求 w = 8 (cos 150° + i sin 150°) 的所有复立方根

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。

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。

Given: $z^{\frac{1}{n}} = r^{\frac{1}{n}}(cosθ + i\ sinθ)$, then its roots are:

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

Where, 

k lies between 0 and n – 1 and n is the exponent or radical. 

求 w = 8 (cos 150° + i sin 150°) 的所有复立方根

解决方案：

w = 8(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 8(cos(150° + 360n) + i sin(150° + 360n).

$\sqrt[3]{w} = \sqrt[3]{8[(cos(150° + 360n) + i sin(150° + 360n)]}$

w^1/3 = {8[(cos(150° + 360n) + i sin(150° + 360n)]}^1/3

= 2^(3)(1/3) [(cos(150° + 360n) + i sin(150° + 360n)]^1/3

As per DeMoivre’s Theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx). 

= $2[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}$

Substitute n = 0, 1, 2 to find the roots.

• For n = 0, w₁ = $2[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}$ = 2(cos 50° + i sin 50°)
• For n = 1, w₂ = $2[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}$ = 2(cos 170° + i sin 170°)
• For n = 2, w₃ = $2[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}$ = 2(cos 290° + i sin 290°)

类似问题

问题 1. 求 w = 125(Cos 150° + i sin 150°) 的所有复立方根。用 theta 以度为单位写出极坐标形式的根。

解决方案：

w = 125(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 125 (cos(150° + 360n) + i sin(150° + 360n).

$\sqrt[3]{w} = \sqrt[3]{125[(cos(150° + 360n) + i sin(150° + 360n)]}$

w^1/3 = ${125[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}$

= 5^(3)(1/3) $[(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}$

As per DeMoivre’s Theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx). 

= $5[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}$

Substitute n = 0,1,2 to find the roots.

• For n = 0, w₁ = $5[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}$ = 2(cos 50° + i sin 50°)
• For n = 1, w₂ = $5[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}$ = 2(cos 170° + i sin 170°)
• For n = 2, w₃ = $5[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}$ = 2(cos 290° + i sin 290°)

问题 2. 求 w = 27(Cos 150° + i sin 150°) 的所有复立方根。用 theta 以度为单位写出极坐标形式的根。

解决方案：

w = 27(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 27(cos(150° + 360n) + i sin(150° + 360n).

$\sqrt[3]{w} = \sqrt[3]{27[(cos(150° + 360n) + i sin(150° + 360n)]}$

w^1/3 = ${27[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}$

= 3^(3)(1/3) $[(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}$

As per DeMoivre’s Theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx). 

= $3[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}$

Substitute n = 0,1,2 to find the roots.

• For n = 0, w₁ = $3[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}$ = 2(cos 50° + i sin 50°)
• For n = 1, w₂ = $3[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}$ = 3(cos 170° + i sin 170°)
• For n = 2, w₃ = $3[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}$ = 3(cos 290° + i sin 290°)

问题 3. 求 w = 64 (Cos 150° + i sin 150°) 的所有复立方根。用 theta 以度为单位写出极坐标形式的根。

解决方案：

w = 64(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 64(cos(150° + 360n) + i sin(150° + 360n).

$\sqrt[3]{w} = \sqrt[3]{64[(cos(150° + 360n) + i sin(150° + 360n)]}$

w^1/3 = ${64[(cos(150° + 360n) + i sin(150° + 360n)]}1/3$

= 4^(3)(1/3) $[(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}$

As per DeMoivre’s Theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx). 

= $4[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}$

Substitute n = 0,1,2 to find the roots.

• For n = 0, w1 = $4[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}$ = 4(cos 50° + i sin 50°)
• For n = 1, w2 = $4[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}$ = 4(cos 170° + i sin 170°)
• For n = 2, w3 = $4[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}$ = 4(cos 290° + i sin 290°)

问题 4. 求 w = 343 (Cos 150° + i sin 150°) 的所有复立方根。用 theta 以度为单位写出极坐标形式的根。

解决方案：

w = 343(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 343(cos(150° + 360n) + i sin(150° + 360n).

$\sqrt[3]{w} = \sqrt[3]{343[(cos(150° + 360n) + i sin(150° + 360n)]}$

w^1/3 = ${343[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}$

= 7^(3)(1/3) $[(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}$

As per DeMoivre’s Theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx). 

= $7[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}$

Substitute n = 0,1,2 to find the roots.

• For n = 0, w1 = $7[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}$= 7(cos 50° + i sin 50°)
• For n = 1, w2 = $7[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3}$ = 7(cos 170° + i sin 170°)
• For n = 2, w3 = $7[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}$ = 7(cos 290° + i sin 290°)

问题 5. 求 w = 729 (Cos 150° + i sin 150°) 的所有复立方根。用 theta 以度为单位写出极坐标形式的根。

解决方案：

w = 729(Cos 150° + i sin 150°)

The above complex number can also be expressed as w = 729(cos(150° + 360n) + i sin(150° + 360n).

$\sqrt[3]{w} = \sqrt[3]{729[(cos(150° + 360n) + i sin(150° + 360n)]}$

w^1/3 = ${729[(cos(150° + 360n) + i sin(150° + 360n)]}^{1/3}$

= 9^(3)(1/3)$[(cos(150° + 360n) + i sin(150° + 360n)]^{1/3}$

As per DeMoivre’s Theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx). 

= $9[(cos(\frac{150°+360n}{3}) + i sin(\frac{150°+360n}{3})]^{1/3}$

Substitute n = 0,1,2 to find the roots.

• For n = 0, w1 = $9[(cos(\frac{150°+360(0)}{3}) + i sin(\frac{150°+360(0)}{3})]^{1/3}$= 9(cos 50° + i sin 50°)
• For n = 1, w2 = $9[(cos(\frac{150°+360(1)}{3}) + i sin(\frac{150°+360(1)}{3})]^{1/3 }$ = 9(cos 170° + i sin 170°)
• For n = 2, w3 = $9[(cos(\frac{150°+360(2)}{3}) + i sin(\frac{150°+360(2)}{3})]^{1/3}$ = 9(cos 290° + i sin 290°)