Edexcel Further Maths May 2025 CP1 Q3: Complex Numbers (Purely Imaginary Proof)
Comprehensive worked solution and examiner mark-scheme analysis by Tutor Sheefa. Focuses on mark allocation, algebraic traps, and step-by-step mathematical reasoning.
The Exam Archetype: The Purely Imaginary Conjugate Identity
Examiners do not invent new mathematics each year — they test consistent structural archetypes. In this paper, Q3 tests your ability to navigate the boundary between conceptual algebra and precise numerical computation.
$$w \in i\mathbb{R} \iff w + w^* = 0 \iff \text{Re}(w) = 0$$
$$z = x + iy \implies z z^* = |z|^2 = x^2 + y^2$$
Examiner Traps & Common Mark-Scheme Penalties
- Assuming $z$ is real instead of expressing $z = x + iy$.
- Sign errors during conjugate expansion $(a + ib)(a - ib) = a^2 + b^2$.
Formal Mathematical Solution
Below is the full step-by-step derivation meeting official mark-scheme criteria for method (M) and accuracy (A) marks:
Algebraic Substitution and Conjugate Multiplication
If $|z| = 1$, the numerator becomes $2(1 - 1) = 0$, proving that $\text{Re}(w) = 0$ and therefore $w$ is purely imaginary.
🎙️ Read Spoken Video Explanation (111 segments, 678 words) ▾ Expand
Unedited transcript of Tutor Sheefa's spoken audio instructions during the walkthrough:
Question number three. We have a complex number Z equals to A + B I where A and B are real constants. Given that Z over Z conjugate is purely imaginary. For part A, we must show that A² is equal to B². For part B, given also that Z * Z conjugate is equals to 50. Determine the possible complex numbers for Z. Okay, let's begin with part A. Okay, Z is equals to A + B I. Therefore, Z conjugate would be A minus B I. Z over Z conjugate it will be A + B I over A minus B I to rationalize this okay we will multiply both our denominator and numerator with a + b I let's expand For our numerator we will get a² + a b i + a b i + b² i² over a² + a b i - a b i - b² i². Okay, we do know that. Okay, note I² is equ= to -1. So that is what will be replaced here and here. So for our numerator we will get a² + 2 a b i + oh wait multiply by -1 so it will be - b² over a² a b i - a b i those terms will eliminate each other so here it b + b². What we're going to do is we're going to rearrange okay the real numbers together and the imaginary numbers together. So we will get for our real part a² - b² over a² + b² and then we + 2 a b over a² + b² i. What did they say? They say that Z over Z conjugate is purely imaginary. That means the real part here would be equal to zero. So that would be our next step. Okay. The real part for Z over Z conjugate is equal to zero. a² - b² over a² + b² = 0. Okay, I'll move this up a bit here. When we multiply our denominator and bring it to the other side, it will be zero. So, we'll get a² - b² = 0. Therefore, a² is equal to b². We got exactly what we wanted to show. Okay, moving on to part B. Okay, I'm going to erase this so that we have space for the working for part B. given also that Z ult* by Z conjugate is 50. So B A + B I multiply okay a - b I is equals to 50. So let's expand this. We will get oh we've done this before a + b i it will be a² + b². So here it will be a² + b² is equ= to 50. Okay. So we have a quadratic equation. A² is equ= to B² and another one is A² + B square is equals to 50. Okay. So let's name this. So a² is equ= to b². We'll name this as equation one. And here would be equation number two. So let 1 into 2. So we will get a² + a² is = 50. 2 a² is = 50. And then a² is = 25. A is equ= to + - 5 because of A² is equ= to B² that means B will also be b² is also 25 and B is also + - 5. Okay. So therefore we have two values for a and two values for b. So what are the possible equation? It will be when a is positive 5 it will be a oh wait it will be 5 + 5 i. That's one option. It can also be 5 - 5 i. And then when a is negative it will be -5 + 5 I and then -5 - 5 I. So these four are all positive options for Z's. So Z1, Z2, Z3 and Z number four. Okay, with that we are done with this part. We managed to show that a square is equals to b² and we also managed to find all four possible complex numbers for z.