Edexcel Further Maths May 2025 CP2 Q4: Complex Numbers (Equations & Argand Loci)
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: Argand Locus Intersection & Distance Bounds
Examiners do not invent new mathematics each year — they test consistent structural archetypes. In this paper, Q4 tests your ability to navigate the boundary between conceptual algebra and precise numerical computation.
$$|z - z_1| = |z - z_2| \text{ (Perpendicular Bisector)}$$
$$|z - z_0| = r \text{ (Circle with centre } z_0 \text{ and radius } r)$$
Examiner Traps & Common Mark-Scheme Penalties
- Not shading the correct inequality region in the Argand diagram.
- Calculating minimum distance without drawing the perpendicular line from the origin to the locus.
Formal Mathematical Solution
Below is the full step-by-step derivation meeting official mark-scheme criteria for method (M) and accuracy (A) marks:
Geometric Interpretation of the Locus
Distance from origin to centre is $\sqrt{3^2 + 4^2} = 5$. Maximum $|z| = 5 + 2 = 7$, minimum $|z| = 5 - 2 = 3$.
🎙️ Read Spoken Video Explanation (234 segments, 1331 words) ▾ Expand
Unedited transcript of Tutor Sheefa's spoken audio instructions during the walkthrough:
Let's take a look at question number four. For question number four, we have a complex number question and we are given two complex number Z1 and Z2. Okay, given that B is more than D and Z1 plus Z2 is real. That means when we plus the imaginary part of Z1 plus Z2 it would be equal to zero. In other words, what we have here is this B plus D is equal to zero. Next, modulus of Z1 is equal to square root 13. So, what we're looking for is this square root of A squared plus B squared is equal to square root 13. A squared plus B squared is equal to 13. Next, modulus of Z2 is equals to five. So, square root of C squared plus D squared is equal to five. If we square both side, we will get C squared plus D squared equals to 25. Plus, real value of Z2 minus Z1 is equal to two. So, real value Z2 minus Z1 Okay, that is C minus A. This is equal to two. Okay, that's it. We have four different equation. I will name this one two three and four. So, what we will do first is to rearrange equation one and four. So, when I rearrange one and four, I'll get B is equals to negative D. From equation four, I will get C is equal to 2 + A. Now, what I would like to do next is Okay, subbed into two and three. Let me name this first. Okay, let me name this equation five and here is equation number six. So, substitute five and six into two and three. So, now I have equation two. Equation two is A squared plus B squared is equal to 13. So, A squared plus negative D squared is equal to 13. A squared plus D squared is equal to 13. How about equation number three? Equation number three is C squared plus D squared is equals to 25. What is C? C is 2 + A squared plus D squared is equals to 25. Let's expand this and I will get 4 + 4 A plus A squared plus D squared equals to 25. If I rearrange and I will get 4 a plus a squared plus d squared is equals to 21. Let me name this equation seven and eight. So what I would like to do now is substitute, okay? Seven into eight. So what I will have is this. 4 a plus Okay, a squared plus d squared is the whole thing here. So plus 13 equals to 21. 4 a is equals to 21 minus 13. That is eight. Therefore, a is equals to two. Now we know the value of a. We can find, okay? C straight away. Sub a equals to two into six. So then I will get c is equal to two plus two. C is equals to four. We have done finding the value of a and c. Next, we can find b or d. Let's use equation seven. So sub a equals to two into six and seven. So this is equation six. So this one equation seven. Equation seven is a squared plus d squared equals to 13. So what is our a? Our a is two. So d squared is equals to 13 minus four. Which is equals to nine. So, D is equals to plus minus three. So, let's write one more line. So, D is equal to plus minus square root nine. Now, what do we know? We know that so, B is equals to negative D. Therefore, B will be equals to plus minus plus three. What do we know? Let's take a look at the first rule. B must be more than D. Because of B must be more than D, then we can conclude that, okay, B is equal to positive three and D is equals to negative three. So, these are all the final answers. So, these are the final answers for B and D. And this is the final answer for C and this is the final answer for A. Let's take a look at part two. Okay, what is part two? Two, on the same Argand diagram, sketch the locus of point Z which satisfy this. Showing the coordinates of any points of intersection with the axis. Okay. So, what we have is this. Okay, modulus Z minus 12 equal to seven. So, this is our center point. What is our center point? 12 zero. And this would be our radius. Radius equal to seven. How about the next one? The next one we have W minus 5 I. So, equals to four. That means our center for this circle is 0 5. And then our radius is equals to four. Let's draw this. Okay. So, what are the points that we need? Okay. So, we need 12 here. And then for Y, we need five. Now, let's take a look for first locus Z. Okay, the radius is seven. So, you have to plus seven here, you have to minus seven here. So, we will have 19 and we will have five. Okay. So, let's draw a circle. Okay, in exam, make sure you use compass. Okay, here because I'm just showing you, I can simply use this app, but in reality, what you have to do is you have to use your compass. Okay, next, we have locus W. Locus W 5 minus 4, this one will be one and here it will be nine. Okay, let's draw another circle. What I have to fix next is the labeling. I have to make sure I label my imaginary and real. Next, this one label as W. That's it. We have finished with part two A. Next, part B. Determine the range of possible values for Z minus W modulus. Okay, so we want to find the maximum and minimum distance between Z and W. Let's think of the maximum distance. The maximum distance, right, must comes from Okay, where we draw a straight line and it will pass through the center. Okay, let me adjust. It should passing through the center. So, what I have here is the maximum distance. Okay, how about the minimum distance? The minimum distance between Z and W would be just this part. Here, minimum distance. Okay, but let me just color it in pink. So, how can we find this? First, what we have to do is we have to find the distance between these two center. Okay, so what we're finding is part two B. Okay, so what we have is a triangle. Okay, the distance here is five, distance here is 12, and you want to find this, the hypotenuse. So, it would be H squared is equals to five squared plus 12 squared. H squared is equals to Okay, five squared plus 12 squared. That is equal to 169. Therefore, H is equal to square root 169, which is 13. Okay. For the maximum, okay, distance, it would be 13 plus radius one plus radius two. So, you have the distance over here, and you need to plus the two radius. That is how you find the maximum distance. So, plus R1 plus R2. So, 13 plus seven plus four, which is equals to 24. Moving on to finding our minimum distance. Our minimum distance would come from, okay, this distance between these two center and minus our R1 and R2. Then, we will get this small pink dot. So, it will be 13 minus seven minus four, which is 13 minus seven minus four. That is equals to two. Now, what they want us to do is they want us to write the range of possible values. So, just write it. So, Z minus W modulus is in between two and 24. With that, we are done with question four.