Complex Numbers And Equations

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Cambridge International AS & A Level Further Mathematics (9231), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

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Original Cambridge 2027 concept practice for source-backed gaps. Indicative self-marking only, not an official paper or complete syllabus coverage.

Selected subtopics: De Moivre and roots of unity · Trigonometric series from complex numbers · Integrating factors and transformed differential equations

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Question 1Prove de Moivre's theorem for positive integer powers by induction, using the angle-addition formulae. Then find all fourth roots of unity and verify that their sum is zero and their squared moduli sum to 4.5 indicative marks

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Prove de Moivre's theorem for positive integer powers by induction, using the angle-addition formulae. Then find all fourth roots of unity and verify that their sum is zero and their squared moduli sum to 4.

Cambridge 2027: Direct conceptual match

Checked the induction multiplication and evaluated all four fourth powers and both requested root sums.

9231: 2.5 Complex numbers; candidate bullet 2 (bullet ordinal, not an official subcode): Prove positive-integer de Moivre by induction
9231: 2.5 Complex numbers; candidate bullet 3, unity subitem (bullet ordinal, not an official subcode): Find and use roots of unity

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Do not assume the multiple-angle result inside the induction step: derive it from the cosine/sine addition identities.
For $z^{4}$ =1, the modulus is 1 and arguments are k*pi/2. Four distinct values exhaust the roots of a quartic.

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For n=1, (cos theta+i sin theta)^1=cos theta+i sin theta.
Assume the formula for n=k; multiplying by cos theta+i sin theta and using angle addition yields cos((k+1)theta)+i sin((k+1)theta).
The base and implication prove the formula for all positive integer n.
The fourth roots are exp(i*k*pi/2), k=0,1,2,3, namely 1,i,-1,-i.
Their sum is 0 and each has squared modulus 1, giving total 4.

Examiner tip

Finding roots of unity means solving $z^{n}$ =1, not simply an arbitrary non-unit complex-root problem.

Question 2(a) Use z=e^(i theta) to express sin^4 theta in terms of cos(2theta) and cos(4theta). (b) Sum sin(k*theta) for k=0,...,n by taking the imaginary part of a finite complex geometric series, assuming theta is not a multiple of 2pi. Evaluate that sum for n=3,theta=pi/3.5 indicative marks

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(a) Use z=e^(i theta) to express sin^4 theta in terms of cos(2theta) and cos(4theta). (b) Sum sin(k*theta) for k=0,...,n by taking the imaginary part of a finite complex geometric series, assuming theta is not a multiple of 2pi. Evaluate that sum for n=3,theta=pi/3.

Cambridge 2027: Direct conceptual match

Expanded the fourth power independently, checked the trigonometric identity numerically at several angles, and directly summed the four sine terms.

9231: 2.5 Complex numbers; candidate bullet 3, powers subitem (bullet ordinal, not an official subcode): Express trigonometric powers through multiple angles
9231: 2.5 Complex numbers; candidate bullet 3, series subitem (bullet ordinal, not an official subcode): Sum trigonometric series with complex numbers

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The fourth power of 2i is 16, not -16. The $z^{m}$ +z^-m pairs equal 2cos(m theta).
For the sum, factor each 1-e^(ia) as -2i*e^(ia/2)*sin(a/2), cancel, and then take the imaginary part. The formula has a removable limit at excluded full-turn angles, but no such limit is needed here.

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From sin theta=(z- $z^{-1}$ )/(2i), expand sin^4 theta=( $z^{4}$ -4z^2+6-4z^-2+ $z^{-4}$ )/16.
Pairing conjugate powers gives sin^4 theta=3/8-(1/2)cos(2theta)+(1/8)cos(4theta).
The geometric series is sum e^(ik theta)=(1-e^(i(n+1)theta))/(1-e^(i theta)), with nonzero denominator under the stated restriction.
Taking its imaginary part gives sum sin(k theta)=sin(n theta/2)*sin((n+1)theta/2)/sin(theta/2).
At n=3,theta=pi/3 the sum is $\sqrt{3}$ , agreeing with 0+ $\sqrt{3}$ /2+ $\sqrt{3}$ /2+0.

Examiner tip

State the denominator restriction before dividing by 1-e^(i theta).

Question 3Solve y'+2y=x with y(0)=1 using an integrating factor. Verify the result in the equation and describe the difference between its complementary function and particular integral.5 indicative marks

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Solve y'+2y=x with y(0)=1 using an integrating factor. Verify the result in the equation and describe the difference between its complementary function and particular integral.

Cambridge 2027: Direct conceptual match

Verified the integrating factor product derivative, the parts primitive and direct substitution of y=x/2-1/4+(5/4)e^-2x.

9231: 2.6 Differential equations; candidate bullet 1 (bullet ordinal, not an official subcode): Solve first-order linear equations with integrating factors
9231: 2.6 Differential equations; candidate bullet 2 (bullet ordinal, not an official subcode): Distinguish complementary function and particular integral
9231: 2.6 Differential equations; candidate bullet 6 (bullet ordinal, not an official subcode): Use initial conditions and interpret a differential-equation model

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The factor is exp(integral 2 dx), multiplying every term of the differential equation.
Differentiate the final expression to get 1/2-(5/2)e^(-2x); adding twice y cancels the exponential and constants, leaving x. Both the equation and initial condition are required checks.

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An integrating factor is e^(2x).
The left side becomes d(e^(2x)y)/dx, so e^(2x)y=integral x e^(2x) dx.
Integration by parts gives e^(2x)(x/2-1/4)+C.
Thus y=x/2-1/4+C e^(-2x); y(0)=1 gives C=5/4.
The complementary function is C e^(-2x), the particular integral x/2-1/4; substitution gives y'+2y=x.

Examiner tip

Integrating factor and complementary function are not the same object.

Question 4For x>0 solve

x^{2}

*y''-x*y'+y=0, with y(1)=2 and y'(1)=3, by the given substitution t=ln x and Y(t)=y(

e^{t}

). Derive the transformed derivatives rather than replacing x by t directly.5 indicative marks

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For x>0 solve $x^{2}$ *y''-x*y'+y=0, with y(1)=2 and y'(1)=3, by the given substitution t=ln x and Y(t)=y( $e^{t}$ ). Derive the transformed derivatives rather than replacing x by t directly.

Cambridge 2027: Direct conceptual match

Derived both chain-rule identities and independently substituted the final y, y' and y'' with both x=1 conditions.

9231: 2.6 Differential equations; candidate bullet 3 (bullet ordinal, not an official subcode): Find constant-coefficient complementary functions for all root types
9231: 2.6 Differential equations; candidate bullet 5 (bullet ordinal, not an official subcode): Reduce equations using a given substitution
9231: 2.6 Differential equations; candidate bullet 6 (bullet ordinal, not an official subcode): Use initial conditions and interpret a differential-equation model

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Since dt/dx=1/x, differentiate Y'/x carefully to obtain (Y''-Y')/ $x^{2}$ .
Direct verification gives y'=3+ln x and y''=1/x; x-(3x+x ln x)+(2x+x ln x)=0. The x>0 domain makes the logarithm and transformation valid.

Self-marking points

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The chain rule gives x*y'=Y' and $x^{2}$ *y''=Y''-Y'.
The equation becomes Y''-2Y'+Y=0.
The repeated auxiliary root is 1, so Y=(A+B*t) $e^{t}$ .
Returning to x gives y=x(A+B ln x).
y(1)=2 fixes A=2, while y'=A+B ln x+B and y'(1)=3 give B=1; hence y=x(2+ln x).

Examiner tip

The derivative correction -Y' is essential when transforming the second derivative.

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