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...my cat's breath smells like cat food

oh come on, you should know excellence is our school motto well, "pursuit of excellence" in such context i see excellence as the betterment of oneself through the upkeep of human endeavours; progress my point is that, excelling is not about being better than others, beating others or...

I believe: "Real excellence and humility are not incompatible one with the other, on the contrary they are twin sisters." -Jean Baptiste Lacordaire

:)

yep my preferences are 1 511718 Bachelor of Engineering (Aeronautical) (Space) at University of Sydney 2 511801 Bachelor of Combined Law at University of Sydney 3 511770 Bachelor of Engineering/Bachelor of Science at University of Sydney 4 511733...

98 raw in trial yes.

aerospace engineering at usyd or aero eng / law at usyd or maybe eng / advanced science (physics)

didn't complete a couple of solutions in q4,5 and 7 110 max that paper was piss easy compared to other years i mean come on, who didn't finish question 8 that's so wrong

nb: we're dealing with ordered arrangements i) you can have 4xx 5xx 6xx... 9xx ie 6 different 1st digits then 8 * 7 ways of choosing the last 2 digits ie 8P3 total combinations = 9P3 thus, P = 8P3/9P3 = 2/3 ii) do each case selectively for first digit 9, you have 8...

Here we go, algebraic proof #2 (proof #4) let z1 = a+bi and z2=c+di square both sides collect like terms square both sides again you end up with the ratio ac=bd ie b/a = d/c since argZ1 = arctan (b/a) and argZ2 = arctan (d/c) therefore argZ1 = argZ2 (as b/a = d/c) elegant...

Spicegirl: technically, your proof is geometric as you use geometric properties (angles, parallel lines, cointerior angles of parallel lines) to give geometric representations of your algebraic constructs. try again :)

IS THAT YOU, DENG?

Well yes, the question asks to prove the result geometrically as well as algebraically. obvious as the statement is, it was the algebraic method that was beyond me.

A friend of mine also proved this result by showing Z1/Z2 is equal to its conjugate, the algebra being quite complex (heh, no pun intended). It'd be interesting to see how many methods people come up with

let A be point Z1 B be point Z2 and C be point Z2 + Z1 x is the angle OAC angle AOB = argZ1 - argZ2 (by observation) and OAC = pi - AOB (AC||OB, cointerior angles of parallel lines supplimentary) thus x = pi - [argZ1 - argZ2]

cool, thanks ...who are you?

Hi, I was wondering if someone could do this for me if |Z1 + Z2| = |Z1| + |Z2|, show algebraically, that arg(Z1) = arg(Z2) What I've done so far is if arg(Z1) = arg(Z2) then, arg(Z1) - arg(Z2) = 0 then, arg(Z1/Z2) = 0 ie. must show that Z1/Z2 ( R I think this can be done by...

in the inside cover it says we're required to learn _one_ of the texts.

the real him

woah woah is that how johnson looks like what a sex beast *growl* pity he looks gay