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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.

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...

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.)

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...

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}}...

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...

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...

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\%.

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).

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...

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...

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...

1. Math degree 2. Sure, PM me.

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.

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...

Hint: Make the substitutions a = e^\alpha, b = e^\beta, c = e^\gamma.

If you plot this using software the graph you get is not the same as the original parametric equation. (You only get the upper half of the graph.) This theorem requires f to be invertible wherever x is defined. The input \theta + \arctan(1/4) varies over an interval of length 2\pi , where...

Observe that x + y = 9 \cos \theta 5x - 4y = -9 \sin \theta so \left(x+y\right)^2+\left(5x-4y\right)^2=81.

Are we talking about the UNSW course? If so then I didn't really like it. Out of the seven math courses I've done so far I'd rank this near the bottom, tied with or just above MATH2221 Differential Equations. It suffers from the same problem as the other first year math courses where you don't...

That isn't quite what I said - I said if a,b,c were constants then the expression is a polynomial. What sort of polynomial it is, we can't tell without further information. For the polynomial to be quadratic we require a -2 \ne 0 , which cannot be the case since the polynomial is the zero...