Complex Roots of Polynomial Equations with Real Coefficients | 实系数多项式方程的复数根

📚 Complex Roots of Polynomial Equations with Real Coefficients | 实系数多项式方程的复数根

In A-Level mathematics, once complex numbers are introduced, we can solve every polynomial equation. However, when the coefficients of the polynomial are real, a remarkable symmetry appears: any non-real complex root must be accompanied by its complex conjugate. This result, known as the Conjugate Root Theorem, greatly simplifies solving and factoring polynomials.

在 A-Level 数学中，一旦引入复数，我们就能解所有多项式方程。然而，当多项式的系数是实数时，会出现一个显著的对称性：任何非实复数根必然伴随着它的共轭复数。这个结论称为“共轭根定理”，它极大地简化了多项式的求解与因式分解。

1. Complex Conjugates and Basic Operations | 共轭复数与基本运算

For a complex number \( z = a + bi \), where \( a \) and \( b \) are real, its complex conjugate is defined as \( \bar{z} = a – bi \). The product of a complex number and its conjugate is always a non-negative real number: \( z\bar{z} = a^2 + b^2 \).

对于复数 \( z = a + bi \)（其中 \( a \)、\( b \) 为实数），其共轭复数定义为 \( \bar{z} = a – bi \)。一个复数与其共轭的乘积总是非负实数：\( z\bar{z} = a^2 + b^2 \)。

Conjugation also respects addition and multiplication. If \( z_1 \) and \( z_2 \) are complex numbers, then \( \overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2 \) and \( \overline{z_1 z_2} = \bar{z}_1 \bar{z}_2 \). These properties are essential for proving the Conjugate Root Theorem.

共轭运算还保持加法和乘法：若 \( z_1 \)、\( z_2 \) 是复数，则 \( \overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2 \)，且 \( \overline{z_1 z_2} = \bar{z}_1 \bar{z}_2 \)。这些性质对证明共轭根定理至关重要。

2. The Conjugate Root Theorem | 共轭根定理

Statement: If a polynomial \( P(z) \) has real coefficients and \( \alpha \) is a complex root, then its conjugate \( \bar{\alpha} \) is also a root of \( P(z) \).

定理表述： 若多项式 \( P(z) \) 具有实系数，且 \( \alpha \) 是其一个复数根，则它的共轭 \( \bar{\alpha} \) 也是 \( P(z) \) 的根。

Proof idea: Write \( P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_0 \), where every \( a_k \) is real. If \( P(\alpha) = 0 \), then take conjugates of both sides. Since conjugation preserves addition and multiplication, and \( \bar{a}_k = a_k \), we obtain \( P(\bar{\alpha}) = 0 \).

证明思路： 设 \( P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_0 \)，其中每个 \( a_k \) 都是实数。若 \( P(\alpha) = 0 \)，对等式两边取共轭。由于共轭保持加法与乘法，且 \( \bar{a}_k = a_k \)，便得到 \( P(\bar{\alpha}) = 0 \)。

3. Solving a Quadratic with Real Coefficients | 求解实系数二次方程

When the discriminant of a quadratic equation is negative, the roots are a complex conjugate pair. For example, consider \( z^2 – 4z + 13 = 0 \). Using the quadratic formula:

当一元二次方程的判别式为负时，根是一对共轭复数。例如，考虑 \( z^2 – 4z + 13 = 0 \)。用求根公式：

\( z = \frac{4 \pm \sqrt{16 – 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i \)

The two roots are \( 2 + 3i \) and \( 2 – 3i \). Notice that they are conjugates of each other. Their sum is \( 4 \), and their product is \( (2+3i)(2-3i) = 4 + 9 = 13 \), matching the coefficients of the original equation.

两根为 \( 2 + 3i \) 和 \( 2 – 3i \)。注意它们互为共轭。它们的和为 \( 4 \)，积为 \( (2+3i)(2-3i) = 4 + 9 = 13 \)，与原方程的系数吻合。

If \( b^2 – 4ac < 0 \), the roots are \( \alpha \) and \( \bar{\alpha} \).
若 \( b^2 – 4ac < 0 \)，则根为 \( \alpha \) 与 \( \bar{\alpha} \)。

4. Cubic Polynomials: One Real Root and One Conjugate Pair | 三次多项式：一个实根与一对共轭复数根

A cubic equation with real coefficients has degree 3, so it has three roots (counting multiplicities). Because non-real roots come in conjugate pairs, a cubic must have at least one real root, and if it has non-real roots, they must be exactly one conjugate pair and one real root.

实系数三次方程的次数为 3，因此有三个根（按重数计）。因为非实根成对出现，三次方程至少有一个实根；若它含有非实根，则恰好是一对共轭复数根加上一个实根。

Suppose a cubic polynomial is known to have a root \( z = 1 + 2i \). By the theorem, \( 1 – 2i \) is also a root. The third root can be found by dividing the polynomial by the quadratic factor corresponding to the conjugate pair.

假设已知某三次多项式有一个根 \( z = 1 + 2i \)。根据定理，\( 1 – 2i \) 也是根。第三个根可通过将多项式除以这对共轭根所对应的二次因式来求得。

Example: Let \( P(z) = z^3 – 5z^2 + 11z – 15 \). Verify that \( z = 1 + 2i \) is a root, then find all roots.

示例： 设 \( P(z) = z^3 – 5z^2 + 11z – 15 \)。验证 \( z = 1 + 2i \) 是根，并求所有根。

First check: \( (1+2i)^3 – 5(1+2i)^2 + 11(1+2i) – 15 \). Compute step by step: \( (1+2i)^2 = -3 + 4i \), \( (1+2i)^3 = (-3+4i)(1+2i) = -11 – 2i \). Then \( P(1+2i) = (-11-2i) -5(-3+4i) + (11+22i) -15 = 0 \). So it is a root.

先验证：\( (1+2i)^3 – 5(1+2i)^2 + 11(1+2i) – 15 \)。逐步计算：\( (1+2i)^2 = -3 + 4i \)，\( (1+2i)^3 = (-3+4i)(1+2i) = -11 – 2i \)。于是 \( P(1+2i) = (-11-2i) -5(-3+4i) + (11+22i) -15 = 0 \)。所以它是根。

Since \( 1+2i \) and \( 1-2i \) are roots, the quadratic factor is \( (z – (1+2i))(z – (1-2i)) = (z-1)^2 + 4 = z^2 – 2z + 5 \). Now divide \( P(z) \) by \( z^2 – 2z + 5 \):

由于 \( 1+2i \) 与 \( 1-2i \) 是根，二次因式为 \( (z – (1+2i))(z – (1-2i)) = (z-1)^2 + 4 = z^2 – 2z + 5 \)。现在用 \( P(z) \) 除以 \( z^2 – 2z + 5 \)：

\( z^3 – 5z^2 + 11z – 15 \div (z^2 – 2z + 5) = z – 3 \)

The quotient is \( z – 3 \), so the real root is \( z = 3 \). The full set of roots is \( \{1+2i, 1-2i, 3\} \).

商为 \( z – 3 \)，所以实根为 \( z = 3 \)。完整根集为 \( \{1+2i, 1-2i, 3\} \)。

5. Factoring a Cubic into Real Linear and Quadratic Factors | 将三次多项式分解为实线性因式与实二次因式

When a cubic has one real root \( r \) and a conjugate pair \( a \pm bi \), it can be written as:

当三次多项式有一个实根 \( r \) 和一对共轭复根 \( a \pm bi \) 时，它可以写作：

\( P(z) = (z – r)\left(z^2 – 2az + (a^2+b^2)\right) \)

Here \( z^2 – 2az + (a^2+b^2) \) is a quadratic with real coefficients. This form is often required in exam questions: first find the real root, then factorise the remaining quadratic. If the discriminant of the quadratic is negative, it cannot be factorised further over the reals.

其中 \( z^2 – 2az + (a^2+b^2) \) 是实系数二次式。考试题常要求这种形式：先求实根，再分解剩余二次式。若该二次式的判别式为负，则在实数范围内不能再分解。

For example, \( z^3 – 3z^2 + 4z – 2 \) has a root \( z = 1 \). Dividing gives \( z^2 – 2z + 2 \), whose roots are \( 1 \pm i \). Hence, \( z^3 – 3z^2 + 4z – 2 = (z-1)(z^2 – 2z + 2) \).

例如，\( z^3 – 3z^2 + 4z – 2 \) 有一个根 \( z = 1 \)。除以 \( z-1 \) 得 \( z^2 – 2z + 2 \)，其根为 \( 1 \pm i \)。因此，\( z^3 – 3z^2 + 4z – 2 = (z-1)(z^2 – 2z + 2) \)。

6. Quartic Polynomials: Possible Root Patterns | 四次多项式：可能的根模式

A quartic with real coefficients has four roots. The possibilities are:

实系数四次多项式有四个根。可能的情况有：

Four real roots.
Two real roots and one complex conjugate pair.
Two complex conjugate pairs.
四实根。
两实根与一对共轭复根。
两对共轭复根。

There cannot be one real root and three non-real roots, because non-real roots must occur in conjugate pairs, which would make the total number of non-real roots even.

不可能出现一个实根加三个非实根的情况，因为非实根必须成对出现，这使得非实根的总数为偶数。

Example: Find all roots of \( P(z) = z^4 – 4z^3 + 14z^2 – 36z + 45 \), given that \( z = 3i \) is a root.

示例： 已知 \( z = 3i \) 是 \( P(z) = z^4 – 4z^3 + 14z^2 – 36z + 45 \) 的一个根，求所有根。

Since \( z = 3i \) is a root, \( z = -3i \) is also a root. Their quadratic factor is \( (z – 3i)(z + 3i) = z^2 + 9 \). Divide \( P(z) \) by \( z^2 + 9 \):

因为 \( z = 3i \) 是根，所以 \( z = -3i \) 也是根。对应的二次因式为 \( (z – 3i)(z + 3i) = z^2 + 9 \)。用 \( P(z) \) 除以 \( z^2 + 9 \)：

\( (z^4 – 4z^3 + 14z^2 – 36z + 45) \div (z^2 + 9) = z^2 – 4z + 5 \)

Now solve \( z^2 – 4z + 5 = 0 \). The roots are \( 2 \pm i \). Therefore, the four roots are \( 3i, -3i, 2+i, 2-i \).

再解 \( z^2 – 4z + 5 = 0 \)，其根为 \( 2 \pm i \)。因此四个根为 \( 3i, -3i, 2+i, 2-i \)。

7. Constructing a Polynomial from Its Roots | 由根构造多项式

If we know the roots of a polynomial with real coefficients, we can build it by multiplying linear factors. For each complex conjugate pair \( a \pm bi \), the product \( (z – (a+bi))(z – (a-bi)) = z^2 – 2az + (a^2+b^2) \) is a real quadratic factor. For each real root \( r \), we include \( (z – r) \).

如果我们知道一个实系数多项式的根，就可以通过相乘线性因式来构造它。对于每一对共轭复根 \( a \pm bi \)，乘积 \( (z – (a+bi))(z – (a-bi)) = z^2 – 2az + (a^2+b^2) \) 是实二次因式。对于每个实根 \( r \)，包含 \( (z – r) \)。

Example: Find a cubic polynomial with real coefficients whose roots are \( -1 \), \( 2 + i \), and \( 2 – i \).

示例： 求一个实系数三次多项式，其根为 \( -1 \)、\( 2 + i \) 和 \( 2 – i \)。

The conjugate pair gives the factor \( z^2 – 4z + 5 \). The real root gives \( z + 1 \). Thus the polynomial is \( (z+1)(z^2 – 4z + 5) = z^3 – 3z^2 + z + 5 \).

共轭对给出因式 \( z^2 – 4z + 5 \)。实根给出 \( z + 1 \)。因此多项式为 \( (z+1)(z^2 – 4z + 5) = z^3 – 3z^2 + z + 5 \)。

8. Graphical Interpretation: Why Complex Roots Do Not Appear on the Real Axis | 图形解释：为什么复数根不出现在实轴上

On the Argand diagram, the roots of a polynomial with real coefficients are symmetric with respect to the real axis. A real root lies on the horizontal axis, while a non-real root and its conjugate are reflected symmetrically across the real axis.

在阿甘图上，实系数多项式的根关于实轴对称。实根位于水平轴上，而非实根与其共轭关于实轴成镜像对称。

The graph of a real polynomial \( y = P(x) \) only shows the real roots as x-intercepts. Non-real roots are invisible on the Cartesian graph, but they still control the shape of the curve through the quadratic factors \( (x-a)^2 + b^2 \), which are always positive and never touch the x-axis.

实系数多项式 \( y = P(x) \) 的图像只把实根显示为与 x 轴的交点。非实根在笛卡尔图上不可见，但它们通过二次因式 \( (x-a)^2 + b^2 \) 控制曲线的形状，该因式恒为正，永远不会触碰 x 轴。

For a quadratic with no real roots, such as \( y = x^2 + 4x + 13 \), the graph is a parabola that lies entirely above the x-axis. Its complex roots \( -2 \pm 3i \) are not visible, but their sum and product are encoded in the coefficients of the parabola.

对于无实根的二次函数，如 \( y = x^2 + 4x + 13 \)，图像是位于 x 轴上方的抛物线。它的复数根 \( -2 \pm 3i \) 不可见，但它们的和与积都蕴含在抛物线的系数中。

9. Common Exam Pitfalls | 常见考试陷阱

Here are frequent mistakes students make when dealing with complex roots:

以下是学生在处理复数根时常犯的错误：

Forgetting that if \( a+bi \) is a root, then \( a-bi \) is also a root when coefficients are real.
Using the wrong sign for the conjugate: \( a-bi \), not \( -a+bi \).
When dividing by a quadratic factor, making arithmetic errors with signs.
Forgetting to check whether a given root satisfies the equation before applying the theorem.
Failing to write the final answer in the form \( a \pm bi \) or as a fully factorised expression.
忽略实系数条件下，\( a+bi \) 是根则 \( a-bi \) 也是根。
共轭的符号写错：应为 \( a-bi \)，而不是 \( -a+bi \)。
除以二次因式时，符号运算出现算术错误。
在应用定理之前，没有检查给定根是否确实满足方程。
没有将最终答案写成 \( a \pm bi \) 的形式或完全因式分解的形式。

In addition, when constructing a polynomial, remember that multiplying by a non-zero constant produces another polynomial with the same roots. Thus the answer is not unique unless a leading coefficient is specified.

另外，在构造多项式时，记住乘以非零常数会得到另一个具有相同根的多项式。因此除非指定首项系数，否则答案并不唯一。

10. Worked Exam-Style Question | 考试风格例题精讲

Question: A quartic polynomial \( P(z) \) with real coefficients has roots \( 2 – i \) and \( 1 + 3i \). Find the polynomial in the form \( z^4 + az^3 + bz^2 + cz + d \).

题目： 一个实系数四次多项式 \( P(z) \) 的根为 \( 2 – i \) 和 \( 1 + 3i \)。求该多项式，形式为 \( z^4 + az^3 + bz^2 + cz + d \)。

Solution: Since the coefficients are real, \( 2 + i \) and \( 1 – 3i \) are also roots. The four roots are \( 2 \pm i \) and \( 1 \pm 3i \).

解答： 由于系数为实数，\( 2 + i \) 与 \( 1 – 3i \) 也是根。四个根为 \( 2 \pm i \) 与 \( 1 \pm 3i \)。

First, form the quadratic factor for \( 2 \pm i \): \( (z – (2+i))(z – (2-i)) = z^2 – 4z + 5 \). For \( 1 \pm 3i \): \( (z – (1+3i))(z – (1-3i)) = z^2 – 2z + 10 \).

先构造 \( 2 \pm i \) 的二次因式：\( (z – (2+i))(z – (2-i)) = z^2 – 4z + 5 \)。再构造 \( 1 \pm 3i \) 的二次因式：\( (z – (1+3i))(z – (1-3i)) = z^2 – 2z + 10 \)。

Multiply the two quadratics:

将两个二次式相乘：

\( (z^2 – 4z + 5)(z^2 – 2z + 10) = z^4 – 6z^3 + 23z^2 – 50z + 50 \)

Therefore the polynomial is \( P(z) = z^4 – 6z^3 + 23z^2 – 50z + 50 \).

因此多项式为 \( P(z) = z^4 – 6z^3 + 23z^2 – 50z + 50 \)。

11. Summary of Key Procedures | 关键步骤总结

To handle complex roots of real-coefficient polynomials, follow these steps:

处理实系数多项式的复数根，请按以下步骤：

If one complex root \( a+bi \) is given, immediately write down \( a-bi \) as another root.
Form the real quadratic factor for each conjugate pair: \( z^2 – 2az + (a^2+b^2) \).
Divide the original polynomial by the quadratic factor(s) to reduce its degree.
Solve the remaining equation, which may be real quadratic or lower-degree polynomial.
Check that the total number of roots (counting multiplicity) equals the degree of the polynomial.
若已知一个复数根 \( a+bi \)，立即写出另一个根 \( a-bi \)。
为每一对共轭复根构造实二次因式：\( z^2 – 2az + (a^2+b^2) \)。
用原多项式除以二次因式来降次。
解剩下的方程，它可能是实二次或更低次多项式。
检查根的总数（按重数计）是否等于多项式的次数。

When constructing a polynomial from given roots, always include conjugate pairs automatically. If the leading coefficient is not specified, an arbitrary real constant multiplier may be introduced.

从给定根构造多项式时，务必自动包含共轭对。若未指定首项系数，可乘以任意非零实常数。

12. Practice Problems | 练习与巩固

Try the following questions to test your understanding.

尝试以下问题来检验你的理解。

1. Solve \( z^2 + 6z + 25 = 0 \).

1. 解 \( z^2 + 6z + 25 = 0 \)。

2. Given that \( z = 2 – i \) is a root of \( z^3 – 6z^2 + 13z – 10 = 0 \), find all roots.

2. 已知 \( z = 2 – i \) 是 \( z^3 – 6z^2 + 13z – 10 = 0 \) 的一个根，求所有根。

3. Find a cubic polynomial with real coefficients having roots \( -2 \), \( i \), and \( -i \).

3. 求一个实系数三次多项式，其根为 \( -2 \)、\( i \) 和 \( -i \)。

4. A quartic with real coefficients has roots \( 1 + 2i \) and \( 3 \). Find the fourth root and the polynomial.

4. 一个实系数四次多项式有根 \( 1 + 2i \) 和 \( 3 \)。求第四个根及多项式。

Answers: 1) \( -3 \pm 4i \); 2) \( 2 – i, 2 + i, 2 \); 3) \( (z+2)(z^2 + 1) = z^3 + 2z^2 + z + 2 \); 4) The roots are \( 1 \pm 2i \) and \( 3 \) with multiplicity two, or if only one real root \( 3 \) is known, then the fourth root must be another real root; a minimal polynomial is \( (z^2 – 2z + 5)(z-3)^2 \).

答案：1) \( -3 \pm 4i \)；2) \( 2 – i, 2 + i, 2 \)；3) \( (z+2)(z^2 + 1) = z^3 + 2z^2 + z + 2 \)；4) 根为 \( 1 \pm 2i \) 以及重数为 2 的实根 \( 3 \)；若只已知一个实根 \( 3 \)，则第四个根必为另一个实根。最小多项式为 \( (z^2 – 2z + 5)(z-3)^2 \)。

With the Conjugate Root Theorem, solving polynomial equations with real coefficients becomes systematic. Always look for conjugate pairs, use polynomial division to reduce degree, and verify your root count matches the degree. Master this pattern, and complex roots will no longer be a mystery.

掌握共轭根定理后，解实系数多项式方程就变得系统化。始终寻找共轭对，用多项式除法降次，并验证根的总数与次数一致。熟练掌握这一套路，复数根将不再是难题。

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Complex Roots of Polynomial Equations with Real Coefficients | 实系数多项式方程的复数根