Complex Numbers Past Papers Deep Dive | OCR进阶数学复数历年真题深度解析

📚 Complex Numbers Past Papers Deep Dive | OCR进阶数学复数历年真题深度解析

Mastering complex numbers is essential for success in OCR Year 12 Further Mathematics. This article provides an in-depth analysis of real past paper questions, highlighting common themes, techniques, and examiner expectations. Each section pairs an English explanation with its Chinese counterpart, ensuring you grasp every nuance needed to excel.

掌握复数是OCR Year 12进阶数学成功的关键。本文深度解析历年真题，突出常见主题、解题技巧与评分标准。每个部分均以中英文对照讲解，帮助你全面把握每一个得分要点。

1. Complex Conjugates in Quadratic Equations | 二次方程中的共轭复数

the form x² + px + q = 0, where p and q are real. Examiners expect you to immediately recognise that the second root must be the complex conjugate 2+5i.

一道经典真题：实系数二次方程的一个根是2−5i，求另一个根，并确定方程为x²+px+q=0的形式，其中p,q为实数。考官希望你立即意识到第二个根必定是共轭复数2+5i。

refore the quadratic is x² − (sum)x + product = 0, giving x² − 4x + 29 = 0. Many students mistakenly write the sum with an imaginary part; remember the sum of conjugates is always purely real.

由于系数为实数，复根总是成对共轭出现。根之和=(2−5i)+(2+5i)=4；根之积=(2−5i)(2+5i)=4+25=29。因此二次方程为x²−(根之和)x+根之积=0，即x²−4x+29=0。许多同学会错误地保留虚部，请记住共轭复数的和永远为实数。

A more demanding variant involves a cubic equation with one complex root given. Use polynomial division once you identify the conjugate pair, then deduce the real factor.

更具挑战性的变体是给定一个复根的三次方程。识别出共轭对后使用多项式除法，进而推出实因子。

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2. De Moivre's Theorem in Action | 棣莫弗定理的应用

OCR frequently asks you to simplify expressions like (1 + i√3)ⁿ or to derive trigonometric identities using de Moivre's theorem. For instance, express (1 + i√3)⁸ in the form a + ib, where a and b are real. Begin by writing 1 + i√3 in modulus-argument form: modulus = √1+3 = 2, argument = arctan(√3) = π/3. Thus 1 + i√3 = 2 e^iπ/3.

OCR常要求化简(1+i√3)ⁿ的形式或利用棣莫弗定理推导三角恒等式。例如，将(1+i√3)⁸表示为a+ib，其中a,b为实数。首先将1+i√3写为模-辐角形式：模=√1+3=2，辐角=arctan(√3)=π/3。因此1+i√3=2 e^iπ/3。

Then (1 + i√3)⁸ = 2⁸ e^i8π/3 = 256 e^(i(2π + 2π/3)) = 256 e^i2π/3 cause the argument reduces modulo 2π. Finally, 256(cos(2π/3) + i sin(2π/3)) = 256(−1/2 + i√3/2) = −128 + 128i√3. Always check that your final argument lies within the principal range (−π, π] if specified.

那么(1+i√3)⁸ = 2⁸ e^i8π/3=256 e^(i(2π+2π/3))=256 e^i2π/3，因为辐角可模2π约化。最后，256(cos(2π/3)+i sin(2π/3))=256(−1/2+i√3/2)=−128+128i√3。若题目要求主值范围，请确保最终辐角落在(−π,π]内。

For trigonometric identities, expand (cos θ + i sin θ)⁵ by the binomial theorem, then equate real parts to obtain cos 5θ = 16cos⁵θ − 20cos³θ + 5cos θ. This technique is a favourite in exams.

对于三角恒等式，用二项式定理展开(cos θ + i sin θ)⁵，然后比较实部得到cos 5θ = 16cos⁵θ − 20cos³θ + 5cos θ。这种技巧在考试中非常受青睐。

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3. Roots of Unity and Geometric Patterns | 单位根与几何模式

A standard past-paper request: Solve the equation z⁵ ions on an Argand diagram. Find the sum of all the solutions. The roots are given by z = e^2kπi/5 3, 4. These five points lie on the unit circle equally spaced, forming a regular pentagon.

一道标准的真题要求：解方程z⁵=1，在阿贡图上表示解，并求所有解的和。根为z=e^2kπi/5，k=0,1,2,3,4。这五个点均匀分布在单位圆上，构成正五边形。

³ + ω⁴ = (1−ω^5)/(1−ω) = 0. Occasionally the question asks for the product: it equals (−1)⁵⁻¹ = 1. Make sure you can also find the sum of squares or the sum of pairwise products.

所有根之和为0，因为n次单位根（n>1）的和恒为零。可利用几何级数求和验证：1+ω+ω²+ω³+ω⁴=(1−ω^5)/(1−ω)=0。有时题目会问求积：它等于(−1)⁵⁻¹=1。请确保你也能求平方和或两两乘积之和。

Another common twist: Solve z⁴ 2)). These form a square rotated by 45°. The sum is still 0, a fact often tested under time pressure.

另一种常见变化：解z⁴=−16，根为2 e^(i(π/4 + kπ/2))。这些点构成旋转了45°的正方形，和依然为0，这一事实常在时间压力下被考查。

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4. Argument and Modulus Loci | 辐角与模的轨迹

Past paper question: Sketch the locus of points representing z satisfying |z − 3i| = 2. Given that |z − 1| = |z + i| sian equation of the locus of z, and determine the points of intersection of the two loci.

真题示例：画出满足|z−3i|=2的点轨迹；已知|z−1|=|z+i|，求z轨迹的笛卡尔方程，并确定两轨迹的交点。

locus is a circle with centre at (0, 3) and radius 2. For the second, let z = x + iy, then |(x−1) + iy| = |x + i(y+1)| is the perpendicular bisector of the line joining (1,0) and (0,−1), i.e. the line y = −x.

第一个轨迹是以(0,3)为圆心、2为半径的圆。第二个轨迹，令z=x+iy，则|(x−1)+iy|=|x+i(y+1)|。两边平方得(x−1)²+y²=x²+(y+1)²，化简得−2x+1=2y+1，即y=−x。因此第二个轨迹是连接(1,0)与(0,−1)线段的垂直平分线，即直线y=−x。

To find intersections, substitute y = −x into the circle' Always check if the points lie inside the region required by any additional inequalities.

为求交点，将y=−x代入圆的笛卡尔方程x²+(−x−3)²=4，得2x²+6x+5=0。用求根公式解得坐标。若题目附加了不等式条件，务必验证交点是否在所求区域内。

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5. Solving Equations with Complex Coefficients | 求解含复数系数的方程

ficients. For example, solve z² − (2+i)z + (3−i) = 0. Use the quadratic formula carefully.

尽管AS进阶数学通常使用实系数，但部分试卷含复数系数的二次方程。例如，解z²−(2+i)z+(3−i)=0。仔细使用求根公式。

p² − q² = −9 and 2pq = 8. Solving gives p = 1, q = 4 or p = −1, q = −4. Thus √Δ = ±(1+4i).

计算判别式Δ=(2+i)²−4(3−i)=(4+4i−1)−12+4i=(3+4i)−12+4i=−9+8i。接着求Δ的平方根：设(p+qi)²=−9+8i，得p²−q²=−9且2pq=8。解得p=1,q=4或p=−1,q=−4，因此√Δ=±(1+4i)。

.5 − 1.5i. Many students forget to halve both real and imaginary parts; always divide the entire complex number.

于是z=[(2+i)±(1+4i)]/2，得z=(3+5i)/2=1.5+2.5i或z=(1−3i)/2=0.5−1.5i。许多同学忘记将实部和虚部同时除以2；务必对整个复数进行除法。

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6. Trigonometric Identities via Complex Numbers | 用复数推导三角恒等式

in terms of powers of sin θ and cos θ. Hence find ∫ sin⁴θ dθ. Start with (cos θ + i sin θ)⁴ os⁴θ + 4i cos³θ sin θ − 6 cos²θ sin²θ − 4i cos θ sin³θ + sin⁴θ (since i²=−1, i³=−i, i⁴=1).

一道典型真题：用棣莫弗定理将sin 4θ表示为sin θ和cos θ的幂次。并求∫ sin⁴θ dθ。由(cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ出发，二项式展开左边：cos⁴θ+4i cos³θ sin θ−6 cos²θ sin²θ−4i cos θ sin³θ+sin⁴θ（因i²=−1,i³=−i,i⁴=1）。

rm is fine. For the integral, use identities to express sin⁴θ in terms of cos 4θ, cos 2θ, etc., then integrate term by term. The complex method provides a powerful shortcut.

比较虚部：sin 4θ=4 cos³θ sin θ−4 cos θ sin³θ。可改写为sin 4θ=4 sin θ cos θ(cos²θ−sin²θ)=2 sin 2θ cos 2θ，但直接形式亦可。对于积分，利用恒等式将sin⁴θ表示为cos 4θ, cos 2θ等，然后逐项积分。复数方法提供了强大的捷径。

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7. Finding nth Roots of a Complex Number | 求复数的 n 次方根

egative imaginary axis, its modulus is 64 and its argument is −π/2 (or equivalently 3π/2). Thus −64i = 64 e^−iπ/2.

真题：求−64i的立方根，用指数形式表示。先将−64i写为模-辐角形式。因−64i位于负虚轴上，模为64，辐角为−π/2（或3π/2）。故−64i=64 e^−iπ/2。

The cube roots are given by 64^1/3 e^(i(−π/6 + 2kπ/3)) for k = 0, 1, 2. So the roots are 4 e^−iπ/6, 4 e^iπ/2 = 4i, and 4 e^i7π/6. On the Argand diagram, they form an equilateral triangle. Note that the sum of these roots is 0, which is a useful check.

立方根为64^1/3 e^(i(−π/6+2kπ/3)), k=0,1,2。因此根为4 e^−iπ/6、4 e^iπ/2=4i 和 4 e^i7π/6。在阿贡图上它们构成等边三角形。注意这些根之和为0，可作为验算。

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8. Summation of Series Using Complex Numbers | 利用复数求级数和

example, evaluate S = cos θ + cos 3θ + cos 5θ + ... + cos(2n−1)θ. Consider C + iS_sum = e^iθ + e^3iθ + ... + e^((2n−1)iθ). This is a geometric series with first term e^iθ and common ratio e^2iθ.

常见题型要求利用复数几何级数求三角级

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