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Can you post a little hint as for direction for your marathon question? I am really doubting my complex numbers abilities now :(

Re: HSC 2013 4U Marathon Trying my best using complex numbers (otherwise its quite simple to prove that property, just prove that the 4 triangles that are created with diagonals of rhombus are all congruent) z_1=x+iy z_2=u+iv Now when we form the rhombus fixing one end at the origin, the...

Re: HSC 2013 4U Marathon Is a_k real?

Re: HSC 2013 4U Marathon $When a polynomial $ P(x) $ is divided by $ x-3 $ the remainder is 5 and when it is divided by $ x-4 $ the remainder is 9. Find the remainder when $ P(x) $ is divided by $ (x-3)(x-4)

Re: HSC 2013 4U Marathon (z-1)^n+(z+1)^n=0 \\ \\ 1+(\frac{z+1}{z-1})^n=0 \\ \\ 1+(-i\cot \frac{\theta}{2})^n=0?? $ (not purely imaginary?$)

Bump

Oh I have already have all the HSC papers since 1995, but Ill have a look at other school task 1s as well.

Spiral I have task 1 for MX2 this Friday, any tips on how to maximise marks?

Re: HSC 2013 4U Marathon \frac{1+\cos \theta +i\sin \theta}{1+\cos\theta -i\sin\theta}=\frac{(1+\cos \theta + i\sin \theta)^2}{(1+\cos \theta)^2+\sin^2 \theta } \\ \\ =\frac{1+2\cos \theta +\cos^2 \theta -\sin^2\theta+2i(\sin \theta(1+\cos \theta)}{2+2\cos \theta}=\frac{2\cos \theta(1+\cos...

Are you sure your result for the first part of your marathon question is correct? I keep getting something different (cis theta)

Re: HSC 2013 4U Marathon Oh wow, I cant believe I could not see that :/ 1+z+z^2+...+z^{2n}=\frac{z^{2n+1}-1}{z-1}=\frac{(z-1)(z-z_1)(z-\bar{z_1})...}{z-1}=(z^2-2\cos \frac{2\pi}{2n+1}+1)(z^2-2\cos \frac{4\pi}{2n+1}+1)...(z^2-2\cos \frac{2\pi n}{2n+1}+1)

Re: HSC 2013 4U Marathon Assume that the factorisation is: (something that probably was wrong in the first place) (z^2+bz+1)^n=1+z+z^2+...+z^{2n} Equate co-efficient of z on LHS we can get z an n number of times (z^2+bz+1)(z^2+bz+1)...(z^2+bz+1)=1+z+... nbz=z \\ \therefore b=\frac{1}{n}...

Re: HSC 2013 4U Marathon Is it: 1+z+z^2+...+z^{2n}=(z^2+\frac{z}{n}+1)^n Making sure before I continue

Re: HSC 2013 4U Marathon Nice. Also bob for part b do we need to factorise the whole thing into quadratic factors or only 1 quadratic factor?

Re: HSC 2013 4U Marathon Very nice solution.

Re: HSC 2013 4U Marathon Also for the 957 question, I am currently trying to find pythagorean triads that have multiples as 957 lol. Like for the triple 3, 4, 5 957 is divisible by 3 into 319 times Multiply 4 by 319 =/= square number And it didnt work for the next couple of triples that...

Re: HSC 2013 4U Marathon Assuming a, and b real: Construct: (\sqrt{a}-\sqrt{b})^2 \geq 0 a-2\sqrt{ab}+b \geq 0 \\ \therefore \frac{a+b}{2} \geq \sqrt{ab} ii) \ \ \ \ $If both roots are complex discriminant less than zero$ \\ \\ b^2-4ac < 0 \\ \\ b^2 < 4ac \\ b<2\sqrt{ac} \ \ \ \ \ \...

Re: HSC 2013 4U Marathon Solution

Re: HSC 2013 4U Marathon I just found a mistake in my question, when I did it I divided the an inequality by b, which is not allowed since that is assuming b>0 there is indeed two domains for b such that there are 3 real roots. Will do the question now

Re: HSC 2013 4U Marathon Err yeah sure I guess, it isnt my strength but practise makes perfect.