| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Arithmetic |
| Type | Complex conjugate properties and proofs |
| Difficulty | Easy -1.2 This question tests basic definitions of complex conjugate and modulus with a straightforward algebraic verification. Part (i) requires simple recall of standard notation, and part (ii) involves direct substitution showing zz* = a² + b² = |z|², requiring minimal algebraic manipulation and no problem-solving insight. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z) |
| Answer | Marks | Guidance |
|---|---|---|
| \( | z | = \sqrt{a^2 + b^2}\) |
| \(z^* = a - b\text{j}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(zz^* = (a + b\text{j})(a - b\text{j}) = a^2 + b^2\) | M1 | |
| \( | z | ^2 = a^2 + b^2\) |
| \(zz^* - | z | ^2 = a^2 + b^2 - (a^2 + b^2) = 0\) |
# Question 2:
**(i)**
$|z| = \sqrt{a^2 + b^2}$ | B1 |
$z^* = a - b\text{j}$ | B1 |
**(ii)**
$zz^* = (a + b\text{j})(a - b\text{j}) = a^2 + b^2$ | M1 |
$|z|^2 = a^2 + b^2$ | A1 |
$zz^* - |z|^2 = a^2 + b^2 - (a^2 + b^2) = 0$ | A1 | conclusion clearly stated
---2 (i) Given that $z = a + b \mathrm { j }$, express $| z |$ and $z ^ { * }$ in terms of $a$ and $b$.\\
(ii) Prove that $z z ^ { * } - | z | ^ { 2 } = 0$.
\hfill \mbox{\textit{OCR MEI FP1 2006 Q2 [5]}}