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HKALE textbooks :P

compare HSC harder 3U to cambridge basically, people who set the HSC papers wont be stupid enough to set questions that are similar to textbook questions, so you cant just study off a textbook and expect to get high marks. HSC tend to make questions trickier and require more skill - while...

cambridge questions are too easy and straightforward - they lack the sophistication and skill required in HSC papers

i wonder where you got this question from da ;)

do you mean arg [(z-1)/(z+1)] = pi/2 ?? in that case it is arg (z-1) - arg (z+1) = pi/2 <=> arg (z+1) - arg (z-1) = -pi/2 which is a semicircle below the x axis, if you use sum of interior angles equals opposite exterior in the triangle if you join up z with -1 and 1 see also...

|z| means the modulus, ie. the distance from the complex number to the origin so letting z= x + iy by pythagoras, |z| = rt (x² + y²) then |z|² is just wot CrashOveride said

IMO, Conquering Chemistry >>>> Chemistry Pathways, which doesn't even stick to the syllabus (the latter)

no worries anytime :)

i think this answers your question :)

the first and third line are the same in his proof and he went and wrote those were < LOL =/ his book isnt that bad, but i think his steps in maths inductions are a bit brief...

ah i see what you mean - didnt really check the proof haha hmmm quite strange... his proof is quite all over the place haha - the inequality signs are wrong..probably a printing error?

well usually by convention you read from top line to bottom line: ie. for your case, 2 - 1/k + 1/(k+1)^2 = 2 - (1/k - 1/[k(k+1)]) as they are equal not one greater than other NOT 1 + 1/4 + 1/9... 1/k^2 + 1/(k+1)^2 = 2 - (1/k - 1/[k(k+1)]) thus the next line (in the proof) means 2 -...

the locus can be interpret geomtrically: you can probably try algebraically: arg [ (z - 1) / (z + 1) ] = -pi/3 let z = x+iy etc, but thats probably too messy and long not if it is arg (z-1) - arg (z+1) = pi/3 , the locus is in the second and first quadrants