Question 4 - A-Level Maths

CAIE P3 2023 March — Question 4 5 marks

Exam Board	CAIE
Module	P3 (Pure Mathematics 3)
Year	2023
Session	March
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex numbers 2
Type	Solve polynomial equations with complex roots
Difficulty	Standard +0.8 This question requires manipulating complex conjugates, rationalizing a complex denominator, and solving a quadratic equation in complex form. While the individual steps are standard A-level techniques, combining them in a non-routine way with the conjugate term zz* = \|z\|² elevates this above typical textbook exercises, making it moderately challenging but accessible to well-prepared students.
Spec	4.02e Arithmetic of complex numbers: add, subtract, multiply, divide

4 Solve the equation $$\frac { 5 z } { 1 + 2 \mathrm { i } } - z z ^ { * } + 30 + 10 \mathrm { i } = 0$$ giving your answers in the form $x + \mathrm { i } y$, where $x$ and $y$ are real.

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Question 4:

Answer	Marks	Guidance
Answer	Marks	Guidance
Substitute $z = x + iy$ and $z^* = x - iy$ to obtain a correct equation, horizontal or with $\frac{(1-2i)}{(1-2i)}$ seen, in $x$ and $y$	B1	$5(x+iy) - (x+iy)(x-iy)(1+2i) + (30+10i)(1+2i) = 0$; leads to $x - 2ix + iy + 2y - x^2 - y^2 + 30 + 10i = 0$
Use $i^2 = -1$ at least once and equate real and imaginary parts to zero	*M1	OE for their horizontal equation
Obtain two correct equations e.g. $x + 2y - x^2 - y^2 + 30 = 0$ and $-2x + y + 10 = 0$	A1	$5x - (x^2+y^2) + 10 = 0$; $5y - 2(x^2+y^2) + 70 = 0$; Allow $-2ix + iy + 10i = 0$
Solve quadratic equation for $x$ or $y$	DM1	$x^2 - 9x + 18 = (x-3)(x-6) = 0$; $y^2 + 2y - 8 = (y+4)(y-2) = 0$; DM0 if $x$ or $y$ imaginary
Obtain answers $3 - 4i$ and $6 + 2i$	A1
Total	5

## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $z = x + iy$ and $z^* = x - iy$ to obtain a correct equation, horizontal or with $\frac{(1-2i)}{(1-2i)}$ seen, in $x$ and $y$ | **B1** | $5(x+iy) - (x+iy)(x-iy)(1+2i) + (30+10i)(1+2i) = 0$; leads to $x - 2ix + iy + 2y - x^2 - y^2 + 30 + 10i = 0$ |
| Use $i^2 = -1$ at least once and equate real and imaginary parts to zero | ***M1** | OE for their horizontal equation |
| Obtain two correct equations e.g. $x + 2y - x^2 - y^2 + 30 = 0$ and $-2x + y + 10 = 0$ | **A1** | $5x - (x^2+y^2) + 10 = 0$; $5y - 2(x^2+y^2) + 70 = 0$; Allow $-2ix + iy + 10i = 0$ |
| Solve quadratic equation for $x$ or $y$ | **DM1** | $x^2 - 9x + 18 = (x-3)(x-6) = 0$; $y^2 + 2y - 8 = (y+4)(y-2) = 0$; **DM0** if $x$ or $y$ imaginary |
| Obtain answers $3 - 4i$ and $6 + 2i$ | **A1** | |
| **Total** | **5** | |

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4 Solve the equation

$$\frac { 5 z } { 1 + 2 \mathrm { i } } - z z ^ { * } + 30 + 10 \mathrm { i } = 0$$

giving your answers in the form $x + \mathrm { i } y$, where $x$ and $y$ are real.\\

\hfill \mbox{\textit{CAIE P3 2023 Q4 [5]}}

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Q1 3 Q2 5 Q3 5 Q4 5 Q5 6 Q6 7 Q7 9 Q8 9 Q9 7 Q10 9 Q11 10