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Re: HSC 2016 Complex Numbers Marathon Use same method to prove it as paradoxica did above with complex numbers.

Re: HSC 2016 Complex Numbers Marathon NEW Q: sketch arg(z-2)=arg(z+i)

Re: HSC 2016 Complex Numbers Marathon That method is great cause you didn't even need the assumption ow lies on sum of u1 and u2 as you followed it using its construction

Re: HSC 2016 Complex Numbers Marathon The best approach to these harder questions is to use the answer and think given what you have what you need to prove, since OW is diameter prove |w|=2/|u1+u2| and a more advanced method is play with that result and you get prove |w(u1+u2)|=2 which looks...

Re: HSC 2016 Complex Numbers Marathon Sub z=w into each tangent equation since that's the intersection point. Then add each equation and you'll notice if you factorise correctly you get sum of two conjugates

Re: HSC 2016 Complex Numbers Marathon Another problem is the centre of the circle is not the midpoint of the chord, it's impossible. I'll give you a hint to solve it really easily in a few seconds. The intersection of the tangents is the point w so if you let OW be the complex number w you can...

Re: HSC 2016 Complex Numbers Marathon A kites diagonals do not bisect each other

Re: HSC 2016 Complex Numbers Marathon NEW If z= cis(theta) solve z^4-2z^3+3z^2-2z+1=0

Re: HSC 2016 Complex Numbers Marathon NEW Prove that if a, b and c are concyclic and the circle passes through the origin then 1/a, 1/b and 1/c are collinear. Where a, b and c are complex and in quadrant one.

Re: HSC 2016 Complex Numbers Marathon Note those are ones on top and bottom if you can't see clearly

Math Man is back and seeing the different marathons thought different topic themed marathons seems like a good idea. I would love to do mechanics but it's too early for that and I love complex numbers! Rules: 1. Once a question has been answered you can ask a new question. 2. A new...

Can you restate the question?

working out is, complete the square on the bottom as: (x^{2}+2x+1)cos^{2}x-2(1+x)sinxcosx +sin^{2}x -cos^{2}x-sin^{2}x= \\ \\ ((x+1)cosx-sinx)^{2}-1 and we note that \frac{d}{dx} (x+1)cosx-sinx= -(1+x)sinx so the integral is in the form \int \frac{f'(x)}{f^{2}(x)-1}dx...

Then the answer is -\frac{1}{2}log[\frac{(x+1)cosx-sinx-1}{(x+1)cosx-sinx+1}] +c

i can integrate if there is a minus sign on denom, not plus

better way is: we know w is a solution to z^{n}=1 therefore w^{n}=1 Now sub w^{m} into w^{n}=1 we get (w^{m})^{n} = (w^{n})^{m}=1^{m}=1 and that's it.

it's a shit trial, really easy, not worth having

i have the paper for this year. It's the easiest paper they have written imo, not worth doing. I think the early 2000-2003 papers for hsc were decent

b is correct, just draw the curve

well, you didnt get that raped, you did very well considering how overboard i went, and you did show signs of brilliance through some parts.