Complex Numbers and Polynomials | 复数与多项式

📚 Complex Numbers and Polynomials | 复数与多项式

ional in many areas of mathematics, but particularly essential for understanding polynomials. In IB Mathematics, we explore the algebra and geometry of complex numbers and their intimate relationship with polynomial equations, including the Fundamental Theorem of Algebra.

复数将实数系统扩展，使得像 x² + 1 = 0 这样的方程有了解。它们是数学许多领域的基础，但对于理解多项式尤为关键。在IB数学中，我们探究复数的代数和几何性质，以及它们与多项式方程的紧密联系，包括代数基本定理。

---

1. Introduction to Complex Numbers | 复数简介

z can be written in Cartesian form as z = a + bi, where a and b are real numbers, and i = √–1 e real part is Re(z) = a, and the imaginary part is Im(z) = b. If b = 0, z is purely real.

复数 z 可以写成直角坐标形式 z = a + bi，其中 a 和 b 是实数，i = √–1。实部为 Re(z) = a，虚部为 Im(z) = b。若 b = 0，则 z 为实数。

The imaginary unit i satisfies i² = –1, i³ = – and higher powers cycle with period 4. This cyclic behaviour simplifies many calculations.

虚数单位 i 满足 i² = –1，i³ = –i，i⁴ = 1，更高次幂以4为周期循环。这一循环规律简化了许多计算。

qual if and only if their real and imaginary parts are respectively equal: a + bi = c + di ⇔ a = c and b = d.

我们称两个复数相等，当且仅当它们的实部和虚部分别相等：a + bi = c + di ⇔ a = c 且 b = d。

---

2. Operations with Complex Numbers | 复数的四则运算

wise: (a + bi) ± (c + di) = (a ± c) + (b ± d)i. This mirrors vector addition in the plane.

加法和减法按分量进行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。这与平面上的向量加法相对应。

Multiplication uses the distributive law and the identity i² = –1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i. No special rule is required beyond expanding brackets.

乘法运用分配律及 i² = –1：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i。除了展开括号外没有特殊规则。

Division involves the complex conjugate. The conjugate of z = a + bi is z* = a – bi. To compute (a+bi)/(c+di), multiply numerator and denominator by c – di, giving ((a+bi)(c–di))/(c²+d²) = (ac+bd)/(c²+d²) + (bc–ad)/(c²+d²) i.

除法需用到共轭复数。z = a + bi 的共轭为 z* = a – bi。计算 (a+bi)/(c+di) 时，分子分母同乘 c – di，得到 ((a+bi)(c–di))/(c²+d²) = (ac+bd)/(c²+d²) + (bc–ad)/(c²+d²) i。

₂*, and (z₁/z₂)* = z₁*/z₂*. Moreover, z z* = a² + b² is always a non-negative real number.

共轭的常用性质：(z*)* = z，(z₁ ± z₂)* = z₁* ± z₂*，(z₁z₂)* = z₁* z₂*，(z₁/z₂)* = z₁*/z₂*。此外，z z* = a² + b² 总是非负实数。

---

广告 / Sponsored

3. The Complex Plane: Modulus and Argument | 复平面：模与辐角

Every complex number a + bi can be represented as a point (a, b) on the Argand diagram, with the x-axis as the real axis and the y-axis as the imaginary axis.

每个复数 a + bi 可以在 Argand 图上表示为点 (a, b)，其中 x 轴为实轴，y 轴为虚轴。

The modulus |z| = √a² + b² is the distance from the origin to the point. It satisfies |z₁z₂| = |z₁| |z₂| and |z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0).

模长 |z| = √a² + b² 是原点到该点的距离，满足 |z₁z₂| = |z₁| |z₂| 和 |z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0)。

, with appropriate quadrant adjustment. The principal argument is usually taken in the interval (–π, π].

辐角，记作 arg(z) = θ，是正实轴与原点至 (a,b) 线段之间的角度，满足 tan θ = b/a，并根据象限调整。主辐角通常取区间 (–π, π]。

The pair (r, θ) where r = |z| enabling a geometric interpretation of multiplication: multiply moduli and add arguments.

由 r = |z| 和 θ = arg(z) 组成的对 (r, θ) 给出了复数的极坐标表示，使得乘法具有几何解释：模长相乘，辐角相加。

---

4. Polar Form and Euler's Formula | 极坐标形式与欧拉公式

sin θ), often abbreviated as r cis θ. This representation is unique up to adding 2πk to θ.

利用 r 和 θ，我们写出极坐标（或三角）形式：z = r (cos θ + i sin θ)，常简写为 r cis θ。这种表示在相差 2πk 的意义下唯一。

Euler's formula e^iθ one of the most remarkable results in mathematics. It condenses the polar form to z = r e^iθ, which greatly streamlines multiplication, division, and exponentiation.

欧拉公式 e^iθ = cos θ + i sin θ 是数学中最美妙的结果之一。它将极坐标形式浓缩为 z = r e^iθ，极大简化了乘法、除法和乘方运算。

In exponential form, z₁ z₂ = r₁ r₂ e^(i(θ₁+θ₂)) and z₁ / z₂ = (r₁/r₂) e^(i(θ₁–θ₂)). For a power, zⁿ = rⁿ e^inθ, as long as we interpret the exponent carefully.

在指数形式下，z₁ z₂ = r₁ r₂ e^(i(θ₁+θ₂))，z₁ / z₂ = (r₁/r₂) e^(i(θ₁–θ₂))。对于幂次，zⁿ = rⁿ e^inθ，只要我们恰当理解指数。

---

5. De Moivre's Theorem | 棣莫弗定理

De Moivre' any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). This directly follows from Euler's formula and the index laws, and can also be proven by induction.

棣莫弗定理指出，对任意整数 n，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。这可由欧拉公式和指数律直接得出，也可用归纳法证明。

The theorem is especially powerful for deriving multiple-angle identities, such as expressing cos 3θ in terms of cos θ, and for computing high powers of complex numbers without expanding binomials.

该定理对推导多倍角恒等式（例如将 cos 3θ 用 cos θ 表达）和计算复数的高次幂非常强大，无需进行二项式展开。

To find the n distinct n-th roots of a complex number w = ρ e^iφ, we solve zⁿ = w. The roots are given by z_k = ρ^(1/n) e^(i(φ + 2πk)/n), for k = 0, 1, ..., n–1. Geometrically, they lie at the vertices of a regular n-gon centred at the origin.

要求复数 w = ρ e^iφ 的 n 个不同 n 次方根，我们解 zⁿ = w。根由 z_k = ρ^(

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）