求 5(1 + i√3) 的第四个根

实数和虚数结合形成复数。虚部 I (iota) 表示 -1 的平方根。复数的虚部是 i。 a + ib 是矩形或标准形式的复数的典型表示。例如，100 + 25i 是一个复数，其中 100 代表实部，25i 代表虚部。

复数的极坐标形式表示

为了表示一个复数，这里写了实部和虚部的极坐标。表示数轴与实轴即x轴的倾斜角度。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。

毕达哥拉斯定理将用于得到长度 r。三角比可用于计算参数。 z = a + ib 类型的复数的极坐标形式表示如下：

r = 模数[cos（参数）+ isin（参数）]

或者，z = r[cosθ + isinθ]

在这种情况下，r = $\sqrt{a^2+b^2}$和 θ = tan ^-1 {b/a}。

计算复数的根

DeMoivre 定理可用于简化高阶复数。它可用于确定复数的根以及根据复数的指数展开复数。

鉴于： $z^{\frac{1}{n}} = r^{\frac{1}{n}}(cosθ + isinθ)$ ，那么它的根是：

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$

其中，k 介于 0 和 n – 1 之间，n 是指数或根。

求 5(1 + i√3) 的第四个根。以三角函数形式离开。

解决方案：

Modulus of the given number = $\sqrt{5^2+(5\sqrt{3})^2}$ = 10

Argument = tan-1[5√3/ 5] = π/3.

Thus, the polar form of 5 + 5√3i = $10 [\cos{(\frac{\pi}{3})} + i \sin{(\frac{\pi}{3}})]$

According to DeMoivre’s formula, all the nth roots of a complex number are given by:

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$, where  k lies between 0 and n – 1 and n is the exponent or radical. 

Here, r = 10, θ = π/3 and n = 4.

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

• For k = 0, z = $\sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(0)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(0)}{4})$

= $\sqrt[4]{10}[cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})]$

• For k = 1, z = $\sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(1)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(1)}{4})$

= $\sqrt[4]{10}[cos(\frac{7\pi}{12})+i\sin(\frac{7\pi}{12})]$

• For k = 2, z = $\sqrt[4]{10} (\cos{(\frac{\frac{\pi}{3} + 2\cdot \pi\cdot 2}{4} )} + i \sin{(\frac{\frac{\pi}{3} + 2\cdot \pi\cdot 2}{4})})$

= $\sqrt[4]{10}[cos(\frac{13\pi}{12})+i\sin(\frac{13\pi}{12})]$

• For k = 3, z = $\sqrt[4]{10}[cos(\frac{\frac{\pi}{3}+2\pi(3)}{4})+i\sin(\frac{\frac{\pi}{3}+2\pi(3)}{4})$

= $\sqrt[4]{10}[cos(\frac{19\pi}{12})+i\sin(\frac{19\pi}{12})]$

Thus, the four roots of 5(1 + i√3) are $\sqrt[4]{10}[cos(\frac{\pi}{12})+i\sin(\frac{\pi}{12})], \sqrt[4]{10}[cos(\frac{7\pi}{12})+i\sin(\frac{7\pi}{12})], \sqrt[4]{10}[cos(\frac{13\pi}{12})+i\sin(\frac{13\pi}{12})]$ and $\sqrt[4]{10}[cos(\frac{19\pi}{12})+i\sin(\frac{19\pi}{12})]$.

类似问题

问题 1：求 -2 – 2√3i 的立方根。

解决方案：

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

According to DeMoivre’s formula, all the nth roots of a complex number are given by:

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$, where k lies between 0 and n – 1 and n is the exponent or radical.

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

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

= 0.27 + 1.56i

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

= −1.49 − 0.54i

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

= 1.21 −1.02i      

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

问题 2：求 32 + 0i 的第五个根。

解决方案：

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

Argument = θ = tan-1(0/ 32) = 0.

According to DeMoivre’s formula, all the nth roots of a complex number are given by:

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$, where k lies between 0 and n – 1 and n is the exponent or radical. 

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

• For m = 0,  z = $[cos(\frac{2\pi(0)}{5})+i\ sin(\frac{2\pi(0)}{5})$ = 2
• For m = 1,  z = $[cos(\frac{2\pi(1)}{5})+i\ sin(\frac{2\pi(1)}{5})$ = 0.62 + 1.9i
• For m = 2,  z = $[cos(\frac{2\pi(2)}{5})+i\ sin(\frac{2\pi(2)}{5})$ = −1.62 + 1.18i
• For m = 3,  z = $[cos(\frac{2\pi(3)}{5})+i\ sin(\frac{2\pi(3)}{5})$  = −1.62 − 1.18i
• For m = 4,  z = $[cos(\frac{2\pi(4)}{5})+i\ sin(\frac{2\pi(4)}{5})$  = 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.            

问题 3：求 -8√3 + 8i 的四次根。

解决方案：

Polar form = $16[cos(\frac{5\pi}{6})+isin(\frac{5\pi}{6})]$

We have k = 2, n = 4 and θ = 5π/ 6.

According to DeMoivre’s formula, all the nth roots of a complex number are given by:

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$, where k lies between 0 and n – 1 and n is the exponent or radical. 

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

• For m = 0, z = $2[cos(\frac{\frac{5\pi}{6}+2\pi(\theta)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(0)}{4})$ = 1.58 + 1.21i.
• For m = 1, z = $2[cos(\frac{\frac{5\pi}{6}+2\pi(1)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(1)}{4})$ = −1.21 + 1.58i.
• For m = 2, z = $2[cos(\frac{\frac{5\pi}{6}+2\pi(2)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(2)}{4})$ = −1.58 −1.21i.
• For m = 3, z = $2[cos(\frac{\frac{5\pi}{6}+2\pi(3)}{4})+i\ sin(\frac{\frac{5\pi}{6}+2\pi(3)}{4})$ = 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.

问题 4：求 -27i 的第六个根。以三角函数形式离开。

解决方案：

Modulus = $\sqrt{x^2+y^2} = \sqrt{(-32)^2} = 27.$

Argument = θ = tan^-1(-27/ 0) = π/2.

Polar form = $27[cos(\frac{-π}{2})+isin(\frac{-π}{2})]$

According to DeMoivre’s formula, all the nth roots of a complex number are given by:

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$, where k lies between 0 and n – 1 and n is the exponent or radical. 

Find the 6 roots by substituting the values of m as 0, 1, 2, 3, 4 and 5.

• For m = 0,  z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2π(\theta)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(\theta)}{6})$

= $\sqrt{3}[cos(\frac{-\pi}{12})+i\ sin(\frac{-\pi}{12})]$

• For m = 1,  z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(1)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(1)}{6})$

= $\sqrt{3}[cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]$

• For m = 2,  z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(2)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(2)}{6})$

= $\sqrt{3}[cos(\frac{7\pi}{12})+i\ sin(\frac{7\pi}{12})]$

• For m = 3,  z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(3)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(3)}{6})$

= $\sqrt{3}[cos(\frac{11\pi}{12})+i\ sin(\frac{11\pi}{12})]$

• For m = 4,  z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(4)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(4)}{6})$

= $\sqrt{3}[cos(\frac{5\pi}{4})+i\ sin(\frac{5\pi}{4})]$

• For m = 5, z = $\sqrt[6]{27}[cos(\frac{\frac{-\pi}{2}+2\pi(5)}{6})+i\ sin(\frac{\frac{-\pi}{2}+2\pi(5)}{6})$

= $\sqrt{3}[cos(\frac{19\pi}{12})+i\ sin(\frac{19\pi}{12})]$

问题 5：求 81i 的第四个根。以三角函数形式离开。

解决方案：

Polar form = $81[cos(\frac{\pi}{2})+isin(\frac{\pi}{2})]$

We have k = 81, n = 4 and θ = π/ 2.

According to DeMoivre’s formula, all the nth roots of a complex number are given by:

$r^{\frac{1}{n}}[\cos{(\frac{\theta+2\pi k}{n})}+i\sin({\frac{\theta+2\pi k}{n})}]$, where k lies between 0 and n – 1 and n is the exponent or radical.

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

• For m = 0, z = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(\theta)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(\theta)}{4})$ = $3[cos(\frac{\pi}{8})+i\ sin(\frac{\pi}{8})]$
• For m = 1, z = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(1)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(1)}{4})$ = $3[cos(\frac{5\pi}{8})+i\ sin(\frac{5\pi}{8})]$
• For m = 2, z = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(2)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(2)}{4})$ = $3[cos(\frac{9\pi}{8})+i\ sin(\frac{9\pi}{8})]$
• For m = 3, z = $\sqrt[4]{81}[cos(\frac{\frac{\pi}{2}+2\pi(3)}{4})+i\ sin(\frac{\frac{\pi}{2}+2\pi(3)}{4}) = 3[cos(\frac{13\pi}{8})+i\ sin(\frac{13\pi}{8})]$