The Roots of zⁿ = α for Non-Real α | 非实数α时方程zⁿ=α的根

📚 The Roots of zⁿ = α for Non-Real α | 非实数α时方程zⁿ=α的根

In this article we study how to find all complex numbers z that satisfy zⁿ = α, where α is a non-real complex number. This is equivalent to finding all n-th roots of a non-real complex number. The solution relies on the polar form of a complex number and De Moivre’s theorem.

在本文中，我们研究如何求出满足 zⁿ = α 的所有复数 z，其中 α 是一个非实数复数。这等价于求一个非实数复数的全部 n 次方根。求解过程需要用到复数的极坐标形式以及棣莫弗定理。

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1. Polar Form of a Non-Real Complex Number | 非实数复数的极坐标形式

The starting point is to write α in polar form. If α = a + bi with b ≠ 0, then α is non-real. Let ρ be its modulus and φ its argument:

出发点是先把 α 写成极坐标形式。若 α = a + bi 且 b ≠ 0，则 α 是非实数。设 ρ 为它的模，φ 为它的辐角：

ρ = |α| = √(a² + b²),   φ = arg α

Then we may write α = ρ(cos φ + i sin φ) = ρ cis φ. For a non-real number, φ is not 0 or π; typically we choose the principal value satisfying -π < φ ≤ π.

于是可以写成 α = ρ(cos φ + i sin φ) = ρ cis φ。对于非实数，φ 不等于 0 或 π；通常选择主值满足 -π < φ ≤ π。

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2. Setting Up the Equation zⁿ = α | 建立方程 zⁿ = α

We look for a complex number z = r(cos θ + i sin θ) = r cis θ, with r > 0 and θ real. We want zⁿ = α. By De Moivre’s theorem, raising z to the power n gives:

我们寻找复数 z = r(cos θ + i sin θ) = r cis θ，其中 r > 0，θ 为实数。要求 zⁿ = α。由棣莫弗定理，对 z 取 n 次方得：

zⁿ = rⁿ [cos(nθ) + i sin(nθ)] = rⁿ cis(nθ)

Since zⁿ = α = ρ cis φ, the moduli must be equal and the arguments must differ by a multiple of 2π:

由于 zⁿ = α = ρ cis φ，模必须相等，辐角必须相差 2π 的整数倍：

rⁿ = ρ,    nθ = φ + 2kπ   (k ∈ ℤ)

Because r > 0 and ρ > 0, we take the unique positive real n-th root: r = ρ^(1/n).

因为 r > 0 且 ρ > 0，我们取唯一的正实数 n 次方根：r = ρ^(1/n)。

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3. The General Formula for the Roots | 根的通用公式

From nθ = φ + 2kπ, we obtain θ = (φ + 2kπ)/n. Although k can be any integer, only k = 0, 1, …, n-1 give distinct roots; increasing k by n adds 2π to θ, which returns to the same complex number.

由 nθ = φ + 2kπ，得到 θ = (φ + 2kπ)/n。虽然 k 可以取任意整数，但只有 k = 0, 1, …, n-1 给出不同的根；k 每增加 n，θ 就增加 2π，对应的复数不变。

z_k = ρ^(1/n) [ cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n) ],   k = 0, 1, …, n-1

Equivalently, we may write z_k = ρ^(1/n) cis((φ + 2kπ)/n). This formula gives all n distinct solutions of zⁿ = α.

等价地，可以写成 z_k = ρ^(1/n) cis((φ + 2kπ)/n)。该公式给出 zⁿ = α 的所有 n 个不同解。

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4. Why Does the Formula Work? | 公式为什么成立？

De Moivre’s theorem states that [r cis θ]ⁿ = rⁿ cis(nθ). The sine and cosine functions have period 2π, so if two complex numbers have arguments differing by 2πk, they are identical. Therefore nθ = φ + 2kπ is exactly the condition that ensures the n-th power of r cis θ equals ρ cis φ.

棣莫弗定理指出 [r cis θ]ⁿ = rⁿ cis(nθ)。正弦和余弦函数以 2π 为周期，所以如果两个复数的辐角相差 2πk，则这两个复数相同。因此 nθ = φ + 2kπ 正是保证 r cis θ 的 n 次方等于 ρ cis φ 的条件。

There are exactly n distinct values of k modulo n, so the equation zⁿ = α has exactly n complex roots.

k 对 n 取模恰好有 n 个不同值，因此方程 zⁿ = α 恰好有 n 个复数根。

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5. Example 1: Cube Roots of 8i | 例1：8i 的立方根

Solve z³ = 8i, where α = 8i is non-real. Here ρ = 8 and φ = π/2. Hence r = 8^(1/3) = 2, and the roots are:

解方程 z³ = 8i，其中 α = 8i 是非实数。这里 ρ = 8，φ = π/2。因此 r = 8^(1/3) = 2，根为：

z_k = 2 cis((π/2 + 2kπ)/3),   k = 0, 1, 2

Evaluate each value of k:

分别代入 k 的各个值：

k	θ_k	z_k in Cartesian form
0	π/6	√3 + i
1	5π/6	-√3 + i
2	3π/2	-2i

Check for k = 0: (√3 + i)³ = (√3)³ + 3(√3)²i + 3(√3)i² + i³ = 3√3 + 9i – 3√3 – i = 8i. The other two roots can be verified similarly.

验证 k = 0：(√3 + i)³ = (√3)³ + 3(√3)²i + 3(√3)i² + i³ = 3√3 + 9i – 3√3 – i = 8i。另外两个根可以类似验证。

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6. Geometric Interpretation | 几何意义

All n roots of zⁿ = α lie on a circle centred at the origin, with radius ρ^(1/n). The angles are spaced by 2π/n, so the roots form the vertices of a regular n-gon inscribed in that circle.

zⁿ = α 的所有 n 个根都位于以原点为圆心、半径为 ρ^(1/n) 的圆上。相邻根的辐角相差 2π/n，所以这些根构成该圆内接正 n 边形的顶点。

Because α is non-real, φ is not 0 or π. The polygon is therefore rotated by φ/n compared with the standard n-th roots of unity. In particular, none of the roots is real, since a real number raised to the power n is always real and cannot equal a non-real α.

因为 α 是非实数，φ 不等于 0 或 π。因此这个正 n 边形相对于单位根的图形旋转了 φ/n。特别地，这些根中没有实数根，因为实数的 n 次方总是实数，不可能等于非实数 α。

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7. Conjugate Roots and Symmetry | 共轭根与对称性

If z is a root of zⁿ = α, then taking complex conjugates gives \overline{zⁿ} = \overline{α}. Since \overline{zⁿ} = (\bar{z})ⁿ, it follows that \bar{z} is a root of zⁿ = \bar{α}. Thus the roots for α and the roots for \bar{α} are mirror images with respect to the real axis.

若 z 是 zⁿ = α 的根，则两边取共轭得到 \overline{zⁿ} = \overline{α}。由于 \overline{zⁿ} = (\bar{z})ⁿ，可知 \bar{z} 是 zⁿ = \bar{α} 的根。因此 α 的根与 \bar{α} 的根关于实轴对称。

Within a single equation zⁿ = α, the sum of all n roots is zero for n > 1, because the coefficient of z^(n-1) in zⁿ – α is zero. This can be a useful check when calculating roots.

在同一个方程 zⁿ = α 中，当 n > 1 时，所有 n 个根之和为零，因为 zⁿ – α 中 z^(n-1) 的系数为零。这个性质可用于检验所求根是否正确。

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8. Example 2: Cube Roots of -8i | 例2：-8i 的立方根

Solve z³ = -8i. Now α = -8i, so ρ = 8, and the principal argument is φ = -π/2. The formula gives:

解方程 z³ = -8i。此时 α = -8i，所以 ρ = 8，主辐角 φ = -π/2。由公式得：

z_k = 2 cis((-π/2 + 2kπ)/3),   k = 0, 1, 2

Calculate each root:

逐个计算各个根：

k	θ_k	z_k
0	-π/6	√3 – i
1	π/2	2i
2	7π/6	-√3 – i

Notice that these are the negatives of the roots from Example 1? Not exactly: the set is different, but both sets form regular triangles on the circle of radius 2.

注意这些根与例1中的根并不相同，但两组根都在半径为 2 的圆上构成正三角形。

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9. Common Pitfalls | 常见错误

• Forgetting to include 2kπ. Solving only nθ = φ gives one root instead of n roots.

  忘记加上 2kπ。只解 nθ = φ 只能得到一个根，而不是 n 个根。

• Using the wrong sign or value of the argument. Always write α in polar form before applying the formula.

  使用错误的正负号或辐角值。应用公式前务必把 α 写成极坐标形式。

• Choosing r incorrectly. Remember r is the positive real n-th root of ρ, not just any root.

  错误计算 r。记住 r 是 ρ 的正实数 n 次方根，而不是任意根。

• Forgetting that α is non-real, so no real roots exist. Do not attempt to find a real solution.

  忘记 α 是非实数，因此不存在实数根。不要试图寻找实数解。

• Mixing degrees and radians. The formula is most natural in radians; if you use degrees, convert consistently.

  混用角度制和弧度制。公式在弧度制下最自然；如果使用角度制，请统一转换。

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10. Summary | 总结

To solve zⁿ = α for a non-real complex number α, first write α = ρ cis φ. Then the n roots are given by the formula

要求解非实数复数 α 的方程 zⁿ = α，首先写出 α = ρ cis φ。那么 n 个根由以下公式给出：

z_k = ρ^(1/n) cis((φ + 2kπ)/n),   k = 0, 1, …, n-1

These roots are equally spaced on a circle of radius ρ^(1/n) centred at the origin. Because α is non-real, the whole set is rotated by φ/n and contains no real numbers.

这些根均匀分布在以原点为圆心、半径为 ρ^(1/n) 的圆上。由于 α 是非实数，整组根旋转了 φ/n，并且不包含任何实数。

Always remember to use De Moivre’s theorem, include the 2kπ term, and check your answers by substituting back into the original equation.

始终记住使用棣莫弗定理、包含 2kπ 项，并通过代回原方程来检验答案。

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