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Monday, July 03, 2006


I am extremely sorry for not being able to post anything. But I just can't help it. My college has reopened & I have absolutely no time. Well, got some time at last & here I am writing a post.

Alright, if I remember correctly there were some problems in the previous posts regarding raising a no. to a complex power. Well not too sure but doing so will result in a complex no. Take this for an ex.-

let y=(1+tanA)^(1+i)
(i is imaginary ofcourse)
taking ln of both sides

lny=(1+i)lntanA ---------(1)

let y=r*(e^if)
(let f=theta, don,t know how to type theta)
taking ln of both sides

lny=lnr+ if --------------(2)

equating (1) & (2)

lnr+ if= lntanA+ ilntanA
thus equating real & imaginary parts we get
r=f= lntanA
Now if ur wondering as to why r=f ,well it's simply bcoz I had taken the question like that.

Now the only problem I have with method is can we take the ln of both sides at the outset.
Well the way I think is that we have already gone out of real nos. so the question of violating the domain of ln function doesn't arise. What do make of of this? Let me know about what ur thinkin. Before I end this post here is a question I would like to share with u.

let f(r)= (x^r)sinrA + (y^r)sinrB + (z^r)sinrC
where x,y,z are real nos. , r is anatural no. & A,B,C are angles of a triangle.
if f(1)=0 , f(2)=0, then prove that
f(r)=0 for all r beongs to natural nos.


Gr81 said...

Well got a bit of what you were saying - but as I had said before in my post that these functions have a proper definition (like domain etc) where they are valid. now as Arun had just posted, we tend to violate these when we start doing complicated things. Stuff like ln(a^m)=m(lna) seems valid for m being real, but dunno about imaginery - Thats where our knowledge fails!

Still I think this will satisfy me for the time being (coz I can't think much as I hav my exams!)
Keep Posting!

Arun Chaganty said...

This is how I believe mathematicians get around this stuff (no offense to any budding mathematical geniuses).

A: Hmm.... Let's see, ln(a^m) = mln(a) practically for real numbers (with good proof). but complex?

B: Ahh.. How bout this? Let ln(a^m) = m ln(a) be a DEFINITION. Clearly this applies for real numbers, so we can VISUALISE reality to be such that complex numbers also follow this.

I've seen in many cases, (gamma function (look it up) for factorials, non-euclidean geometry, fractal dimensions (also Hausdorff dimensions) for normal dimensions (this is a freaky definition of dimension that says you can have figure with FRACTIONAL dimensions)) that we start out with well defined, and naturally occuring property (got through just plain sense), then to get around discontnuties, we fill in that space with crud, and as long as it satifies the initial case, we assume it represents the SAME natural phenomenon.
But here in this case, to the best of my knowledge, and intuition, we will get infite solutions, but we consider only a small segment of it. so we consider:

x^(a+ib) = x^a(e^(lnx)*ib)
= x^a (cos(b*ln(x)) + i (sin (b*ln x))
They fall under complex logirithms (again look it up).

Gr81 said...

Very well said Arun!!
I'll surely looks them up when I free my comp from these virus!
Keep Posting!