如何使用 DeMoivre 定理化简 z ⁴ + 8√3 – 8i = 0？

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。

复数的极坐标形式

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。

显然，毕达哥拉斯定理可以用于计算长度 r。可以使用三角比计算参数。因此，对于 z = p + iq 形式的复数，其极坐标形式如下：

r = Modulus[cos(argument) + isin(argument)]

Or, z = r[cosθ + isinθ]

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

如何使用 DeMoivre 定理化简 z⁴ + 8√3 – 8i = 0？

解决方案：

DeMoivre’s Theorem can be employed to simplify complex numbers of higher order. It can be used for expansion of complex numbers as per their exponent as well as to calculate the roots of complex numbers. 

If a complex number z is of the form zⁿ = kⁿ(cosθ + isinθ), then its roots are:

$z = k[cos(\frac{θ+2πm}{n})+isin(\frac{θ+2πm}{n})]$

where 0 ≤ m ≤ n – 1 and n is the root of the given complex number.

Given: z⁴ + 8√3 – 8i = 0

⇒ z⁴ = 8(–√3 + 1)

⇒ $z^4 = 8(2)[\frac{−√3}{2}+i\frac{1}{2})$

⇒ $z^4 = 2^4[\frac{−√3}{2}+i\frac{1}{2}]$

Comparing this with zⁿ = kⁿ(cosθ + isinθ), we have k =2, n = 4 and θ = 5π/ 6.

Find the 4 roots by substituting the values of m as 0, 1, 2 and 3 respectively.

• For m = 0, $z = 2[cos(\frac{\frac{5π}{6}+2π(0)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(0)}{4})$

                               = 1.58 + 1.21i

• For m = 1, $z = 2[cos(\frac{\frac{5π}{6}+2π(1)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(1)}{4})$

                               = −1.21 + 1.58i

• For m = 2, $z = 2[cos(\frac{\frac{5π}{6}+2π(2)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(2)}{4})$

                               = −1.58 −1.21i

• For m = 3, $z = 2[cos(\frac{\frac{5π}{6}+2π(3)}{4})+i\ sin(\frac{\frac{5π}{6}+2π(3)}{4})$

                               = 1.21 − 1.58i

Thus, the four roots of z are 1.58 + 1.21i, −1.21 + 1.58i, −1.58 −1.21i and 1.21 − 1.58i.

类似问题

问题1.如果z ³ + 2 + 2√3i = 0；应用 DeMoivre 定理求解 z。

解决方案：

Given: z³ + 2 + 2√3i = 0

r = $\sqrt{x^2+y^2}$ = √(16) = 4, θ = 4π/ 3.

Find the 3 roots by substituting the values of m as 0, 1 and 2 respectively.

• For m = 0, z = $\sqrt[3]4.[cos(\frac{\frac{4π}{3}+2π(0)}{3})+i\ sin(\frac{\frac{4π}{3}+2π(0)}{3})]$

                              = 0.27 + 1.56i

• For m = 1, z = $\sqrt[3]4.[cos(\frac{\frac{4π}{3}+2π(1)}{3})+i\ sin(\frac{\frac{4π}{3}+2π(1)}{3})]$

                    = −1.49 − 0.54i

• For m = 2, z = $\sqrt[3]4.[cos(\frac{\frac{4π}{3}+2π(2)}{3})+i\ sin(\frac{\frac{4π}{3}+2π(2)}{3})]$

                              = 1.21 −1.02i

Thus, the roots are 0.27 + 1.56i, −1.49 − 0.54i and 1.21 −1.02i.       

问题 2. 使用 DeMoivre 定理计算 z ² − 2i = 0。

解决方案：

Given: z² − 2i = 0.

⇒ z² = 2i

Clearly, n = 2 and θ =  90°.

Find the 2 roots by substituting the values of m as 0 and 1.

For m = 0,  z = $[cos(\frac{\frac{π}{2}+2π(0)}{2})+i\ sin(\frac{\frac{π}{2}+2π(0)}{2})$

                     = 1 + i

For m = 1, z = $[cos(\frac{\frac{π}{2}+2π(1)}{2})+i\ sin(\frac{\frac{π}{2}+2π(1)}{2})$

                    = −1 − i

Thus, the roots are 1 + i and  −1 − i.

问题 3. 计算 x ⁵ − 32 = 0。

解决方案：

Given: x⁵ − 32 = 0

⇒ x⁵ = 32 + 0i

Modulus = $\sqrt{x^2+y^2} = \sqrt{(32)^2}$= 32.

Argument = θ = tan^-1(0/ 32) = 0.

Find the 5 roots by substituting the values of m as 0, 1, 2, 3 and 4.

For m = 0,  $z = [cos(\frac{2π(0)}{5})+i\ sin(\frac{2π(0)}{5})$

                     = 2

For m = 1,  z = $[cos(\frac{2π(1)}{5})+i\ sin(\frac{2π(1)}{5})$

                     = 0.62 + 1.9i

For m = 2,  z = $[cos(\frac{2π(2)}{5})+i\ sin(\frac{2π(2)}{5})$

                     = −1.62 + 1.18i

For m = 3,  z = $[cos(\frac{2π(3)}{5})+i\ sin(\frac{2π(3)}{5})$

                     = −1.62 − 1.18i

For m = 4,  z = $[cos(\frac{2π(4)}{5})+i\ sin(\frac{2π(4)}{5})$

                     = 0.62 − 1.9i

Thus, the roots are 2, 0.62 + 1.9i, −1.62 + 1.18i, −1.62 − 1.18i and 0.62 − 1.9i.