• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
  • Like us on facebook here

Recent content by fan96

  1. fan96

    complex number question help!!!

    In my solution, the z for which the substitution \cos \theta + i \sin \theta occurs are the eighth roots of -1 , not the actual solutions of ii). This is a valid substitution, just not the typical one that would be expected here.
  2. fan96

    complex number question help!!!

    One possible solution: Define f(z) = \frac{z-1}{z+1}, g(z) = \frac{1+z}{1-z}. Note that these functions are inverses when z \ne -1, 1 . Let w_n = \exp(i \cdot ((2n+1)\pi)/8) be the solutions to w^8 = -1. Now if \left(\frac{z-1}{z+1} \right)^8 = f(z)^8 = -1, then we know that any...
  3. fan96

    complex number question help!!!

    There appears to be a sign error here: (The first attempt at a solution fails because it assumes all solutions to part ii) lie on the unit circle.)
  4. fan96

    inequality

    My bad, I didn't check that last step correctly. This may not quite be within the MX2 syllabus, but if you differentiate the expression \frac{1 - a^2b^2}{(1+a^2)(1+b^2)} + \frac {a+b}{\sqrt{(1+a^2)(1+b^2)}} twice, as a function of a and then as a function of b , it can be shown that...
  5. fan96

    inequality

    First rearrange the inequality: \frac{1-a^2}{1+a^2} + \frac{1-b^2}{1+b^2} \le 2 \sqrt 2 - \frac 2 {\sqrt{1+c^2}} \iff 1 - \frac{2a^2}{1+a^2} + 1 - \frac{2b^2}{1+b^2} \le 2 \sqrt 2 - \frac 2 {\sqrt{1+c^2}} \iff 1 -\frac{a^2+b^2+2a^2b^2}{(1+a^2)(1+b^2)} \le \sqrt 2 - \frac 1 {\sqrt{1+c^2}}...
  6. fan96

    Complex number questions

    For Q3: Let |z - w|^2 = r^2 describe a circle on the complex plane with centre w . Substitute v = 1/z, and expand to get |1 - vw|^2 = |v|^2r^2 (1 - vw)\overline{(1-vw)} = |v|^2r^2 1 - 2 \text{ Re}(vw) + |v|^2|w|^2 = |v|^2r^2 1 - 2 \text{ Re}(vw) = |v|^2(r^2 - |w|^2). If the circle...
  7. fan96

    Complex number questions

    Here's a solution for Q1. If z_1 + z_2 + z_3 = 0 and |z_1| = |z_2| = |z_3| = 1, then for any -\pi < \theta \le \pi we also have z_1e^{i\theta} + z_2e^{i\theta} + z_3e^{i\theta} = 0 and |z_1e^{i\theta}| = |z_2e^{i\theta}| = |z_3e^{i\theta}| = 1. Therefore, without loss of generality, let...
  8. fan96

    Help!!!! How does this work?

    The question is poorly worded, but if each ticket is for a separate raffle drawing then binomial probability gives \binom{12}{2} \left(\frac 1{10}\right)^2 \left(\frac 9{10}\right)^{10} \approx 23\%.
  9. fan96

    Solve this mathematical induction question and youll get 1000 years of good luck!!!

    Hint: we may write 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} as 1 - \left( \frac{2 \cdot \frac{n+2}{n+1}}{(n+2)2^{n+1}} - \frac{\frac{n+3}{n+1}}{(n+2)2^{n+1}}\right).
  10. fan96

    Application of vectors

    One more thing: Tension is a property of taut strings/ropes etc. that pull attached objects towards each other. Suppose you attach a ball to a string and then hold the string up midair, suspending the ball. The ball is not moving, despite the pull of gravity. That's because there's a force...
  11. fan96

    Application of vectors

    The question only says the object is accelerating a m/s up the plane. There is no positive direction defined here at all. Regardless, it doesn't matter which direction is chosen to be positive. The only thing that matters is that you clearly state which convention you have chosen and that...
  12. fan96

    Application of vectors

    The block accelerates in one dimension - either up or down the plane. The numbers we use to describe acceleration are also one dimensional - either positive or negative. It's simply a matter of choosing which plane direction is "positive" and which one is "negative". The choice itself doesn't...
  13. fan96

    discrete mathematics

    1. Math degree 2. Sure, PM me.
  14. fan96

    Polynomial Question Help Plz

    We have x^3 P(x^{-1}) = P(x), so if P(\alpha) = 0 then \alpha^3 P(\alpha^{-1}) = 0 . Clearly \alpha cannot be zero, so \alpha^{-1} is another root of P . Also since \alpha \ne \pm 1 , \alpha and \alpha^{-1} are indeed distinct roots.
  15. fan96

    Hard Proofs Question

    I think there's a possibility that there's some algebraic trick you can do (independent of the previous parts) to find C once you know that it actually exists. Part v) specifically states that the limit exists (with a DO NOT PROVE attached). That would be unnecessary and possibly misleading...
Top