Question 3 - A-Level Maths

Edexcel CP1 2023 June — Question 3 10 marks

Exam Board	Edexcel
Module	CP1 (Core Pure 1)
Year	2023
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex Numbers Argand & Loci
Type	Region shading with multiple inequalities
Difficulty	Standard +0.3 This is a standard Core Pure 1 question testing routine complex number conversions and basic loci. Part (a) requires finding modulus-argument form (straightforward with r=4√2, θ=3π/4). Part (b) involves standard division and De Moivre's theorem. Part (c)(ii) requires recognizing that \|z-z₁\|<\|z-z₂\| represents a half-plane (perpendicular bisector), which is a standard textbook locus. All techniques are routine for CP1 with no novel problem-solving required, making it slightly easier than average.
Spec	4.02b Express complex numbers: cartesian and modulus-argument forms 4.02f Convert between forms: cartesian and modulus-argument 4.02k Argand diagrams: geometric interpretation 4.02m Geometrical effects: multiplication and division

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

$$z _ { 1 } = - 4 + 4 i$$

Express $\mathrm { z } _ { 1 }$ in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$, where $r \in \mathbb { R } , r > 0$ and $0 \leqslant \theta < 2 \pi$ $$z _ { 2 } = 3 \left( \cos \frac { 17 \pi } { 12 } + i \sin \frac { 17 \pi } { 12 } \right)$$
Determine in the form $a + \mathrm { i } b$, where $a$ and $b$ are exact real numbers,
1. $\frac { Z _ { 1 } } { Z _ { 2 } }$
2. $\left( z _ { 2 } \right) ^ { 4 }$
Show on a single Argand diagram
1. the complex numbers $z _ { 1 } , z _ { 2 }$ and $\frac { z _ { 1 } } { z _ { 2 } }$
2. the region defined by $\left\{ z \in \mathbb { C } : \left| z - z _ { 1 } \right| < \left| z - z _ { 2 } \right| \right\}$

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Question 3:

Part (a):

Answer	Marks	Guidance
\(	z_1	= \sqrt{(-4)^2+4^2}\) or $\arg z_1 = \pi - \frac{\pi}{4}$
$(z_1 =)\, 4\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$ or e.g. $(z_1=)\sqrt{32}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$	A1	Correct expression. The "$z_1=$" is not required. Correct answer with no working scores both marks.

Part (b)(i):

Answer	Marks	Guidance
$\frac{z_1}{z_2} = \frac{4\sqrt{2}}{3}\left(\cos\left(\frac{3\pi}{4}-\frac{17\pi}{12}\right)+i\sin\left(\frac{3\pi}{4}-\frac{17\pi}{12}\right)\right)$ or via exponential form or via Cartesian multiplication by conjugate	M1	Employs correct method: divides moduli and subtracts arguments, or uses exponential form, or converts $z_2$ to Cartesian and multiplies by conjugate. Correct answer with no working scores no marks.
$= -\frac{2\sqrt{2}}{3} - \frac{2\sqrt{6}}{3}i$ or $-\frac{2\sqrt{2}}{3} - i\frac{2\sqrt{6}}{3}$ or $-\frac{2\sqrt{2}}{3}+i\left(-\frac{2\sqrt{6}}{3}\right)$	A1	Correct exact answer in required form. Do not allow e.g. $-\frac{2}{3}(\sqrt{2}+\sqrt{6}i)$ or $\frac{-2\sqrt{2}-2\sqrt{6}i}{3}$ unless correct form seen previously.

Part (b)(ii):

Answer	Marks	Guidance
$z_2^4 = 3^4\left(\cos\left(4\times\frac{17\pi}{12}\right)+i\sin\left(4\times\frac{17\pi}{12}\right)\right)$ or $(z_2)^4 = \left(3e^{\frac{17\pi}{12}i}\right)^4 = 3^4 e^{\frac{17\pi}{12}\times 4i}$	M1	Applies De Moivre's theorem correctly to $z_2$: modulus is $3^4$ and argument is $4\times\frac{17\pi}{12}$. Correct answer with no working scores no marks.
$= \frac{81}{2} - \frac{81\sqrt{3}}{2}i$ or $\frac{81}{2}-i\frac{81\sqrt{3}}{2}$ or $\frac{81}{2}+i\left(-\frac{81\sqrt{3}}{2}\right)$	A1	Correct exact answer in required form. Do not allow e.g. $\frac{81}{2}\left(1-\frac{81\sqrt{3}}{2}i\right)$ or $\frac{81-81\sqrt{3}i}{2}$ unless correct form seen previously.

Part (c)(i):

Answer	Marks	Guidance
$z_1$ and $z_2$ correctly positioned: $z_1$ approximately on $y=-x$ in quadrant 2, $z_2$ below $y=x$ closer to origin than $z_1$ in quadrant 3	B1	Look for correct quadrants. Points usually labelled but mark positively if clear which is which.
$\frac{z_1}{z_2}$ in correct quadrant following through from (b)(i)	B1ft	Follow through their answer to (b)(i). Point sometimes labelled $z_3$.

Part (c)(ii):

Answer	Marks	Guidance
Draws perpendicular bisector of $z_1 z_2$ (solid or dashed) or draws a line crossing $z_1 z_2$ and shades one side	M1	Line that is perpendicular bisector of $z_1 z_2$ or crosses $z_1 z_2$ with shading.
Perpendicular bisector of $z_1 z_2$ drawn with either side shaded, clear they are not discounting the upper region; $z_1$ in quadrant 2 and $z_2$ in quadrant 3	A1	Note: region may be drawn on separate diagram. You do not need to see a line joining $z_1$ to $z_2$.

# Question 3:

## Part (a):
| $|z_1| = \sqrt{(-4)^2+4^2}$ or $\arg z_1 = \pi - \frac{\pi}{4}$ | M1 | Any correct expression for $|z_1|$ or $\arg z_1$. |
| $(z_1 =)\, 4\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$ or e.g. $(z_1=)\sqrt{32}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$ | A1 | Correct expression. The "$z_1=$" is not required. Correct answer with no working scores both marks. |

## Part (b)(i):
| $\frac{z_1}{z_2} = \frac{4\sqrt{2}}{3}\left(\cos\left(\frac{3\pi}{4}-\frac{17\pi}{12}\right)+i\sin\left(\frac{3\pi}{4}-\frac{17\pi}{12}\right)\right)$ or via exponential form or via Cartesian multiplication by conjugate | M1 | Employs correct method: divides moduli and subtracts arguments, or uses exponential form, or converts $z_2$ to Cartesian and multiplies by conjugate. Correct answer with no working scores no marks. |
| $= -\frac{2\sqrt{2}}{3} - \frac{2\sqrt{6}}{3}i$ or $-\frac{2\sqrt{2}}{3} - i\frac{2\sqrt{6}}{3}$ or $-\frac{2\sqrt{2}}{3}+i\left(-\frac{2\sqrt{6}}{3}\right)$ | A1 | Correct exact answer in required form. Do not allow e.g. $-\frac{2}{3}(\sqrt{2}+\sqrt{6}i)$ or $\frac{-2\sqrt{2}-2\sqrt{6}i}{3}$ unless correct form seen previously. |

## Part (b)(ii):
| $z_2^4 = 3^4\left(\cos\left(4\times\frac{17\pi}{12}\right)+i\sin\left(4\times\frac{17\pi}{12}\right)\right)$ or $(z_2)^4 = \left(3e^{\frac{17\pi}{12}i}\right)^4 = 3^4 e^{\frac{17\pi}{12}\times 4i}$ | M1 | Applies De Moivre's theorem correctly to $z_2$: modulus is $3^4$ and argument is $4\times\frac{17\pi}{12}$. Correct answer with no working scores no marks. |
| $= \frac{81}{2} - \frac{81\sqrt{3}}{2}i$ or $\frac{81}{2}-i\frac{81\sqrt{3}}{2}$ or $\frac{81}{2}+i\left(-\frac{81\sqrt{3}}{2}\right)$ | A1 | Correct exact answer in required form. Do not allow e.g. $\frac{81}{2}\left(1-\frac{81\sqrt{3}}{2}i\right)$ or $\frac{81-81\sqrt{3}i}{2}$ unless correct form seen previously. |

## Part (c)(i):
| $z_1$ and $z_2$ correctly positioned: $z_1$ approximately on $y=-x$ in quadrant 2, $z_2$ below $y=x$ closer to origin than $z_1$ in quadrant 3 | B1 | Look for correct quadrants. Points usually labelled but mark positively if clear which is which. |
| $\frac{z_1}{z_2}$ in correct quadrant following through from (b)(i) | B1ft | Follow through their answer to (b)(i). Point sometimes labelled $z_3$. |

## Part (c)(ii):
| Draws perpendicular bisector of $z_1 z_2$ (solid or dashed) or draws a line crossing $z_1 z_2$ and shades one side | M1 | Line that is perpendicular bisector of $z_1 z_2$ or crosses $z_1 z_2$ with shading. |
| Perpendicular bisector of $z_1 z_2$ drawn with either side shaded, clear they are not discounting the upper region; $z_1$ in quadrant 2 and $z_2$ in quadrant 3 | A1 | Note: region may be drawn on separate diagram. You do not need to see a line joining $z_1$ to $z_2$. |

Show LaTeX source

\begin{enumerate}
  \item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}

$$z _ { 1 } = - 4 + 4 i$$

(a) Express $\mathrm { z } _ { 1 }$ in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$, where $r \in \mathbb { R } , r > 0$ and $0 \leqslant \theta < 2 \pi$

$$z _ { 2 } = 3 \left( \cos \frac { 17 \pi } { 12 } + i \sin \frac { 17 \pi } { 12 } \right)$$

(b) Determine in the form $a + \mathrm { i } b$, where $a$ and $b$ are exact real numbers,\\
(i) $\frac { Z _ { 1 } } { Z _ { 2 } }$\\
(ii) $\left( z _ { 2 } \right) ^ { 4 }$\\
(c) Show on a single Argand diagram\\
(i) the complex numbers $z _ { 1 } , z _ { 2 }$ and $\frac { z _ { 1 } } { z _ { 2 } }$\\
(ii) the region defined by $\left\{ z \in \mathbb { C } : \left| z - z _ { 1 } \right| < \left| z - z _ { 2 } \right| \right\}$

\hfill \mbox{\textit{Edexcel CP1 2023 Q3 [10]}}

This paper (8 questions)

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Q1 5 Q2 6 Q3 10 Q4 5 Q5 12 Q6 12 Q7 12 Q10

Answer	Marks	Guidance
\(	z_1	= \sqrt{(-4)^2+4^2}\) or \(\arg z_1 = \pi - \frac{\pi}{4}\)
\((z_1 =)\, 4\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\) or e.g. \((z_1=)\sqrt{32}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)\)	A1	Correct expression. The "\(z_1=\)" is not required. Correct answer with no working scores both marks.

Answer	Marks	Guidance
\(\frac{z_1}{z_2} = \frac{4\sqrt{2}}{3}\left(\cos\left(\frac{3\pi}{4}-\frac{17\pi}{12}\right)+i\sin\left(\frac{3\pi}{4}-\frac{17\pi}{12}\right)\right)\) or via exponential form or via Cartesian multiplication by conjugate	M1	Employs correct method: divides moduli and subtracts arguments, or uses exponential form, or converts \(z_2\) to Cartesian and multiplies by conjugate. Correct answer with no working scores no marks.
\(= -\frac{2\sqrt{2}}{3} - \frac{2\sqrt{6}}{3}i\) or \(-\frac{2\sqrt{2}}{3} - i\frac{2\sqrt{6}}{3}\) or \(-\frac{2\sqrt{2}}{3}+i\left(-\frac{2\sqrt{6}}{3}\right)\)	A1	Correct exact answer in required form. Do not allow e.g. \(-\frac{2}{3}(\sqrt{2}+\sqrt{6}i)\) or \(\frac{-2\sqrt{2}-2\sqrt{6}i}{3}\) unless correct form seen previously.

Answer	Marks	Guidance
\(z_2^4 = 3^4\left(\cos\left(4\times\frac{17\pi}{12}\right)+i\sin\left(4\times\frac{17\pi}{12}\right)\right)\) or \((z_2)^4 = \left(3e^{\frac{17\pi}{12}i}\right)^4 = 3^4 e^{\frac{17\pi}{12}\times 4i}\)	M1	Applies De Moivre's theorem correctly to \(z_2\): modulus is \(3^4\) and argument is \(4\times\frac{17\pi}{12}\). Correct answer with no working scores no marks.
\(= \frac{81}{2} - \frac{81\sqrt{3}}{2}i\) or \(\frac{81}{2}-i\frac{81\sqrt{3}}{2}\) or \(\frac{81}{2}+i\left(-\frac{81\sqrt{3}}{2}\right)\)	A1	Correct exact answer in required form. Do not allow e.g. \(\frac{81}{2}\left(1-\frac{81\sqrt{3}}{2}i\right)\) or \(\frac{81-81\sqrt{3}i}{2}\) unless correct form seen previously.

Answer	Marks	Guidance
\(z_1\) and \(z_2\) correctly positioned: \(z_1\) approximately on \(y=-x\) in quadrant 2, \(z_2\) below \(y=x\) closer to origin than \(z_1\) in quadrant 3	B1	Look for correct quadrants. Points usually labelled but mark positively if clear which is which.
\(\frac{z_1}{z_2}\) in correct quadrant following through from (b)(i)	B1ft	Follow through their answer to (b)(i). Point sometimes labelled \(z_3\).

Answer	Marks	Guidance
Draws perpendicular bisector of \(z_1 z_2\) (solid or dashed) or draws a line crossing \(z_1 z_2\) and shades one side	M1	Line that is perpendicular bisector of \(z_1 z_2\) or crosses \(z_1 z_2\) with shading.
Perpendicular bisector of \(z_1 z_2\) drawn with either side shaded, clear they are not discounting the upper region; \(z_1\) in quadrant 2 and \(z_2\) in quadrant 3	A1	Note: region may be drawn on separate diagram. You do not need to see a line joining \(z_1\) to \(z_2\).