📚 Expressing Complex Numbers in the Form x + iy | 将复数表示为 x + iy 形式
AQA A-Level Maths often asks you to give an answer "in the form x + iy", where x and y are real numbers. This phrasing is a reminder that a complex number has two independent components: a real part x and an imaginary part y. Once you can confidently add, subtract, multiply and divide complex numbers, writing an expression in the form x + iy becomes a routine but high-mark skill.
AQA A-Level 数学经常要求将答案写成 x + iy 的形式，其中 x 和 y 是实数。这一表述提醒我们，复数有两个独立组成部分：实部 x 和虚部 y。只要熟练掌握复数的加、减、乘、除，将表达式写成 x + iy 的形式就会成为一项常规但高分值的技能。
1. The Standard Form x + iy | 标准形式 x + iy
x + iy, where x is the real part Re(z) and y is the imaginary part Im(z). Both x and y must be real numbers, so an answer like 2 + 3i is in the correct form, while 2 + i√3 also counts because it can be written as 2 + (√3)i.
虚数单位 i 由 i² = -1 定义。复数 z 写作 z = x + iy，其中 x 是实部 Re(z)，y 是虚部 Im(z)。x 和 y 都必须是实数，因此像 2 + 3i 这样的答案形式正确，而 2 + i√3 也算正确，因为它可以写成 2 + (√3)i。
z = x + iy, x = Re(z), y = Im(z)
In an exam answer, the coefficient of i should be written as a single real number or a simplified fraction. For example, 3 + i√7 is acceptable, but 3 + √-7 is not yet in the required form.
在考试答案中，i 的系数应写成一个实数或最简分数。例如，3 + i√7 是可以接受的，但 3 + √-7 还不符合要求。
2. Adding and Subtracting Complex Numbers | 复数的加法与减法
Addition and subtraction are straightforward: collect real terms and imaginary terms separately. This works because the real axis and imaginary axis are independent directions.
加法和减法非常简单：分别合并实部和虚部。这是因为实轴和虚轴是相互独立的方向。
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
6 − 2i. Subtracting needs care with signs: (4 + 3i) − (2 − 5i) = 4 + 3i − 2 + 5i = 2 + 8i.
例如，(4 + 3i) + (2 − 5i) = (4 + 2) + (3 − 5)i = 6 − 2i。减法需要注意符号：(4 + 3i) − (2 − 5i) = 4 + 3i − 2 + 5i = 2 + 8i。
When the expression contains brackets, remove them first and then group the real and imaginary parts. Never try to combine a real term with an imaginary term into one number.
当表达式中含有括号时，先去括号，再分别合并实部和虚部。切勿尝试把实部项和虚部项合并为一个数。
3. Multiplying Complex Numbers | 复数的乘法
Multiplication uses the normal expansion of brackets, followed by replacing i² with −1. This is the most common source of mark-winning simplification in AQA papers.
复数乘法使用通常的括号展开方法，然后将 i² 替换为 −1。这是 AQA 试卷中最常见的得分化简步骤。
(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i
(3 + 2i)(1 − 4i) = 3 − 12i + 2i − 8i². Since i² = −1, this becomes 3 − 10i + 8 = 11 − 10i.
例如，(3 + 2i)(1 − 4i) = 3 − 12i + 2i − 8i²。因为 i² = −1，所以变为 3 − 10i + 8 = 11 − 10i。
If you obtain an i² term, always convert it before giving your final real and imaginary parts. Leaving −8i² in the answer would not be in the form x + iy.
如果得到 i² 项，一定要先转换，再给出最终的实部和虚部。把 −8i² 留在答案中就不符合 x + iy 的形式要求。
4. Dividing by a Complex Number | 除以复数
Division is not possible while the denominator is complex, so you multiply the numerator and denominator by the complex conjugate of the denominator. This makes the denominator real.
当分母为复数时无法直接进行除法，因此需要将分子和分母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)] / [(c + di)(c − di)]
= [(ac + bd) + (bc − ad)i] / (c² + d²)
the form x + iy. Multiply top and bottom by 2 − i: 5(2 − i)/(4 + 1) = (10 − 5i)/5 = 2 − i.
例如，将 5/(2 + i) 表示成 x + iy 的形式。分子分母同时乘以 2 − i：5(2 − i)/(4 + 1) = (10 − 5i)/5 = 2 − i。
(6 + 3i)/3 = 2 + i, but (6 + 3i)/3i is not finished because i remains in the denominator.
最后一步应把分式拆成实部和虚部。例如，(6 + 3i)/3 = 2 + i，但 (6 + 3i)/3i 还没有完成，因为 i 仍留在分母中。
5. Powers of i and Simplification | i 的幂与化简
i, i² = −1, i³ = −i, i⁴ = 1. After i⁴ the pattern repeats, so i⁵ = i, i⁶ = −1, and so on.
i 的幂以四为周期循环：i¹ = i，i² = −1，i³ = −i，i⁴ = 1。i⁴ 之后规律重复，因此 i⁵ = i，i⁶ = −1，依此类推。
i¹ = i, i² = −1, i³ = −i, i⁴ = 1
= −1. This helps simplify expressions such as 2i³ − 3i² + 5 = 2(−i) − 3(−1) + 5 = 8 − 2i.
对于较高的幂，如 i¹⁰，将指数除以 4 并利用余数：10 = 4 × 2 + 2，所以 i¹⁰ = i² = −1。这有助于化简诸如 2i³ − 3i² + 5 = 2(−i) − 3(−1) + 5 = 8 − 2i 的表达式。
Always reduce any power of i to one of i, −1, −i or 1 before identifying x and y.
在确定 x 和 y 之前，一定要把 i 的任意次幂化简为 i、−1、−i 或 1 之一。
6. Dealing with Square Roots of Negative Numbers | 负数的平方根处理
A negative square root must be converted into an imaginary term before you can write the answer in the form x + iy. The key identity is √−a = i√a for a > 0.
负数的平方根必须先转换为虚数项，才能将答案写成 x + iy 的形式。关键恒等式是：当 a > 0 时，√−a = i√a。
√−9 = 3i, √−4 = 2i, √−a = i√a
For example, express 2 + √−4 in the form x + iy. Since √−4 = 2i, the expression becomes 2 + 2i, so x = 2 and y = 2.
例如，将 2 + √−4 表示成 x + iy 的形式。因为 √−4 = 2i，所以表达式变为 2 + 2i，因此 x = 2，y = 2。
Be careful with products such as √−2 × √−3. You must convert each root to i form first: i√2 × i√3 = i²√6 = −√6. Using √−2(−3) = √6 directly is a common error.
处理 √−2 × √−3 这样的乘积时要小心。必须先把每个根式转换为 i 的形式：i√2 × i√3 = i²√6 = −√6。直接写成 √−2(−3) = √6 是常见错误。
7. Quadratic Equations and Complex Roots | 二次方程与复数根
he discriminant b² − 4ac is negative, the roots are complex conjugates of the form x ± iy.
对于二次方程 ax² + bx + c = 0，如果判别式 b² − 4ac 为负，那么根是形如 x ± iy 的共轭复数。
x = [−b ± i√4ac − b²] / (2a) when b² − 4ac < 0
For example, solve x² + 2x + 5 = 0. The discriminant is 4 − 20 = −16, so x = [−2 ± √−16]/2 = [−2 ± 4i]/2 = −1 ± 2i.
例如，解方程 x² + 2x + 5 = 0。判别式为 4 − 20 = −16，所以 x = [−2 ± √−16]/2 = [−2 ± 4i]/2 = −1 ± 2i。
ith x = −1 and y = 2 or y = −2. This is often tested alongside sums and products of roots.
两个根为 −1 + 2i 和 −1 − 2i，都写成 x + iy 的形式，其中 x = −1，y = 2 或 y = −2。这一考点常与根的和与积一起考查。
8. Modulus and Argument Support | 模与辐角辅助
ked to write it as x + iy. Use z = r(cosθ + i sinθ), then expand to get r cosθ + i r sinθ.
有时复数以模-辐角形式给出，并需要写成 x + iy。使用 z = r(cosθ + i sinθ)，展开得到 r cosθ + i r sinθ。
z = r(cosθ + i sinθ) = r cosθ + i r sinθ
xample, write 6(cos π/3 + i sin π/3) in the form x + iy. Since cos π/3 = 1/2 and sin π/3 = √3/2, the result is 6 × 1/2 + i × 6 × √3/2 = 3 + 3√3 i.
例如，将 6(cos π/3 + i sin π/3) 写成 x + iy 的形式。因为 cos π/3 = 1/2，sin π/3 = √3/2，所以结果是 6 × 1/2 + i × 6 × √3/2 = 3 + 3√3 i。
The modulus and argument can also be used to check an answer. For z = x + iy, |z| = √x² + y² correction. If your x and y values do not match the given modulus and argument, check your expansion.
模和辐角也可用于检验答案。对于 z = x + iy，|z| = √x² + y²，arg z = arctan(y/x) 并需根据象限修正。如果 x 和 y 的值与给定模和辐角不匹配，请检查展开过程。
9. Worked Example: Rationalising a Quotient | 例题：商的有理化
Express (3 − 2i)/(1 + 4i) in the form x + iy, where x and y are real numbers.
将 (3 − 2i)/(1 + 4i) 表示成 x + iy 的形式，其中 x 和 y 为实数。
4i. The numerator becomes (3 − 2i)(1 − 4i) = 3 − 12i − 2i + 8i² = 3 − 14i − 8 = −5 − 14i.
首先将分子和分母同时乘以 1 + 4i 的共轭复数 1 − 4i。分子变为 (3 − 2i)(1 − 4i) = 3 − 12i − 2i + 8i² = 3 − 14i − 8 = −5 − 14i。
(1 + 4i)(1 − 4i) = 1 + 16 = 17. Therefore the quotient is (−5 − 14i)/17 = −5/17 − 14/17 i.
分母变为 (1 + 4i)(1 − 4i) = 1 + 16 = 17。因此商为 (−5 − 14i)/17 = −5/17 − 14/17 i。
(3 − 2i)/(1 + 4i) = −5/17 − 14/17 i
4/17. Both parts are real, simplified fractions, which is exactly what the examiner wants.
所以 x = −5/17，y = −14/17。两个部分都是实数且为最简分数，这正是考官希望看到的结果。
10. Exam Technique and Common Mistakes | 考试技巧与常见错误
In AQA exam questions, always present x and y as integers or simplified fractions. If a denominator can be cancelled, cancel it. For example, (8 + 2i)/2 should be written as 4 + i, not left as a single fraction.
在 AQA 考试题中，始终将 x 和 y 表示为整数或最简分数。如果分母可以约分，就进行约分。例如，(8 + 2i)/2 应写成 4 + i，而不要保留为单个分式。
other is a sign error when subtracting a complex number, especially when a negative imaginary part is involved.
一个常见错误是展开括号后忘记 i² = −1。另一个是在减去复数时出现符号错误，尤其是涉及负虚部时。
Check your final answer by confirming that no i appears in a denominator, no i² remains, and the coefficient of i is a single real number. If the question says x and y are real, your answer must clearly separate the two parts.
检查最终答案时，确认分母中没有 i，没有 i² 残留，并且 i 的系数是单个实数。如果题目说明 x 和 y 是实数，那么你的答案必须清晰地区分这两个部分。
Finally, when working with exact values such as √3 or π, leave them in exact form unless the question asks for a decimal approximation.
最后，当使用 √3 或 π 等精确值时，除非题目要求取小数近似，否则应保留精确形式。
Published by TutorHao | Maths 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