Complex numbers (Further Pure 2) · 复数(Further Pure 2)
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| de Moivre's theorem/də ˈmɔɪvəz ˈθɪərəm/ | 棣莫弗定理 | dì mò fú dìng lǐ |
| polar form/ˈpəʊlə fɔːm/ | 极坐标形式 | jí zuò biāo xíng shì |
| argument/ˈɑːɡjuːmənt/ | 辐角 | fú jiǎo |
| modulus/ˈmɒdjʊləs/ | 模 | mó |
| imaginary part/ɪˈmædʒɪnəri pɑːt/ | 虚部 | xū bù |
| roots of unity/ruːts ɒv ˈjuːnɪti/ | 单位根 | dān wèi gēn |
| unit circle/ˈjuːnɪt ˈsɜːkl/ | 单位圆 | dān wèi yuán |
Raising to a power, the easy way
- To work out $(\cos\theta + i\sin\theta)^{10}$ the long way, you'd multiply ten times.
- De Moivre's theorem 棣莫弗定理 does it in one line — and as a bonus, it splits a circle into equal pieces.
求幂的简单方法
- 要用长方法算出 $(\cos\theta + i\sin\theta)^{10}$,你得乘十次。
- 棣莫弗定理(De Moivre's theorem)一行就做到——而且作为奖励,它把一个圆分成相等的片。
De Moivre's theorem
- A complex number in polar form 极坐标形式: $z = r(\cos\theta + i\sin\theta)$.
- De Moivre's theorem — for any integer $n$:
- Raising to a power multiplies the argument 辐角 by $n$ (and the modulus 模 by $r^n$).
棣莫弗定理
- 极坐标形式(polar form)的一个复数:$z = r(\cos\theta + i\sin\theta)$。
- 棣莫弗定理——对任何整数 $n$:
- 求 $n$ 次幂把辐角乘以 $n$(并把模乘以 $r^n$)。
Complex numbers · 复数
z = x + yi
Plot z as a point: its distance from O is the modulus, its angle the argument. · 把 z 画成一个点:它到 O 的距离是模,它的角度是辐角。
De Moivre's theorem states (cos θ + i sin θ)ⁿ equals: · 棣莫弗定理陈述 (cos θ + i sin θ)ⁿ 等于:
Raising to the power n multiplies the argument by n: cos nθ + i sin nθ. · 求 n 次幂把辐角乘以 n:cos nθ + i sin nθ。
Expanding multiple angles
- Comparing real and imaginary parts 虚部 of $(\cos\theta + i\sin\theta)^n$ gives identities.
- Worked example: the real part of the cube gives $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.
展开多倍角
- 比较 $(\cos\theta + i\sin\theta)^n$ 的实部和虚部给出恒等式。
- 例题:立方的实部给出 $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$。
Using de Moivre, cos 3θ expands to: · 使用棣莫弗,cos 3θ 展开为:
Take the real part of (cosθ + i sinθ)³: cos 3θ = 4cos³θ − 3cosθ. · 取 (cosθ + i sinθ)³ 的实部:cos 3θ = 4cos³θ − 3cosθ。
The roots of unity 单位根
- The equation $z^n = 1$ has exactly $n$ solutions — the $n$th roots of unity.
- They lie on the unit circle 单位圆, equally spaced $\dfrac{360^\circ}{n}$ apart, starting at $1$.
The five solutions of $z^5 = 1$ sit on the unit circle, $72^\circ$ apart — a perfect pentagon.
Why equally spaced? Each root has modulus $1$ and arguments $\dfrac{360^\circ k}{n}$ for $k = 0, 1, \dots, n-1$ — so consecutive roots differ by the same angle.
单位根
- 方程 $z^n = 1$ 恰好有 $n$ 个解——$n$ 次单位根(roots of unity)。
- 它们位于单位圆(unit circle)上,等间距 $\dfrac{360^\circ}{n}$,从 $1$ 开始。
$z^5 = 1$ 的五个解坐落在单位圆上,相隔 $72^\circ$——一个完美的五边形。
为什么等间距? 每个根都有模 $1$ 和辐角 $\dfrac{360^\circ k}{n}$,其中 $k = 0, 1, \dots, n-1$——所以连续的根相差相同的角度。
How many distinct roots of unity does the equation z⁵ = 1 have? · 方程 z⁵ = 1 有多少个不同的单位根?
zⁿ = 1 has exactly n roots, equally spaced on the unit circle; here n = 5. · zⁿ = 1 恰好有 n 个根,等间距在单位圆上;这里 n = 5。
The fifth-roots of unity are equally spaced on the unit circle. How many degrees apart are consecutive roots? · 五次单位根等间距在单位圆上。连续的根相隔多少度?
360°/5 = 72° between consecutive roots. · 连续的根之间 360°/5 = 72°。
The n roots of unity are equally spaced around the unit circle. · n 个单位根等间距在单位圆周围。
They sit at equal angular gaps of 360°/n around the circle. · 它们以 360°/n 的相等角度间隔坐落在圆周围。
Common slip
De Moivre needs polar form first. $(1 + i)^8$ is not $1^8 + i^8$ — convert $1+i$ to $r(\cos\theta+i\sin\theta)$ with $r = \sqrt2$, $\theta = 45^\circ$, then apply the theorem.
- Summing a series of $\cos$/$\sin$ terms uses the complex sum $C + \mathrm{i}S$ of complex numbers.
常见的失误
棣莫弗需要先有极坐标形式。 $(1 + i)^8$ 不是 $1^8 + i^8$——先把 $1+i$ 转换成 $r(\cos\theta+i\sin\theta)$,其中 $r = \sqrt2$,$\theta = 45^\circ$,然后应用定理。
- 对 $\cos$/$\sin$ 级数求和时使用复数(complex numbers)的复合和 $C + \mathrm{i}S$。
Before applying de Moivre to (1 + i)⁸, you must first write 1 + i in ______ form. · 在把棣莫弗应用于 (1 + i)⁸ 之前,你必须先把 1 + i 写成______形式。
Convert to r(cosθ + i sinθ) first, then raise to the power. · 先转换成 r(cosθ + i sinθ),然后求幂。
You've got it
- de Moivre: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
- it expands multiple angles, e.g. $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$
- the $n$ roots of unity are equally spaced $\dfrac{360^\circ}{n}$ apart on the unit circle
你掌握了
- 棣莫弗:$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
- 它展开多倍角,例如 $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$
- $n$ 个单位根等间距 $\dfrac{360^\circ}{n}$ 在单位圆上