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I very much agree, and the stripping of geometry from Advanced, MX1, and MX2 is problematic for this reason. It is highlighted by problems with vector proofs in MX1 and in various parts of MX2 including complex numbers. I have recently posted an algebraic solution to what should be handled...

If \cfrac{z - 1}{z - i} is real, then the locus of z has the ratio of the moduli as a real constant, giving a circle where the line joining the two fixed points (here z = 1 and z = i) is divided internally and its projection divided externally in the ratio of the constant to 1. The centre of...

Put slightly differently... Let A and B be the points such that \overrightarrow{OA} = z_1 and \overrightarrow{OB} = z_2 where O is the origin. The locus \left|z - z_1\right| = \left|z - z_2\right| is the set of points P such that the distance from A to P and the distance from B to P are equal...

Further, if the placement of z in the first quadrant doesn't work, you need to find a suitable placement. For example, \arg{\cfrac{z+2}{z - 1}} = \cfrac{\pi}{4} We need \arg{(z + 2)} = \alpha, the direction of the vector from -2 to z, to be -\pi < \alpha < 0. Drawing the ray for this vector...

I wouldn't try and remember a most-cases solution, derive it from real part equals zero on a case-by-case basis.

\arg{\left(\cfrac{z-1}{z+3}\right)} = \cfrac{\pi}{6} \quad \implies \quad \arg{(z-1)} - \arg{(z+3)} = \cfrac{\pi}{6} Consider the vector from z = 1 to some point in the first quadrant so that the angle between the vector and the real axis in the positive direction is \alpha, an acute angle...

One way to avoid calculator-only answers is to ask questions where decimalised answers won't be acceptable, like \cfrac{2z + iw}{7}

It is the circle x^2 + (y - 1)^2 = 2 but only for the part above the real axis

This type of problem should be solved with circle geometry as the algebra is difficult because it can be easy to miss the inclusion / exclusion criteria. I know the circle geometry is no longer in the syllabus... learn it anyway.

An easy way to establish that \cfrac{z^5 - 1}{z - 1} = 1 + z + z^2 + z^3 + z^4 is to sum the GP on the RHS. 1 + z + z^2 + z^3 + z^4 is a GP of five terms (n = 5) with first term 1 (a = 1) and common ratio of z (r = z) which sums to \begin{align*} S_n = \cfrac{a\left(r^n - 1\right)}{r - 1} &=...

You can use \bar{z} to get \bar{z} but I prefer \overline{} so I can get \overline{z} is \overline{z} and \overline{a+ib} is \overline{a+ib} because \bar{a+ib}, giving \bar{a+ib} looks ridiculous.

Though not the method requested, it is also possible to shift the axis of rotation y = 10 down by 10 units, so that it becomes the x-axis. The shifted parabola is then y = 2x^2 - 10. Then, using usual volumes by revolution: \begin{align*} V_{\text{parabola}} = \pi\int y^2\ dx &=...

One definition of an ellipse that is mathematically equivalent to the focus-directrix definition is: An ellipse is the locus of points P that are related to two distinct fixed points S and S' (its foci) by the relationship \left|\overrightarrow{PS}\right| + \left|\overrightarrow{PS'}\right| =...

This question really relates to the old syllabus where ellipses were studied as part of the conics module. It really isn't reasonable to ask without the coverage of conics.

\begin{align*} wz &= \cfrac{3+4i}{5} \times \cfrac{5 + 12i}{13} \\ &= \left(\cfrac{3}{5} \times \cfrac{5}{13} + \cfrac{4i}{5} \times \cfrac{12i}{13}\right) + \left(\cfrac{3}{5} \times \cfrac{12i}{13} + \cfrac{4i}{5} \times \cfrac{5}{13}\right) \\ &= \cfrac{15 + 48i^2}{65} + \cfrac{(36 +20)i}{65}...

Thank you, I hope that they make sense and help lots of BoS contributors. :)

For question 17, a polynomial P(x) = 0 can inly have a multiple root if there are any x-values for which P(x) = P'(x) = 0, using the multiple root theorem. If you attempt to solve the equation P(x) = P'(x), you should find it has only one solution, x = 0. However, P(0) = 1 \neq 0 and so the...

Some help for (a) in a related thread. Having established that the solutions are: x^2 = \cfrac{-A \pm \sqrt{A^2 - 4B}}{2} = - \cfrac{A \mp \sqrt{A^2 - 4B}}{2} < 0 there are two pairs of solutions, x^2 = -c^2 and x^2 = -d^2 where one of c^2 is one of \cfrac{A + \sqrt{A^2 - 4B}}{2} \qquad...

A complex number has only one principal argument, which is the argument \theta satisfying \theta \in \big(-\pi,\ \pi\big]

The factorised form of the equation must be a\left(x - \alpha\right)\left(x - \overline{\alpha}\right) = ax^2 + bx + c. By expanding the LHS and equating coefficients, you should find that b and c must be real so long as a is real.