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let A(a,0) then m(PA)=bsin@/a(cos@-1) and m(QA)=bsin#/a(cos#-1) and cos@-1=1-2sin^2@/2-1=-2sin^2@/2 and using this for both "m"s we get m(PA)*m(QA)=-1 b^2sin@sin#/a^2(-2sin^2@/2)(-2sin^2#/2)=-1 4sin@/2 cos@/2 sin#/2 cos#/2/(4sin^2@/2sin^2#/2=-a^2/b^2 cot@/2 cot#/2=-a^2/b^2 and so tan@/2...

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Diment won't dux it, it'll be leslie and it is highly doubtful even leslie will get a 100 uai and the reasons i have(before ppl start attacking me) is that 1) Although kostya got 100 uai in 2005, he didn't sit many exams and the guys under him in rankings or in the case of physics just ahead of...

Agree with watatank-as long as u get into ur course i guess it's good enough-i mean someone who wants to study something that doesn't require a high uai doesn't need to work harder for a great uai cos what they get will suffice-the uai is a big thing in the first couple of months then we pretty...

this guy is excellent-congrats on med maung-plus his attitude to english is pretty good lol(jks)

bump- i'd go to tutoring myself for english to this guy

Yep that's what i said(maybe i should have mentioned properties of sine as well )

z1/z2=(r1cisa/r2cisb)=cis(-b)/cis(-b) * (r1cisa/r2cisb) and then expanding and simplifying using the properties of cosine u get the required answr.

Lol yeah(i know i should have done it there then pasted it but i couldn't be bothered)

For the second part, z1- (z1-2iz2)/(1-2i)=(1-2i)z1 -z1 +2iz2/(1-2i) =-2iz1+2iz2/(1-2i)=2i(z2-z1)/1-2i and this vector is perpendicular to the vector z1-2z2/1-2i -z2 i.e one side of the triangle formed by these three points makes right angles with another side, hence it is a right angled...

for the first one, when using the parallelogram method of adding vectors u get a rhombus as the shape with modulus z1=modulus z2, the vectors z1-z2 and z1+z2 are the diagonals-the diagonals of a rhombus are always at right angles thus the ratio in the question is in the form +- ki i.e purely...

Maybe u could look more closely at derivations of certain results i.e the formulas ur learning try and derive them urself and i think it's a case of what affinity said

Well locus q in the form of z=c+it/c-it where c is a constant-often giving the modulus etc isn't enough because z traces out the circle either clockwise or anticlockwise, found by subbing in values-that would probably give a more complete locus-sorry bout that i should stated clearly rather than...

thanx gryphondarks, i reckon i was lucky for modern tho-(Man SGS class of 04 had a lot of smart curries as far as i can see-u, nit,shafqat.) Chichi i've sent u a reply or message

For these type of questions u can also just take the modulus of both sides, so lets denote the modulus symbol with "[" these brackets then z= 2+it/2-it take modulus of both sides [z]=[2+it/2-it] =[2+it]/[2-it] =sqrt(4 +t^2)/sqrt(4+t^2) =1 i.e modulus of z is 1 and so the same answer as jyu said.

what an english beast-anyone who gets 97 is an english beast

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Sure bos1234 and i hope u get a better uai than me and the course u want- i rarely go on the 4u forum, but if i see a few q u post up i'll try them(lol i've become really lazy) Watatank, i have no issues with what u said(actually i thought it was pretty funny) btw what school u go to?(i know...

thanx Jachie, again, for attempting to clarify to the aggro guys that i wasn't hijacking this thread or anythin lol-ppl started attacking my sig and i simply gave my reasons and i don't need people like lampshade or hobbit to congratulate me or anything to make me feel good, i've made it into my...