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Ahh...i get why you're lost Mathematician, Umm, I shouldn't have chosen z. Let's choose t. So since we're looking at a product xyz = (a+b)(a+bw)(a+bw^2), with a linear involving each of the cube roots of unity. The strategy was to recognise that it looks like a polynomial with roots 1...

Umm, you let z be -a/b, giving you the required result. And umm...I'm really busy with uni stuff at the moment. May try once I settle down into the work.

z^3 - 1 = (z - 1)(z - w)(z - w^2) (by definition of w, and the factor theorem) then b^3 * (z^3 - 1) = (bz - b)(bz - bw)(bz - bw^2) .......(1) Sub z = -a/b into (1) b^3 * [ (-a/b)^3 - 1 ] = (-a - b)(-a - bw)(-a - bw^2) .......(2) Multiply both sides of (2) by -1, and thus: (a +...

Using r^2 = |z|^2 = z*zbar z/ (z^2 + r^2) = z / (z^z + z*zbar) = 1 / (z + zbar) = 1 / 2Re(z)

Also, an alternate solution, for people who know how to graph: arg(z-z1) - arg(z-z2) = alpha (where z1, z2 are complex numbers and -pi < alpha <= pi) for those who don't know, it's an arc where the angle subtended to the arc by the interval A(z1)B(z2) is the angle alpha. Anyway, if...

I think (ii) has something to do with z + z(bar) = 2Re(z) Remembering also that sum of conjugates = conjugate of sum, and that if z = cisx then z(bar) = cis(-x)

Ahh this one it took me a while too... anyhow: y = (e^2x + 1)/e^x y*e^x = e^2x + 1 (e^x)^2 - y*e^x + 1 = 0 then using quadratic formula, we find e^x then we log both sides to express x in terms of y.

use compound angles for both trig expressions on the LHS and see wot we get: tan(pi/4 + A) = [tan(pi/4) + tan(A)] / [1 - tan(pi/4)tanA] = (1 + tanA)/(1-tanA) similarly tan(pi/4 - A) = (1-tanA)/(1+tanA) subtract one from the other with common denominator: LHS = [ (1+tanA)^2 +...

Ahh...nice that's heaps similar to the solution i was thinking of. But here's wot I was thinking (it sorta is the same). 1, w, w^2, ..., w^6 are the roots of the polynomial P(z) = z^7 - 1 (-1-1)(-1-w)(-1-w^2)...(-1-w^6) = P(-1) = -2 thus (1+1)(1+w)(1+w^2)...(1+w^6) = 2

Perhaps like so, but it'll take up space: /b | f(x)dx /a

Umm, firstly you have to recognise an elegant solution when you see one, and study how the solution is linked to the question. And, once you finished with a problem, don't stop there; go spend an extra 5 minutes looking for an alternate solution. Since you already have a feel for the...

dodgy question to start off with, Mathematician, is this a definite or an indefinite integral? If indefinite, I dun think the result is true at all...

What's the trigonometric result from this question?

idunno wot u're referring to there about the "4 tests". You can use any of the below to prove a parallelogram: * 2 pairs of opposite sides parallel * 2 pairs of opposite sides equal * one pair of opposite sides both equal and parallel * diagonals bisect each other

How about this one: Let A, B, C, ..., G represent 1, w, w^2, ..., w^6 where w = cis(2pi/7). Let H represent -1 Prove HA*HB*...*HG = 2 there's several answers to this one. I found the good solution only after I did the HSC. it involves polynomials i think...

Oh, btw, the quote "I'm a victim of society" came from Otis of the Monkey Island game

sum of roots to the polynomial z^n - 1 = 0 is 0 (because coefficient of z^(n-1) term is 0

Sorry...sorry... the sub-questions should be reworded to say prove alpha is also a root of [3], [4], [5].... but close enough...

Let m1 = tanP, m2 = tanQ so m = tan((P+Q)/2) for it to be bisector so tan(P - (P+Q)/2) = tan((P+Q)/2 - Q) use compound angles for tan function: (m1-m) / (1+mm1) = (m-m2) / (1+mm2) Cross multiply to get 1st question Expand and regroup to get required quadratic in m remember...

I think I'll let people have a shot at the question, so I'll make up a set of follow-up questions so it's easier: NB: on second thoughts I don't think this is ~elegant~ but it's always nice to have alternatives. Consider x^3 + 3px^2 + 3qx + r = 0 ...[1] has a double root, called alpha...