📚 The Roots of zⁿ = α for Non-Real α | 非实数α时方程zⁿ=α的根
, 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。
1. Polar Form of a Non-Real Complex Number | 非实数复数的极坐标形式
rm. 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
al 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。
2. Setting Up the Equation zⁿ = α | 建立方程 zⁿ = α
mber 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)
= 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)。
3. The General Formula for the Roots | 根的通用公式
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。
zk = ρ^(1/n) [ cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n) ], k = 0, 1, ..., n-1
Equivalently, we may write zk = ρ^(1/n) cis((φ + 2kπ)/n). This formula gives all n distinct solutions of zⁿ = α.
等价地，可以写成 zk = ρ^(1/n) cis((φ + 2kπ)/n)。该公式给出 zⁿ = α 的所有 n 个不同解。
4. Why Does the Formula Work? | 公式为什么成立？
De Moivre' = 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。
actly 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。
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 = 81/3 = 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 = 81/3 = 2，根为：
zk = 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 | zk 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。
6. Geometric Interpretation | 几何意义
). 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。
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 zⁿ̄ = ᾱ. Since zⁿ̄ = (z̄)ⁿ, it follows that z̄ is a root of zⁿ = ᾱ. Thus the roots for α and the roots for ᾱ are mirror images with respect to the real axis.
若 z 是 zⁿ = α 的根，则两边取共轭得到 zⁿ̄ = ᾱ。由于 zⁿ̄ = (z̄)ⁿ，可知 z̄ 是 zⁿ = ᾱ 的根。因此 α 的根与 ᾱ 的根关于实轴对称。
le equation zⁿ = α, the sum of all n roots is zero for n > 1, because the coefficient of zn-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 zn-1 的系数为零。这个性质可用于检验所求根是否正确。
8. Example 2: Cube Roots of -8i | 例2：-8i 的立方根
³ = -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：
zk = 2 cis((-π/2 + 2kπ)/3), k = 0, 1, 2
Calculate each root:
逐个计算各个根：
| k | θ_k | zk |
| 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。
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。
10. Summary | 总结
-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：
zk = ρ^(1/n) cis((φ + 2kπ)/n), k = 0, 1, ..., n-1
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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