Too good!

Haha just heard that Darren got top in class for math. Congrajulations Darren, although its still a bit sad that there are still people from other classes toping him.. Always knew he had the stuff for mathematics. Too bad he's not passionate about it huh.. But I'm sure he'd make a fine mathematician:D

Did a bit of maths with Soomin yesterday. Realized that i'm falling behind them in terms of JC mathematics.. haha.. I guess i just do too little practice.. Oh well.. Its not fair anyway, they got teacher teach one :P

Guess what i dreamt of yesterday? I went to this strange school.. And i was attending math lesson where they had this weirdest math syllabus.. There was a chapter on symmetry, points and lines. And the teacher asked this really complex question, something like tell me what is the @#*$ property that both a group of #$*& points and a @(&# line shares? And i was like O.o And the teacher ordered me to go for extra lessons.. And i went. Then this damn chio maths teacher asked me.. So you dun like maths? And i said "yeah actually i dun really like maths" And then i woke up.. WALAU super sad laaa cannot stand it can!

So i decided its time to do some complex mathematics. To prove to myself that i still can do maths XD

Its common sense in jc mathematics that the expression 1/(1-x^2) can be split into partial fractions.
The denominator splits into (1+x) and (1-x) and thus we have
1/(1-x^2) = A/(1+x) + B/(1-x) and a quick mental sum will reveal A and B to both be half.

What is not so obvious is the expression 1/(1+x^2). While 1+x^2 is not factorable in the real domain, it is factorable in the complex domain. More precisely, 1+x^2 = (1+ix)(1-ix) and so we have 1/(1+x^2)=1/2(1+ix) + 1/2(1-ix).

Lets do some interesting things with this.. It is elementary calculus that the integral of 1/(1+x^2) is artanx

However, with our observation above the integral of 1/(1+x^2) = the integral of 1/2(1+ix) + 1/2(1-ix)= 0.5*(-i*ln(1+ix)+i*ln(1-ix))
=0.5i*[ln(1-ix)-ln(1+ix)]
=0.5i*ln[(1-ix)/(1+ix)]

This must be equal to arctanx since they are the same integral.
Thus we have arctanx = 0.5i*ln[(1-ix)/(1+ix)]

Or equally beautiful, e^arctanx = [sqrt[(1-ix)/(1+ix)]]^i

haha so i do like maths after all!