如何使用 DeMoivre 定理化简 (1 + √3i) ⁶ ？

复数是 a + ib 形式的数字，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 是实部，5i 是虚部。根据 a 和 b 的值，这些可以是纯实数或纯虚数。如果 a + ib 中的 a = 0，则 ib 是纯虚数，如果 b = 0，则有 a，它是纯实数。

复数的模数和极坐标形式

复数的实部和虚部的平方和的非负平方根称为其模。模数表示为 mod(z) 或 |z|或 |x + iy|并为复数 z = a + ib 定义为，

mod(z) 或 |z| = $\sqrt{a^2+b^2}$

在这里，实部和虚部的极坐标被写成描述复数。数轴相对于实轴即x轴倾斜的角度用θ表示。线表示的长度称为其模数，用字母 r 表示。下图将 a 和 b 分别描绘为实部和虚部，OP = r 是模数。

对于 z = p + iq 形式的复数，其极坐标形式如下：

r = 模数[cos（参数）+ isin（参数）]

或者，z = r[cosθ + isinθ]

这里，r = $\sqrt{p^2+q^2}$和 θ = tan ^-1 {q/p}。

如何使用 DeMoivre 定理化简 (1 + √3i) ⁶ ？

解决方案：

In order to expand a complex number as per its given exponent, it first needs to be converted into its polar form, which uses its modulus and argument as its constituents. Then, DeMoivre’s theorem is applied, which states the following,

De Moivre’s Formula: For all real values of say, a number x,

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

Given number: (1 + √3i)⁶

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

Hence, (1 + √3i)⁶ = 64

类似问题

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

Hence, (1 + i)⁵ = -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)

Hence, (2 + 2i)⁶ = −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

Hence, (1 + i)¹⁸ = 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= $[2\sqrt{3}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{31}$

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

Hence, (-√3 + 3i)³¹ = $(2\sqrt{3})^{31}[-\frac{1}{2}+i\frac{3}{2}].$

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)

Hence, (1 – i)¹⁰ = 0 – 32i.