如何使用 De Moivre 定理化简 (1 – i) ¹⁰ ？

复数可以称为实数和虚数的混合，实数或成分是任何分数、有理或无理整数，其虚部表示为与虚数单位 iota 相乘的实数，如图 i 所示。因此，复数表示通过加法和减法这两种算术运算组合的实数和虚数。

实数和虚数

这种既包括有理数又包括无理数的数称为实数。它们基于数轴的概念，零是原点，其右侧的所有数字为正数，原点左侧的所有数字为负数。

数字也可以被描述为数学中负数的平方根。例如， $\sqrt{-100}$是一个虚数，因为它描述了数字 100，它是一个完全平方，是平方根下的负数。这些数字不是有形的，但在数学中使用的意义上仍然是真实的。换句话说，虚数是与实数相反的数字。它们不是基于数轴的概念，因此无法描绘或绘制在数轴上。定义虚数的另一种方式可以是这样的数，当与自身相乘时产生负结果，即平方。

复数的标准形式

标准形式的复数表示为 a + ib，其中 a 和 b 都是实数，但 b 与虚变量 i 相乘，表示整个复数的虚部，可以表示为通过'z'。因此，复数通常写成 z = a + ib 的形式，其中 a 表示实部，ib 或 bi 将是虚部。就此而言，0 + bi 也将被视为一个复数，实部不存在，bi 描绘了它的虚部。例子是，

• 5 + 2i is a complex number, where 5 is the real part and 2i depicts the imaginary part.
• e² + 12i  is a complex number, where e2 is the real part and 12i is the imaginary part.
• √22 – 162i is a complex number, where √22 is the real part and 162i is the imaginary part.

复数的极坐标形式

复数的标准形式也可以称为它的矩形形式。极坐标形式只是表示复数的另一种方式，借助它的模数和参数。极坐标形式使用给定复数的实部和虚部的极坐标来表示。

The equation of polar form of z = x + iy is z = r(cosθ + i sinθ).

Here, z = r(cosθ + i sinθ), where r = |z| = $\sqrt{(x^2+y^2)}$

x = r cosθ, y = r sinθ.

如何使用 De Moivre 定理化简 (1 – i) ¹⁰ ？

解决方案：

De Moivre’s Theorem

This theorem holds utmost importance in the universe of complex numbers as it helps connect the field of trigonometry to the intricacies of complex numerals. It also helps obtain relationships between various trigonometric functions of different angles. It is so-called because this theorem was propounded by one of the most notable mathematicians in history, De Moivre, who contributed a lot to the fields of probability, algebra, etc. The theorem is also referred to as De Moivre’s Formula or De Moivre’s Identity.

Formula

For a number x, such that x ∈ R, or for all real values of x,

(cos x + i sin x)ⁿ = cos nx  + i sin nx 

Or (e^iθ)ⁿ= e^inθ

Here, n is a rational number and i, called iota is the imaginary part.

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.

类似问题

问题 1：使用 De Moivre 定理化简 (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：使用 De Moivre 定理化简 (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：使用 De Moivre 定理化简 (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：使用 De Moivre 定理化简 (-√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 = $(2\sqrt{3})^{31}[-\frac{1}{2}+i\frac{3}{2}]$.