求解以下 x ⁴ = 5 – 5i

复数是 a + ib 形式的数字，其中 a 和 b 是实数，i (iota) 是虚部，表示 √(-1)，它们经常以矩形或标准形式表示。例如，2 + 3i 是一个复数，其中 2 是实部，3i 是虚部。根据 a 和 b 的值，它们可以是完全真实的，也可以是纯虚构的。如果 a + ib 中的 a = 0，则 ib 是一个完全虚数，如果 b = 0，我们得到一个纯实数。

复数的极坐标形式

复整数的模是实部和虚部平方和的非负平方根。模写为 mod(z)、|z| 或 |x + iy|，对于复数 z = a + ib 定义如下：

$|z| = \sqrt{a^2+b^2}$

为了表示一个复数，这里写了实部和虚部的极坐标。符号θ表示数轴相对于实轴即x轴倾斜的角度。线所表示的长度称为其模数，在字母表中用字母 r 表示。在下图中，实部和虚部分别由 a 和 b 表示，而模数由 OP = r 表示。

z = a + ib 类型的复数的极坐标形式表示如下：

z = r[cos θ + i sin θ]

Here, r = $\sqrt{p^2+q^2}$  and θ = $tan^{-1}{q/p}$.

德莫弗公式

要将复数扩展为其所需的指数，必须首先将其转换为其极坐标形式，其中包括模数和参数作为分量。根据 De Moivre 公式，对于一个数字的所有实数值，比如 x，

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

where, n is an integer

The complex number must be converted into its polar form (if not) to use the given formula.

推导

以下是如何通过数学归纳法推导出/证明 DeMoivre 定理：

To prove, (cos x + i sinx)ⁿ = cos(nx) + i sin(nx)     …. (1)

For n = 1, we have,

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

Suppose the formula exists for an integer, say n = k. Then we have,

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

Now all we have to do is show that the formula holds 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 equation (2)]

        = 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}

This proves the DeMoivre’s Theorem.

求解以下 x ⁴ = 5 – 5i

解决方案：

It is given that, x⁴ = 5 – 5i.

The value of x is clearly, x = (5 – 5i)^1/4

Convert the complex number into its polar form.

Here, r = $\sqrt{(5^2+5^2)} = 2\sqrt{5}$, θ = -π/4

The polar form of  (5 – 5i) = $[2\sqrt{5}(cos(\frac{\pi}{4})-i\ sin(\frac{\pi}{4}))]$

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

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

= $[\frac{2\sqrt{5}}{4}(cos(\frac{\pi}{8})-i\ sin(\frac{\pi}{8}))]$

= $[\frac{\sqrt{5}}{2}(cos(\frac{\pi}{8})-i\ sin(\frac{\pi}{8}))]$

= 1.6 − 0.32i

Hence, the value of x is 1.6 − 0.32i.

类似问题

问题 1. 求解 x：x ^(1/10) = 1 – i

解决方案：

It is given that, x^(1/10) = 1 – i

The value of x is clearly, x = (1 – i)¹⁰

Convert the complex number into its polar form.

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)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, the value of x is 0 − 32i.

问题 2. 求解 x：x ^(1/31) = -√3 + 3i

解决方案：

It is given that, x^(1/31) = -√3 + 3i

The value of x is clearly, x = (-√3 + 3i)³¹

Convert the complex number into its polar form.

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) + i sin(nθ).

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

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

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

Hence, the value of x is $(2\sqrt{3})^{31}[-\frac{1}{2}+\frac{3}{2}i]$.

问题 3. 求解 x：x ^(1/18) = 1 + i。

解决方案：

It is given that, x^(1/18) = 1 + i

The value of x is clearly, x = (1 + i)¹⁸

Convert the complex number into its polar form.

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)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, the value of x is 512i.

问题 4. 求解 x：x ^(1/6) = 1 + √3i。

解决方案：

It is given that, x^(1/6) = 1 + √3i

The value of x is clearly, x = (1 + √3i)⁶

Convert the complex number into its polar form.

Here, $r = \sqrt{(1^2+(√3)^2)}$ = 2, θ = π/3

The polar form of (1+i) = $2[cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3})]$

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

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

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

= 64(1 + 0)

= 64

Hence, the value of x is 64.

问题 5. 求解 x：x ^(1/5) = 1 + i。

解决方案：

It is given that, x^(1/5) = 1 + i

The value of x is clearly, x = (1 + i)⁵

Convert the complex number into its polar form.

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, the value of x is -4 – 4i.