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Hmm... Are you sure? e^{z+2 \pi k i}=e^z e^{2k \pi i}=e^z...

Re: 2012 HSC MX1 Marathon Teachers (and other people in general) become lazy when it comes to writing solutions, even to their own examinations... So go figure.

Yes you can treat 'i' as a constant when doing integration. Another interesting point is that the complex exponential function e^(i theta) is periodic with period 2*pi*i. This makes the complex log function a bit more complicated.

Re: 2012 HSC MX1 Marathon Nice, but i think it's worth noting that the first line is only valid if the terms of the series are absolutely convergent.

Re: 2012 HSC MX1 Marathon Did you do the Level 1 course? If so, did you find it difficult at the time when you did it?

Re: 2012 HSC MX1 Marathon Just a question, if don't mind answering... How did you find the Level 1 (or whatever maths course you did) when you did it? And how did you go in it? Just curious -- that's all.

Re: 2012 HSC MX1 Marathon Should really justify the 3rd last line being because the exponential function is continuous.

That's new... :P

I prefer: \sum \alpha_i \alpha_j \alpha_k and \prod \alpha_i...

I know you're into elegant methods etc, but one does not do determinants in high school (or any linear algebra) therefore, although it is elegant, chances are that this student will have no clue what this is. And @ the op: There's a theorem that actually says that given any three points there...

It was in a 2U or 3U HSC paper recently (can't remember which year but I think it was 2010) and these types of questions were my teacher's favourite in exams he'd set.

Lol complex analysis...

As a, b, c are in GP: b = ar, c = ar^2 for some r>0 (as b, c > 0) Now as a>0 and Δ=b^2 - 4ac = (ar)^2 - 4a(ar^2) = -3(ar)^2 < 0 (as (ar)^2 >0) Therefore y=ax^2+bx+c is positive definite and therefore lies entirely above the x axis.

How so? And I think 2011 was quite straightforward. But long... so go figure...

Complex analysis is a very beautiful course. Most of my friends hated it though. You can do things like finding a solution to sin z = 2 (can't remember if this particular case has a solution, but it could have a complex solution) using sin z = (e^z - e^(-z) )/(2i) or by letting z=x+iy and using...

You should really say \text{pv }i^i because of periodicity. By definition, \text{pv } i^i = e^{i Log i} = e^{ i (ln|i| + i Arg(i))}= e^{i( 0 + i \frac{\pi}{2} )} = e^{-\frac{\pi}{2}} = \frac{1}{\sqrt{e^{\pi}}} If you didn't use pv then you'd get something like a + exp(2pi k) term as well...

Not necessarily true. There was a really good maths teacher at my school. He's been there since 1989...

I think that's the same as |arg(z-1+root3)| <= pi/3 because by definition |arg(z-1+root3)|>=0 for all z (abs value of anything is non negative). So the expression becomes -pi/3 <= arg(z-1+root3) <= pi/3.

For y=\pm \frac{1}{2}x \sqrt{1-\frac{4}{x^2}} As x->infinity 4/x^2->0, So y-> +/- (1/2)x This basically means that for large values of x, the curve tends to go near the curve/line y=+/-(1/2)x. Of course it would also approach infinity, but it will do so by staying close to y=+/-(1/2)x.

Well this year's 4U paper was just the stock standard questions, nothing with a twist... Hasn't been like this since 1981 I think. Anyway I think it has heaps to do with the way 4U is delivered. A good teacher for 4U is important.