📚 Exercise 14A.2 | 练习 14A.2
Exercise 14A.2 is designed to reinforce core operations and representations of complex numbers, a fundamental topic in the IB Mathematics: Analysis and Approaches (AA) Higher Level course. In this set of problems, you will work with addition, subtraction, multiplication, division, complex conjugates, modulus and argument calculations, as well as conversions between Cartesian and polar forms. Mastery of these skills is essential for tackling more advanced concepts such as De Moivre's theorem, roots of complex numbers, and applications in polynomial equations.
练习 14A.2 旨在巩固复数的核心运算与表示方法，这是 IB 数学分析与方法（AA）高水平课程的一项基础内容。在这组问题中，你将练习复数的加减乘除、共轭、模与辐角的计算，以及直角坐标形式和极坐标形式之间的转换。熟练掌握这些技能，对于掌握棣莫弗定理、复数的根以及在多项式方程中的应用等更高级概念至关重要。
1. Basic Operations: Addition and Subtraction | 基本运算：加法与减法
d subtract the imaginary parts. For example, (3 + 2i) + (5 - 4i) = (3 + 5) + (2 - 4)i = 8 - 2i. This operation is straightforward and lays the foundation for all subsequent manipulations.
形如 z = a + bi 的复数（其中 a 为实部，b 为虚部）可以通过合并同类项进行加减运算。将两个复数相加时，只需将实部与实部相加，虚部与虚部相加。减法类似：实部相减，虚部相减。例如，(3 + 2i) + (5 - 4i) = (3 + 5) + (2 - 4)i = 8 - 2i。这种运算简单明了，为后续所有操作奠定了基础。
Always express the final answer in standard a + bi form, and be careful with signs when subtracting. A common mistake is to forget to distribute the negative sign to both the real and imaginary parts of the second complex number.
最终答案务必写成标准的 a + bi 形式，并在做减法时注意符号。常见错误是忘记将负号分配给第二个复数的实部和虚部。
2. Multiplication of Complex Numbers | 复数的乘法
+ 3 = 11, and combining imaginary parts yields -2i + 12i = 10i. Therefore, the product is 11 + 10i.
复数乘法需运用分配律（FOIL 方法），然后利用基本恒等式 i² = -1 进行化简。例如，(2 + 3i)(4 - i) = 2*4 + 2*(-i) + 3i*4 + 3i*(-i) = 8 - 2i + 12i - 3i²。由于 i² = -1，-3i² 这一项变为 -3(-1) = +3。实部相加得 8 + 3 = 11，合并虚部得 -2i + 12i = 10i。因此，乘积为 11 + 10i。
eal number. Always check for this simplification to avoid leaving i² in your final answer.
当纯虚数相乘时，例如 (3i)(-2i) = -6i² = 6，结果为实数。务必检查并完成此类化简，避免在最终答案中残留 i²。
3. Complex Conjugate and Division | 复共轭与除法
The complex conjugate of z = a + bi is denoted as z̄ and defined as a - bi. The product z * z̄ 3+4i))/((3-4i)(3+4i)) = (3+4i+6i+8i²)/(9+16) = (3+10i-8)/25 = (-5+10i)/25 = -1/5 + (2/5)i.
复数 z = a + bi 的共轭记为 z̄，定义为 a - bi。乘积 z * z̄ = a² + b² 始终为非负实数。这一性质是对复数进行除法运算的关键。计算 (1 + 2i) ÷ (3 - 4i) 时，分子和分母同乘以分母的共轭 3 + 4i。计算过程为： (3+4i))/((3-4i)(3+4i)) = (3+4i+6i+8i²)/(9+16) = (3+10i-8)/25 = (-5+10i)/25 = -1/5 + (2/5)i。
Division always produces a result in the standard form, provided you rationalise the denominator correctly. Remember that the denominator becomes the sum of squares, eliminating the imaginary part.
只要正确地将分母有理化，除法总会得到标准形式的结果。请记住，分母会变成平方和，从而消去虚部。
4. Modulus of a Complex Number | 复数的模
The modulus of a complex number z = a + bi, written as |z| om the origin to the point (a, b) in the complex plane. It is calculated using the formula |z| = √a² + b². For example, the modulus of 3 + 4i is |3+4i| = √3² + 4² = √9 + 16 = √25 lus is always a non-negative real number and forms the basis for the polar representation.
复数 z = a + bi 的模记作 |z|，表示复平面中原点到点 (a, b) 的距离。其计算公式为 |z| = √a² + b²。例如，3 + 4i 的模为 |3+4i| = √3² + 4² = √9 + 16 = √25 = 5。模始终为非负实数，并构成极坐标表示的基础。
ute value of the real part. Similarly, for a purely imaginary number, it is the absolute value of the imaginary coefficient.
当复数为纯实数（b=0）时，其模就是实部的绝对值。类似地，对于纯虚数，模就是虚部系数的绝对值。
5. Argument and Principal Argument | 辐角与主辐角
n. For a complex number in the first quadrant, θ = arctan(b/a). However, you must adjust for the correct quadrant. For z = -1 + √3 i, a = -1, b = √3. The reference angle is arctan(|√3/-1|) = π/3. Since z is in the second quadrant, the argument is π - π/3 = 2π/3.
复数的辐角，记作 arg(z)，是从原点到点 (a,b) 的连线与正实轴之间形成的角度 θ，以弧度为单位。主辐角通常记作 Arg(z)，根据惯例一般在 (-π, π] 或 [0, 2π) 范围内取值。对于第一象限的复数，θ = arctan(b/a)。但必须根据正确象限进行调整。对于 z = -1 + √3 i，a = -1，b = √3，参考角为 arctan(|√3/-1|) = π/3。由于 z 位于第二象限，辐角为 π - π/3 = 2π/3。
Always sketch the complex number on the Argand diagram to confirm the quadrant. Mistakes in determining the argument often arise from ignoring the signs of a and b.
始终在阿尔冈图上画出复数以确认象限。确定辐角时的错误往往源于忽略了 a 和 b 的符号。
6. Polar Form: r(cosθ + i sinθ) | 极坐标形式
Any complex number can be expressed in polar form as z = r (cos θ + i sin θ), where r = |z| argument. This representation is also written as r cis θ. Using the previous example, -1 + √3 i has modulus 2 and argument 2π/3, so its polar form is 2(cos(2π/3) + i sin(2π/3)). Polar form is particularly useful for multiplication, division, and exponentiation.
任何复数都可以表示为极坐标形式 z = r (cos θ + i sin θ)，其中 r = |z| 为模，θ = arg(z) 为辐角。这种表示法也写作 r cis θ。使用前面的例子，-1 + √3 i 的模为 2，辐角为 2π/3，因此其极坐标形式为 2(cos(2π/3) + i sin(2π/3))。极坐标形式在乘法、除法和乘方运算中尤为有用。
ing the trigonometric functions and distributing: 2(cos(2π/3) + i sin(2π/3)) = 2(-1/2 + i(√3/2)) = -1 + √3 i, returning to the original a+bi format.
从极坐标形式转换回直角坐标形式需要计算三角函数值并展开：2(cos(2π/3) + i sin(2π/3)) = 2(-1/2 + i(√3/2)) = -1 + √3 i，还原为原始的 a+bi 格式。
7. Multiplication and Division in Polar Form | 极坐标形式的乘除法
example, multiplying 2 cis (π/4) and 3 cis (π/6) gives 6 cis (π/4 + π/6) = 6 cis (5π/12).
当复数以极坐标形式表示时，乘法和除法变得极为简便。若 z₁ = r₁ cis θ₁、z₂ = r₂ cis θ₂，则乘积为 z₁z₂ = r₁r₂ cis (θ₁ + θ₂)。商为 z₁/z₂ = (r₁/r₂) cis (θ₁ - θ₂)（前提是 r₂ ≠ 0）。例如，将 2 cis (π/4) 与 3 cis (π/6) 相乘得 6 cis (π/4 + π/6) = 6 cis (5π/12)。
These formulas highlight why polar form is so powerful: multiplication scales the moduli and adds the arguments, while division scales and subtracts. This is particularly useful for repeated multiplication and for finding powers and roots.
这些公式凸显了极坐标形式的强大之处：乘法使模相乘、辐角相加，除法则使模相除、辐角相减。这对于重复乘法和求幂求根尤为有用。
8. De Moivre's Theorem | 棣莫弗定理
De Moivre's theorem extends the multiplication principle to powers: for any integer n, (r cis θ)n = rn cis (nθ). Equivalently, [r(cos θ + i sin θ)]n = rn (cos nθ + i sin nθ). This theorem provides a direct method for computing powers of complex numbers without tedious algebraic expansion. For instance, to compute (1 + i)4, first convert 1 + i to polar form: r = √1²+1² = √2, θ = π/4. Thus (√2 cis π/4)4 = (√2)4 cis (4·π/4) = 4 cis π = 4(cos π + i sin π) = -4.
棣莫弗定理将乘法原理推广至乘方：对任意整数 n，(r cis θ)n = rn cis (nθ)。亦即，[r(cos θ + i sin θ)]n = rn (cos nθ + i sin nθ)。该定理提供了一种直接计算复数乘方的方法，避免了繁琐的代数展开。例如，计算 (1 + i)4，首先将 1 + i 转换为极坐标形式：r = √1²+1² = √2，θ = π/4。于是 (√2 cis π/4)4 = (√2)4 cis (4·π/4) = 4 cis π = 4(cos π + i sin π) = -4。
The theorem is also valid for negative integers, and forms the foundation for finding roots. Always reduce the argument after multiplication to a principal value if required, by adding or subtracting multiples of 2π.
该定理对负整数同样有效，并为求复数根奠定了基础。必要时应通过加减 2π 的整数倍将乘方后的辐角化为主值。
9. Roots of Complex Numbers | 复数的根
To find the n distinct nth roots of a complex number w = r cis θ, we use the formula: zk = ⁿ√r al root corresponds to k = 0, and the roots are equally spaced around a circle of radius ⁿ√r. For example, to find the cube roots of -8, first write -8 in polar form: 8 cis π (since modulus is 8, argument π). The cube roots are zk = ³√8 cis ((π + 2πk)/3) = 2 cis (π/3 + 2πk/3). For k=0: 2 cis π/3 = 1 + √3 i; k=1: 2 cis π = -2; k=2: 2 cis 5π/3 = 1 - √3 i.
为求复数 w = r cis θ 的 n 个互异的 n 次方根，我们使用公式：zk = ⁿ√r cis ((θ + 2πk)/n)，其中 k = 0, 1, ..., n-1。主根对应于 k = 0，所有根均匀分布在以 ⁿ√r 为半径的圆上。例如，求 -8 的立方根，首先将 -8 写成极坐标形式：8 cis π（模为 8，辐角为 π）。立方根为 zk = ³√8 cis ((π + 2πk)/3) = 2 cis (π/3 + 2πk/3)。当 k=0 时：2 cis π/3 = 1 + √3 i；k=1 时：2 cis π = -2；k=2 时：2 cis 5π/3 = 1 - √3 i。
gle on the complex plane. This geometric interpretation often serves as a quick check for algebraic mistakes.
这些根始终展现出对称性。当 n=3 时，它们在复平面上构成一个等边三角形。这一几何意义常可作为快速检查代数错误的手段。
10. Solving Polynomial Equations | 解多项式方程
rewritten as x² = -4, leading to x = ±2i. More generally, the quadratic formula x = [-b ± √b² - 4ac]/(2a) yields complex roots when the discriminant is negative. In Exercise 14A.2, you may encounter equations requiring you to express solutions in a + bi form.
复数使我们能够求解没有实数解的多项式方程。代数基本定理保证，n 次多项式恰有 n 个复数根（计重数）。例如，方程 x² + 4 = 0 可改写为 x² = -4，从而 x = ±2i。更一般地，当判别式为负时，二次公式 x = [-b ± √b² - 4ac]/(2a) 会产生复数根。在练习 14A.2 中，你可能会遇到要求将解表示为 a + bi 形式的方程。
For higher-degree polynomials, factorisation and the use of De Moivre' m might ask: Solve z³ - 8 = 0, which we now know yields the three cube roots of 8: 2, -1 + √3 i, and -1 - √3 i.
对于更高次的多项式，因式分解和棣莫弗定理的应用成为必要工具。一个典型的问题是：解 z³ - 8 = 0，我们现在知道它会产生 8 的三个立方根：2、-1 + √3 i 和 -1 - √3 i。
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