The good news is that pretty much everyone had the right general idea about how to approach every problem. The bad news is that there were many cases of getting into trouble with algebra and then giving up.
Lesson 1: everyone needs more practice. Much more practice.
Lesson 2: everyone makes slips in mathematical working. Everyone. David Hilbert did. The thing to learn is that, when you end up with a bit of working that seems wrong, you look back to see where you’ve made the slip, and fix it, rather than give up.
Q.1: complex roots. Mostly ok. But some of you said z3 = 8 cis (-π/4), and went on to add and subtract “spoke” angles of 2π/3 to z3… instead of calculating the first cube root as 2 cis (-π/12) and then adding and subtracting “spoke” angles of 2π/3 to the cube root.
Q.2: Möbius transformations. Most people were good with the geometric method. You’ll have to learn to add a few more words of explanation for the benefit of the markers before the actual exam, but that’s fine.
A couple of you stated the z-diameter ends in line with the pole as -3 and 3. But the pole is -i (everyone got that right), so the z-diameter ends in line with it are -3i and 3i.
Tobi had the method completely correct, but made a slip in arithmetic and got the answer to be another circle centred on the origin. That should have rung alarm bells. You’ll get another circle centred on the origin if your transformation is w=a/z for some a – i.e. if it’s just an inversion and a rotation and an enlargement, with no translation – and not otherwise.
Everyone had the basic idea on the algebraic method. But most of you messed up the algebra at some point.
There must have been some weakness in my teaching of complex numbers in FP1, because some of you still mix up z2 and |z|2. They are very different!
If z=x+iy, z2 is a complex number multiplied by another complex number to produce yet another complex number, equal to x2-y2+2ixy.
|z|2 is a positive real number, a distance squared, equal to x2+y2, or to r2 if z is written in r cis θ form,
Then there were straighforward slips in algebra, such as we all make. But you know that you must end up with an equation in u and v which has equal coefficients for u2 and for v2 (zero coefficients if the w-locus is a line, non-zero but equal coefficients if the w-locus is a circle).
If your working has led to unequal coefficients for u2 and for v2, you’ve made a slip. Check your working and find it.
Some of you had forgotten the very simple method for finding out whether the inside of the z-locus maps to the outside or the inside of the w-locus. Just pick one point inside the z-locus and check. (Also, z=1 is all right as a point to check, but z=0 is simpler).
Q.3: Inequality. Tobi was the only one of you to do a good diagram. His diagram was perfect. But he didn’t mark clearly where the 3/(x+3) curve was above 1-(4/x). The diagram could have shown him where he had made a slip in working out the critical points (of the type we all make), and on which sides of the critical points the inequality is valid, only he didn’t use it.
You all tried to multiply up by the correct factor, (x+3)2.x2, but then some of you “lost” part of that factor.
3/(x+3) multiplied by (x+3)2.x2 is 3(x+3).x2, isn’t it, not 3x2?
Easy check here: you should find that the singularities in the expressions you’re dealing with (the x-values where the expression involves dividing by zero) “fall out” immediately as factors of the equation you get by multiplying out.
In this question, you end up with an equation
In solving -x2+4x+12=0
a couple of you came up with the roots -6 and -2, where it should be 6 and -2
The lesson, I think, is that it’s easy to make mistakes if you try to solve a quadratic with a negative coefficient on x2
You minimise error by multiplying both sides of the equation by -1 and getting
I’m pretty sure none of you would have made the mistake with the sign of roots if you’d solved the quadratic starting from that. Or if you’d had a good diagram like Tobi’s and used it.
Q.4 – series and method of differences.
Everyone was fine with part (a).
Some of you tried to do part (b) just by showing that it’s true for r=1. It is, but that doesn’t prove it is true for all values of r.
If you have an expression for (2r+1)3, as you all had after doing part (a), then you only need to change some signs to get from it an expression for (2r-1)3. Then subtract, and you have your answer to part (b). It’s really just GCSE level algebraic manipulation, not any sort of “further” maths.
Everyone had the right basic idea with part (c), only a lot of you fumbled the working at some point and then gave up. The same lesson again: Everyone makes slips in working. The good mathematician makes slips, but then doesn’t give up. She or he looks back to see where the slip is, and fixes it.
Q.5 – De Moivre
Everyone had the right basic idea about De Moivre’s theorem, though some of you started your working by writing just sin 5θ instead of cos 5θ + i sin 5θ. But then it was the same story of most of you making a slip somewhere in the working and then giving up because what you saw looked wrong.
Nii did this perfectly except for forgetting that sin θ = 0 has the solution θ = π as well as the solution θ = 0, and sin θ = 0.8891 has the solution θ = 5.188 as well as the solution θ = 1.095, etc.