| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Modulus-argument form conversion |
| Difficulty | Moderate -0.8 This is a straightforward application of modulus-argument form conversion and properties of complex number multiplication. Part (a) requires basic conversion from Cartesian to polar form (assuming z₁ = 2 + 3i, likely a typo). Part (b) uses |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg(z₁) + arg(z₂) to find z₂, then convert back to Cartesian form. These are standard textbook exercises with no novel problem-solving required, making it easier than average. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02d Exponential form: re^(i*theta)4.02e Arithmetic of complex numbers: add, subtract, multiply, divide |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | z_1 | = \sqrt{13}\) and \(\arg z_1 = \tan^{-1}\!\left(\dfrac{3}{2}\right)\) |
| \(z_1 = \sqrt{13}(\cos 0.9828 + \mathrm{i}\sin 0.9828)\) | B1ft | Follow through on \(r= |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find modulus of \(z_2\); e.g. \( | z_1z_2 | = |
| Complete method to find argument of \(z_2\); e.g. \(\arg(z_1z_2)=\arg(z_1)+\arg(z_2)=\dfrac{\pi}{4}\) | M1 | |
| \(\arg(z_2)=\dfrac{\pi}{4}-\tan^{-1}\!\left(\dfrac{3}{2}\right)\) or \(\dfrac{\pi}{4}-0.9828\) or \(-0.1974...\) | A1 | |
| Writes \(z_2\) in form \(r(\cos\theta+\mathrm{i}\sin\theta)\); deduces \(z_2 = 15-3\mathrm{i}\) only | ddM1 A1 | ddM1 dependent on both previous M marks; A1 mark AO 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z_1z_2=(a+b\mathrm{i})(2+3\mathrm{i})=(2a-3b)+(3a+2b)\mathrm{i}\) | ||
| \((2a-3b)^2+(3a+2b)^2=(39\sqrt{2})^2\) or \(3042 \Rightarrow a^2+b^2=234\) | M1 A1 | |
| \(\arg[(2a-3b)+(3a+2b)\mathrm{i}]=\dfrac{\pi}{4} \Rightarrow \tan^{-1}\!\left(\dfrac{3a+2b}{2a-3b}\right)=\dfrac{\pi}{4} \Rightarrow \dfrac{3a+2b}{2a-3b}=1\) | M1 A1 | Note: \(\tan^{-1}\!\left(\dfrac{2a-3b}{3a+2b}\right)=\dfrac{\pi}{4}\) scores M0A0ddM0A0 |
| \(\Rightarrow a=-5b\) | ||
| Solves \(a=-5b\) and \(a^2+b^2=234\) | ddM1 | Dependent on both previous M marks |
| Deduces \(z_2=15-3\mathrm{i}\) only | A1 | AO 2.2a |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|z_1| = \sqrt{13}$ and $\arg z_1 = \tan^{-1}\!\left(\dfrac{3}{2}\right)$ | B1 | Implied by awrt 0.98 or awrt 56.3° |
| $z_1 = \sqrt{13}(\cos 0.9828 + \mathrm{i}\sin 0.9828)$ | B1ft | Follow through on $r=|z_1|$, $\theta = \arg z_1$; $r$ exact, $\theta$ correct to 4 s.f. |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find modulus of $z_2$; e.g. $|z_1z_2|=|z_1|\times|z_2|=39\sqrt{2} \Rightarrow |z_2|=3\sqrt{26}$ or $\sqrt{234}$ | M1 A1 | |
| Complete method to find argument of $z_2$; e.g. $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)=\dfrac{\pi}{4}$ | M1 | |
| $\arg(z_2)=\dfrac{\pi}{4}-\tan^{-1}\!\left(\dfrac{3}{2}\right)$ or $\dfrac{\pi}{4}-0.9828$ or $-0.1974...$ | A1 | |
| Writes $z_2$ in form $r(\cos\theta+\mathrm{i}\sin\theta)$; deduces $z_2 = 15-3\mathrm{i}$ only | ddM1 A1 | ddM1 dependent on both previous M marks; A1 mark AO 2.2a |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z_1z_2=(a+b\mathrm{i})(2+3\mathrm{i})=(2a-3b)+(3a+2b)\mathrm{i}$ | | |
| $(2a-3b)^2+(3a+2b)^2=(39\sqrt{2})^2$ or $3042 \Rightarrow a^2+b^2=234$ | M1 A1 | |
| $\arg[(2a-3b)+(3a+2b)\mathrm{i}]=\dfrac{\pi}{4} \Rightarrow \tan^{-1}\!\left(\dfrac{3a+2b}{2a-3b}\right)=\dfrac{\pi}{4} \Rightarrow \dfrac{3a+2b}{2a-3b}=1$ | M1 A1 | Note: $\tan^{-1}\!\left(\dfrac{2a-3b}{3a+2b}\right)=\dfrac{\pi}{4}$ scores M0A0ddM0A0 |
| $\Rightarrow a=-5b$ | | |
| Solves $a=-5b$ and $a^2+b^2=234$ | ddM1 | Dependent on both previous M marks |
| Deduces $z_2=15-3\mathrm{i}$ only | A1 | AO 2.2a |
---\begin{enumerate}
\item Given that
\end{enumerate}
$$\begin{aligned}
z _ { 1 } & = 2 + 3 \\
\left| z _ { 1 } z _ { 2 } \right| & = 39 \sqrt { 2 } \\
\arg \left( z _ { 1 } z _ { 2 } \right) & = \frac { \pi } { 4 }
\end{aligned}$$
where $z _ { 1 }$ and $z _ { 2 }$ are complex numbers,\\
(a) write $z _ { 1 }$ in the form $r ( \cos \theta + \mathrm { i } \sin \theta )$
Give the exact value of $r$ and give the value of $\theta$ in radians to 4 significant figures.\\
(b) Find $z _ { 2 }$ giving your answer in the form $a + \mathrm { i } b$ where $a$ and $b$ are integers.
\hfill \mbox{\textit{Edexcel CP AS 2020 Q2 [8]}}