin(nθ).

Thus, (-√3 + 3i)³¹= $[2\sqrt{3}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{31}$

= $(\sqrt{2})^{31}[cos(\frac{31\pi}{4})+i\ sin(\frac{31\pi}{4})$

问题 5：展开 (1 – i) ¹⁰ 。

解决方案：

r = $\sqrt{(1^2+(-1)^2)} = \sqrt{2}$, θ = π/4

The polar form of  (1 – i) = $[\sqrt{2}cos(\frac{\pi}{4})+i \ sin(\frac{\pi}{4})]$

According to De Moivre’s Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + isin(nθ).

Thus, (1 – i)¹⁰ = $[\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{10}$

= $(\sqrt{2})^{10}[cos(\frac{10\pi}{4})+i\ sin(\frac{10\pi}{4})]$

= 32 [0 + i(-1)]

= 32 (-i)

= -32i

问题 6. 化简 (1 + √3i) ⁶ .

解决方案：

Modulus of (1 + √3i)⁶ = $\sqrt{1^2+(\sqrt{3})^2} = 2$

Argument = tan^-1(√3/1) = tan^-1(√3) = π/3

⇒ Polar form = $2[cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3})]$ 

Now, (1 + √3i)⁶ = $[2(cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3}))]^6$

As per DeMoivre’s theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx).

⇒ $[2(cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3}))]^6$

= $2^6(cos(\frac{6\pi}{3})+i\ sin(\frac{6\pi}{3}))$

= 64(1 + 0)

= 64

问题 7. 简化 i ^√3 。

解决方案：

Modulus = r = $\sqrt{0^2+1^2}$ = 1

Argument = tan^-1[1/0] = π/2

Polar Form = r[cosθ + isinθ] = $1[cos(\frac{\pi}{2}) +i\ sin(\frac{\pi}{2})]$

Now, i^{√3} = $[cos(\frac{\pi}{2}) + i\ sin(\frac{\pi}{2})]^{\sqrt3}$

As per DeMoivre’s theorem: (cosθ + isinθ)ⁿ = cos(nθ) + isin(nθ).

⇒  $[cos(\frac{\pi}{2}) + i\ sin(\frac{\pi}{2})]^{\sqrt3}$

= $[cos(\frac{\sqrt3\pi}{2}) + i\ sin(\frac{\sqrt3\pi}{2})].$