📚 Working with Complex Numbers | 复数运算
the standard form z = a + bi, where a and b are real numbers and i is the imaginary unit.
复数通过引入虚数单位 i 扩展了实数系，i 满足 i² = -1。每一个复数都可以写成标准形式 z = a + bi，其中 a 和 b 是实数，i 是虚数单位。
i² = -1, and z = a + bi
is called the imaginary part Im(z). If b = 0, z is real; if a = 0, z is purely imaginary.
这里 a 称为实部 Re(z)，b 称为虚部 Im(z)。如果 b = 0，z 是实数；如果 a = 0，z 是纯虚数。
1. The Imaginary Unit and Complex Form | 虚数单位与复数形式
l and their imaginary parts are equal. This means a + bi = c + di implies a = c and b = d.
两个复数相等当且仅当它们的实部相等且虚部相等。也就是说，a + bi = c + di 意味着 a = c 且 b = d。
Complex numbers are closed under addition, subtraction, multiplication and division, so you can perform algebraic operations in the usual way while replacing i² by -1.
复数对加、减、乘、除运算封闭，因此可以按通常的代数方式运算，只需把 i² 替换为 -1。
2. Addition and Subtraction | 加法与减法
real parts and the imaginary parts separately. For example, (3 + 2i) + (1 - 4i) = 4 - 2i.
复数的加法或减法，只需分别合并实部和虚部。例如，(3 + 2i) + (1 - 4i) = 4 - 2i。
(a + bi) ± (c + di) = (a ± c) + (b ± d)i
(5 + 6i) - (2 + 3i) = 3 + 3i. Always simplify the final answer into standard form a + bi.
减法同理：(5 + 6i) - (2 + 3i) = 3 + 3i。最终答案一定要化简为 a + bi 的标准形式。
3. Multiplication and Complex Conjugates | 乘法与共轭复数
brackets and using i² = -1. For example, (2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i² = 14 - 5i.
复数相乘时先展开括号，再利用 i² = -1 化简。例如，(2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i² = 14 - 5i。
(a + bi)(c + di) = (ac - bd) + (ad + bc)i
s z* = a - bi. Multiplying z by z* always gives a non-negative real number: zz* = a² + b².
复数 z = a + bi 的共轭复数是 z* = a - bi。z 与其共轭相乘总是得到非负实数：zz* = a² + b²。
4. Division Using the Conjugate | 利用共轭复数做除法
To divide by a complex number, multiply both numerator and denominator by the conjugate of the denominator. This makes the denominator real.
复数相除时，在分子和分母上同时乘以分母的共轭复数，使分母变为实数。
(a + bi)/(c + di) = [(a + bi)(c - di)]/(c² + d²)
1 - i) = [(1 + i)(1 + i)]/(1² + 1²) = (1 + 2i + i²)/2 = i. Write answers in standard form.
例如，(1 + i)/(1 - i) = [(1 + i)(1 + i)]/(1² + 1²) = (1 + 2i + i²)/2 = i。答案要写为标准形式。
5. Solving Quadratic Equations | 解二次方程
discriminant, its roots are complex. For example, x² + 4x + 13 = 0 has roots x = -2 ± 3i.
当二次方程的判别式为负时，它的根是复数。例如，x² + 4x + 13 = 0 的根为 x = -2 ± 3i。
x = (-b ± √b² - 4ac)/(2a)
For a polynomial with real coefficients, complex roots always occur in conjugate pairs. So if z is a root, z* is also a root.
对于实系数多项式，复数根总是成共轭对出现。因此，如果 z 是根，那么 z* 也是根。
6. The Argand Diagram and Modulus | 阿尔冈图与模
as the point (a, b) in a plane with a horizontal real axis and a vertical imaginary axis.
阿尔冈图将复数 z = a + bi 表示为平面上的点 (a, b)，其中横轴是实轴，纵轴是虚轴。
|z| = √a² + b²
The modulus |z| from the point to the origin. The conjugate z* is the reflection of z in the real axis, so |z*| = |z|.
模 |z| 是该点到原点的距离。共轭复数 z* 是 z 关于实轴的镜像，因此 |z*| = |z|。
7. Argument of a Complex Number | 复数的辐角
al axis to the line joining z to the origin. The principal argument is usually -π < θ ≤ π.
复数 z 的辐角 arg z 是从正实轴到连接 z 与原点的线段的夹角 θ。主辐角通常取 -π < θ ≤ π。
tan θ = b/a, with quadrant check
an alone does not determine the angle. For example, z = -1 + i has arg z = 3π/4, not -π/4.
一定要检查点 (a, b) 所在的象限，因为仅靠 tan 无法确定角度。例如，z = -1 + i 的辐角是 3π/4，而不是 -π/4。
8. Modulus-Argument Form | 模-辐角形式
A complex number can be written in modulus-argument form z = r(cos θ + i sin θ), where r = |z| and θ = arg z. This is also called polar form.
复数可以写成模-辐角形式 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg z。这也称为极形式。
z = r(cos θ + i sin θ)
Multiplication in polar form says: multiply the moduli and add the arguments. Division says: divide the moduli and subtract the arguments.
极形式下的乘法规则是：模相乘，辐角相加。除法规则是：模相除，辐角相减。
z₁z₂ = r₁r₂(cos(θ₁ + θ₂) + i sin(θ₁ + θ₂))
z₁/z₂ = r₁/r₂(cos(θ₁ - θ₂) + i sin(θ₁ - θ₂))
9. De Moivre's Theorem and Powers | 棣莫弗定理与幂
De Moivre' + i sin θ)ⁿ = cos nθ + i sin nθ. This makes finding powers of complex numbers much easier.
棣莫弗定理指出，对于任意整数 n，(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。这使得求复数的幂变得非常容易。
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
To find a power, first write z in polar form. For example, (1 + i)⁴ = [(√2)(cos π/4 + i sin π/4)]⁴ = 4(cos π + i sin π) = -4.
求幂时先把 z 写成极形式。例如，(1 + i)⁴ = [(√2)(cos π/4 + i sin π/4)]⁴ = 4(cos π + i sin π) = -4。
10. Roots of Complex Numbers | 复数的根
e formula below for k = 0, 1, 2, ..., n - 1. The roots are equally spaced around a circle.
求 z = r(cos θ + i sin θ) 的 n 次方根，可用下面公式，k 取 0, 1, 2, ..., n - 1。这些根均匀分布在圆周上。
wₖ = r1/n[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]
For example, the cube roots of 8 are 2, -1 + i√3 and -1 - i√3. Notice that all three roots have modulus 2 and arguments 0, 2π/3 and 4π/3.
例如，8 的三个立方根是 2、-1 + i√3 和 -1 - i√3。注意三个根的模都是 2，辐角分别为 0、2π/3 和 4π/3。
11. Exam Tips and Common Mistakes | 考试技巧与常见错误
Always write final answers in the form a + bi unless the question asks for polar form. When dividing, do not forget to multiply the numerator by the conjugate as well.
除非题目要求极形式，否则最终答案务必写成 a + bi 的形式。做除法时，不要忘记分子也要同时乘以共轭复数。
for the argument, and losing conjugate pairing when solving polynomial equations. Check each step carefully.
常见错误包括忘记 i² = -1、辐角象限判断错误，以及在解多项式方程时忽略共轭成对。每一步都要仔细检查。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图