| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Equations with z/z* or zz* terms |
| Difficulty | Standard +0.3 Part (a) is trivial algebraic verification that zz* = a² + b² is real. Part (b) requires setting up two equations from the given conditions and solving simultaneously, but the algebra is straightforward: from zz* = 18 get a² + b² = 18, from z/z* multiply to get z² = (constant), then solve. Standard Core Pure 1 technique with no novel insight required, slightly easier than average. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02g Conjugate pairs: real coefficient polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(z^* = a - bi\), then \(zz^* = (a+bi)(a-bi) = \ldots\) | M1 | States or implies \(z^* = a-bi\) and finds expression for \(zz^*\). |
| \(zz^* = a^2 + b^2\), therefore a real number | A1 | Achieves \(zz^* = a^2+b^2\) and draws conclusion that \(zz^*\) is a real number. Accept \(\in \mathbb{R}\) as conclusion, but not just "no imaginary part". |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{z}{z^*} = \frac{a+bi}{a-bi} = \frac{(a+bi)(a+bi)}{(a-bi)(a+bi)} = \frac{(a^2-b^2)+2abi}{a^2+b^2} = \frac{7}{9}+\frac{4\sqrt{2}i}{9}\) or \(\frac{z}{z^*} = \frac{z^2}{zz^*} = \frac{z^2}{18} \Rightarrow z^2 = 14+8\sqrt{2}i\) | M1 | Starts process using conjugate to form equation with real denominators, without \(z^*\) or \(i^2\) in equation. |
| Forms two equations from \(a^2+b^2=18\): \(\frac{a^2-b^2}{18}=\frac{7}{9}\) or \(\frac{2ab}{18}=\frac{4\sqrt{2}}{9}\) | M1, A1 | M1: Uses given information to form two equations involving \(a\) and \(b\). A1: Any two correct equations. |
| Solves simultaneously e.g. \(a^2+b^2=18\) and \(a^2-b^2=14\) leading to value for \(a\) or \(b\) | dM1 | Dependent on previous M mark. |
| \(z = \pm(4+\sqrt{2}i)\) | A1 | Deduces correct complex numbers and no extras. Do not accept \(\pm 4 \pm \sqrt{2}i\). |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $z^* = a - bi$, then $zz^* = (a+bi)(a-bi) = \ldots$ | M1 | States or implies $z^* = a-bi$ and finds expression for $zz^*$. |
| $zz^* = a^2 + b^2$, therefore a real number | A1 | Achieves $zz^* = a^2+b^2$ and draws conclusion that $zz^*$ is a real number. Accept $\in \mathbb{R}$ as conclusion, but not just "no imaginary part". |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{z}{z^*} = \frac{a+bi}{a-bi} = \frac{(a+bi)(a+bi)}{(a-bi)(a+bi)} = \frac{(a^2-b^2)+2abi}{a^2+b^2} = \frac{7}{9}+\frac{4\sqrt{2}i}{9}$ or $\frac{z}{z^*} = \frac{z^2}{zz^*} = \frac{z^2}{18} \Rightarrow z^2 = 14+8\sqrt{2}i$ | M1 | Starts process using conjugate to form equation with real denominators, without $z^*$ or $i^2$ in equation. |
| Forms two equations from $a^2+b^2=18$: $\frac{a^2-b^2}{18}=\frac{7}{9}$ or $\frac{2ab}{18}=\frac{4\sqrt{2}}{9}$ | M1, A1 | M1: Uses given information to form two equations involving $a$ and $b$. A1: Any two correct equations. |
| Solves simultaneously e.g. $a^2+b^2=18$ and $a^2-b^2=14$ leading to value for $a$ or $b$ | dM1 | Dependent on previous M mark. |
| $z = \pm(4+\sqrt{2}i)$ | A1 | Deduces correct complex numbers and no extras. Do not accept $\pm 4 \pm \sqrt{2}i$. |
---\begin{enumerate}
\item Given that $z = a + b \mathrm { i }$ is a complex number where $a$ and $b$ are real constants,\\
(a) show that $z z ^ { * }$ is a real number.
\end{enumerate}
Given that
\begin{itemize}
\item $z z ^ { * } = 18$
\item $\frac { z } { z ^ { * } } = \frac { 7 } { 9 } + \frac { 4 \sqrt { 2 } } { 9 } \mathrm { i }$\\
(b) determine the possible complex numbers $z$
\end{itemize}
\hfill \mbox{\textit{Edexcel CP1 2022 Q7 [7]}}