Equating Real and Imaginary Parts | 实部与虚部分别相等法

📚 Equating Real and Imaginary Parts | 实部与虚部分别相等法

In AQA A-level Further Mathematics, one of the most powerful techniques for solving complex equations is to equate real and imaginary parts separately. When two complex numbers are written in the form a + bi, equality between them forces both the real coefficients and the imaginary coefficients to match. This article explains the method, common exam applications, and worked examples you need to master.

在 AQA A-level 进阶数学中，解复数方程最有力的技巧之一就是分别比较实部与虚部。当两个复数都写成 a + bi 的形式时，两者相等意味着实部系数和虚部系数必须分别相等。本文将讲解这一方法、常见考题应用以及你需要掌握的标准例题。

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1. The Core Idea | 核心思想

If two complex numbers z = a + bi and w = c + di are equal, then their real parts must be equal and their imaginary parts must be equal. This gives two real equations from one complex equation.

如果两个复数 z = a + bi 和 w = c + di 相等，那么它们的实部必须相等，虚部也必须相等。这样，一个复数方程就转化为两个实数方程。

a + bi = c + di ⇒ a = c and b = d

This principle is the foundation for solving equations where the unknown is a complex number z = x + yi. By substituting z and expanding, you can collect all real terms and all imaginary terms, then equate coefficients.

这一原则是求解未知数为复数 z = x + yi 的方程的基础。代入 z 并展开后，你可以把所有的实部项和虚部项分别合并，然后比较对应系数。

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2. Standard Form a + bi | 标准形式 a + bi

Before equating real and imaginary parts, always rewrite every complex expression in the form a + bi. For example, expand brackets and replace i² by -1 so that no power of i higher than 1 remains.

在比较实部和虚部之前，一定要先把每个复数表达式化为 a + bi 的标准形式。例如，展开括号并用 -1 替换 i²，使表达式中不再出现高于 1 次的 i 的幂。

Consider the expression (x + yi)(2 – i). Expanding gives 2x + 2yi – xi – yi². Since i² = -1, this becomes (2x + y) + (2y – x)i. The real part is 2x + y and the imaginary part is 2y – x.

考虑表达式 (x + yi)(2 – i)。展开得到 2x + 2yi – xi – yi²。因为 i² = -1，它可化为 (2x + y) + (2y – x)i。实部为 2x + y，虚部为 2y – x。

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3. Solving Linear Complex Equations | 解线性复数方程

A typical question asks you to solve an equation such as (2 + i)z = 1 + 3i, where z is a complex number. Let z = x + yi and substitute it into the equation.

一道典型题目会要求解形如 (2 + i)z = 1 + 3i 的方程，其中 z 是复数。设 z = x + yi 并代入方程。

(2 + i)(x + yi) = 2x + 2yi + xi + yi² = (2x – y) + (x + 2y)i.

Equating real and imaginary parts with 1 + 3i gives two simultaneous equations: 2x – y = 1 and x + 2y = 3.

与 1 + 3i 比较实部和虚部，得到两个联立方程：2x – y = 1 和 x + 2y = 3。

Solving these gives x = 1 and y = 1. Therefore z = 1 + i. Always check by substituting back into the original equation.

解得 x = 1，y = 1。因此 z = 1 + i。一定要代回原方程进行检验。

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4. Equating with Quadratic Terms | 二次项中的实虚部分离

When an equation involves z², expand z = x + yi and simplify using i² = -1. This will produce real and imaginary parts that may each contain both x and y.

当方程中含有 z² 时，将 z = x + yi 代入展开，并利用 i² = -1 化简。这样得到的实部和虚部可能同时含有 x 和 y。

For example, if z² = 5 + 12i, then (x + yi)² = x² + 2xyi + y²i² = (x² – y²) + 2xyi.

例如，若 z² = 5 + 12i，则 (x + yi)² = x² + 2xyi + y²i² = (x² – y²) + 2xyi。

Equating real parts gives x² – y² = 5, and equating imaginary parts gives 2xy = 12. These two real equations can then be solved simultaneously for x and y.

比较实部得到 x² – y² = 5，比较虚部得到 2xy = 12。然后可以联立这两个实数方程求解 x 和 y。

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5. Finding Square Roots of Complex Numbers | 求复数的平方根

The previous method is the standard way to find square roots of a complex number. Solve the system x² – y² = 5 and 2xy = 12 for real x and y.

上述方法是求复数平方根的标准方法。在 x 和 y 为实数的条件下，求解方程组 x² – y² = 5 和 2xy = 12。

From 2xy = 12 we get y = 6/x. Substituting into x² – y² = 5 gives x² – 36/x² = 5, which leads to x⁴ – 5x² – 36 = 0.

由 2xy = 12 得 y = 6/x。代入 x² – y² = 5 得到 x² – 36/x² = 5，进一步化为 x⁴ – 5x² – 36 = 0。

Let t = x². Then t² – 5t – 36 = 0, so t = 9 or t = -4. Since x is real, t = x² ≥ 0, so x² = 9 and x = ±3. The corresponding values are y = ±2, giving the two square roots 3 + 2i and -3 – 2i.

令 t = x²，则 t² – 5t – 36 = 0，所以 t = 9 或 t = -4。由于 x 为实数，t = x² ≥ 0，因此 x² = 9，x = ±3。对应 y = ±2，于是得到两个平方根 3 + 2i 和 -3 – 2i。

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6. Simultaneous Equations with Complex Coefficients | 复数系数方程组

Equating real and imaginary parts can also solve systems of linear equations with complex coefficients. Add or subtract equations to eliminate one complex variable, then compare parts.

比较实部与虚部还可以求解含有复系数的线性方程组。通过相加或相减消去一个复变量，然后再比较实部和虚部。

Suppose z + w = 4 + 3i and z – w = -2 + i. Adding the two equations gives 2z = 2 + 4i, so z = 1 + 2i. Subtracting gives 2w = 6 + 2i, so w = 3 + i.

假设 z + w = 4 + 3i 且 z – w = -2 + i。两式相加得 2z = 2 + 4i，所以 z = 1 + 2i。两式相减得 2w = 6 + 2i，所以 w = 3 + i。

If the coefficients are more complicated, first expand and collect real and imaginary parts, then equate the coefficients of the real and imaginary components separately.

如果系数更复杂，可以先展开并合并实部和虚部，然后分别比较实部和虚部的系数。

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7. Equating Coefficients in Polynomials | 多项式系数比较

For polynomial equations with real coefficients, complex roots occur in conjugate pairs. If a + bi is a root, then a – bi is also a root. You can use this fact together with equating coefficients to factorise polynomials.

对于实系数多项式方程，复根成对出现。如果 a + bi 是一个根，那么 a – bi 也是根。你可以利用这一事实并结合系数比较来分解多项式。

For example, if 1 + i is a root of a quadratic with real coefficients, then the other root must be 1 – i. The quadratic equation can be written as z² – (sum of roots)z + product of roots = 0.

例如，如果 1 + i 是一个实系数二次方程的根，那么另一个根一定是 1 – i。该二次方程可以写成 z² -（根的和）z + 根的积 = 0。

The sum of roots is (1 + i) + (1 – i) = 2, and the product is (1 + i)(1 – i) = 1 – i² = 2. So the quadratic is z² – 2z + 2 = 0.

根的和为 (1 + i) + (1 – i) = 2，根的积为 (1 + i)(1 – i) = 1 – i² = 2。因此二次方程为 z² – 2z + 2 = 0。

This technique is particularly useful when finding unknown coefficients in a polynomial equation from given complex roots.

当需要根据给定的复根求多项式方程中的未知系数时，这一技巧特别有用。

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8. Dealing with Conjugates and Modulus | 处理共轭与模

Some equations involve the complex conjugate z̄ or the modulus |z|. Write z = x + yi. Then z̄ = x – yi and |z| = √(x² + y²). Substitute these into the equation and equate real and imaginary parts.

有些方程会涉及共轭复数 z̄ 或模 |z|。设 z = x + yi，则 z̄ = x – yi，|z| = √(x² + y²)。把这些代入方程，再比较实部和虚部。

For example, solve z + 2z̄ = 6 + i. Let z = x + yi. Then (x + yi) + 2(x – yi) = 3x – yi. Equating with 6 + i gives 3x = 6 and -y = 1, so x = 2 and y = -1. Thus z = 2 – i.

例如，解 z + 2z̄ = 6 + i。设 z = x + yi，则 (x + yi) + 2(x – yi) = 3x – yi。与 6 + i 比较，得 3x = 6 和 -y = 1，所以 x = 2，y = -1。因此 z = 2 – i。

When the modulus appears, be careful because |z| is always real. Equating imaginary parts may directly give a relationship for y, while the real part equation includes the square root term.

当出现模时要注意，因为 |z| 总是实数。比较虚部可能直接给出关于 y 的关系式，而实部方程中会包含平方根项。

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9. Common Mistakes | 常见错误

One common mistake is forgetting that i² = -1. Always replace i² with -1 before separating real and imaginary parts, otherwise the real and imaginary parts will be mixed up.

一个常见错误是忘记 i² = -1。在分离实部和虚部之前，一定要先用 -1 替换 i²，否则实部和虚部会混在一起。

Another mistake is treating i as a variable like x, rather than as a constant with the property i² = -1. Do not try to solve for i or cancel it out like an unknown number.

另一个错误是把 i 当作像 x 一样的变量，而没有把它看成具有 i² = -1 性质的常数。不要试图解出 i，也不要像消去未知数那样把 i 约掉。

Students also sometimes forget that x and y must be real numbers. If you obtain x² = -4, reject that solution because x must be real; otherwise the square root method breaks down.

学生有时还会忘记 x 和 y 必须是实数。如果得到 x² = -4，应舍去该解，因为 x 必须是实数；否则平方根方法就不成立。

Finally, always check your final z by substituting it back into the original equation. This is a quick way to catch arithmetic errors.

最后，一定要把求得的 z 代回原方程进行检验。这是快速发现计算错误的好方法。

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10. Exam-Style Worked Example | 考试型例题

Question: Solve the equation (1 + i)z = 2 + 6i, giving your answer in the form a + bi where a and b are real numbers.

题目：解方程 (1 + i)z = 2 + 6i，并将答案写成 a + bi 的形式，其中 a 和 b 为实数。

Step 1: Let z = x + yi. Then (1 + i)(x + yi) = x + yi + xi + yi² = (x – y) + (x + y)i.

第一步：设 z = x + yi。则 (1 + i)(x + yi) = x + yi + xi + yi² = (x – y) + (x + y)i。

Step 2: Equate real and imaginary parts with 2 + 6i. Real part: x – y = 2. Imaginary part: x + y = 6.

第二步：与 2 + 6i 比较实部和虚部。实部：x – y = 2。虚部：x + y = 6。

Step 3: Solve the simultaneous equations. Adding gives 2x = 8, so x = 4. Substituting back gives y = 2.

第三步：解联立方程。两式相加得 2x = 8，所以 x = 4。代回得 y = 2。

Step 4: Write the final answer z = 4 + 2i. Check: (1 + i)(4 + 2i) = 4 + 2i + 4i + 2i² = 4 – 2 + 6i = 2 + 6i, which is correct.

第四步：写出最终答案 z = 4 + 2i。检验：(1 + i)(4 + 2i) = 4 + 2i + 4i + 2i² = 4 – 2 + 6i = 2 + 6i，正确。

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11. Practice Strategy | 练习策略

Practise expanding brackets with complex numbers until it becomes automatic. Replacing i² with -1 and collecting real and imaginary parts should feel quick and accurate under exam time pressure.

反复练习复数的括号展开，直到形成条件反射。在考试时间压力下，用 -1 替换 i² 并合并实部虚部应当又快又准。

Work through past AQA questions on solving equations involving z and z². Pay attention to whether the question asks for all possible roots or just one specific value, and make sure x and y are real.

多做 AQA 以往关于 z 和 z² 方程求解的真题。注意题目是要求所有可能的根还是只求某个特定值，并确保 x 和 y 为实数。

Draw a quick sketch of the Argand diagram when checking answers. The real and imaginary parts should correspond to the horizontal and vertical coordinates of the point representing z.

检验答案时，可以快速画出阿尔冈图。实部和虚部分别对应表示 z 的点的横坐标和纵坐标。

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12. Summary | 小结

Equating real and imaginary parts converts a single complex equation into two real equations. This method is essential for solving linear and quadratic complex equations, finding square roots, and handling conjugates.

比较实部与虚部能把一个复数方程转化为两个实数方程。这一方法对求解线性与二次复数方程、求平方根以及处理共轭复数都至关重要。

Always rewrite expressions in standard form a + bi first, replace i² by -1, equate the real coefficients, and equate the imaginary coefficients. Then solve the resulting real simultaneous equations.

始终先把表达式化为标准形式 a + bi，用 -1 替换 i²，分别比较实部系数和虚部系数，然后求解得到的实数联立方程。

Mastering this technique will give you a reliable tool for many AQA Further Mathematics questions, and it also reinforces your understanding of complex number structure.

掌握这一方法将为你解答许多 AQA 进阶数学题目提供可靠工具，同时也能加深你对复数结构的理解。

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