# Recent content by fan96

1. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### 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. ### discrete mathematics

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