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God how stupid was I doh... thanks..

Given 1, w & w^2 are the cube roots of unity. How do we show that if z^3-1=0 & pz^5+qz+r=0 have a common root then (p+q+r)(pw^5+qw+r)(pw^10+qw^2+r)=0

thanks CM; That clarifies it much appreciated..

I understand the above but my next question is why are we doing x+y, then x-y and then multiplying the two. Like what is the reason why we did that?

If you are given a complex No. z=t+1/t where t=r(cosa+isina) given a=(pi/4) and r varies and asked to draw the locus of the point P on an argand diagram I get to the point : x=(r+1/r) times 1/(sqrt(2)) y=(r-1/r) times 1/(sqrt(2)) now what I don't get is how do I go from here to the final...

yeah essentially pascals triangle....

stupid stupid stupid..... I did inverse tan of sqrt(3) instead of inverse tan of 1/sqrt(3).. Oh well it was late :).... thanks guys..

I'm trying to find the square roots of: (sqrt(3) +i) and I get: +/- sqrt(2)(cis(pi/6) although according to the answer it should be: +/- sqrt(2)(cis(pi/12) I pretty sure I'm right although just want to make sure, anyone here concur with my answer?

I suppose pascals triangle but I was wondering if there is any easy method (even though pascals triangle is easy) :) Just curious..

is there an easy way of expanding (x+y)^4 without going through the process of tediously doing (x+y)(x+y)(x+y)(x+y) OR Without going (x+y)^2 multiplied by (x+y)^2 ?

yep agree but would like to get past papers from other schools aswell to gauge myself, and also to see the different styles of questions.

Not a bad site for some simple notes but my teacher has written his own notes which are simply awesome, I'm more after a collection of half yearly exams from a variety of schools. But thanks for the help though!!

Does anyone know (or have) where I could get (d/l) a number of half yearly year 12 papers from other schools? I'd like to practice some before my exams in 6 weeks. I have the MLC and Mackillop papers that are available from this site but would like a few others (as many as I could get my hands...