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[Darvi]

Yeah, 'cept usually people won't do i until a couple years after trig. formulas.

Also, why does an irregular number to the power of an imaginary number become a cosinusoidal curve? It's kinda weird.

And apparently, i^i = e^(-pi/2)

Math sure is strange.

[ZetaX]

e^(ix) circles around the unit circle (with constant speed 1 if you wish). Thus it's not that strange  that this gives cos and sin, as these are defined by coordinates of points on that circle.

i^i ist not e^(-pi/). That's just one possible value. The reason is that log(x) is only well defined for all complex x up to addition of a multiple of 2pi·i [note that e^(2pi·i)=1, giving that a and a+2pi·i be the same after exponentiation by e]. Thus you get i^i = e^(i·log(i)) = e^(i·(pi·i/2+n·2pi·i)) = e^(-(4n+1)/2). For n=0, this gives your's.

[Darvi]

No, I mean, why does an imaginary exponential make things go round in circles?

i^i ist not e^(-pi/). That's just one possible value. The reason is that log(x) is only well defined for all complex x up to addition of a multiple of 2pi·i [note that e^(2pi·i)=1, giving that a and a+2pi·i be the same after exponentiation by e]. Thus you get i^i = e^(i·log(i)) = e^(i·(pi·i/2+n·2pi·i)) = e^(-(4n+1)/2). For n=0, this gives your's.

When dealing with pi, I always assume that there's an even number left in the fridge, so I don't mention it usually.

[ZetaX]

Essentially this boils down to "follows from the definition". But some reasons: You have e^(ix)·e^(-ix) = e^0 = 1. And every nice (i.e. holomorphic) function f from IR to IR (reals) if extended to IC (complex numbers) fulfills f(#z)=#f(z) where # i the conjugate [#(a+bi)=a-bi]. Thus you get #(e^(ix)) = e^(#ix)) = e^(-ix). Now you have that the number a=e^(ix) gets it's own inverse 1/a by conjugation, so #a=1/a or in other words a·#a=1. This gives that |a|=1, so that a is on that circle.
So it's somehow a result of e^(x+y)=e^x·e^y and e being a nice function.

[Darvi]

Nevermind I just looked it up on wikipedia >_>

Spoiler: Shit still is weird. (click to show/hide)

Like a roller coaster. Whoooo!

[Tellemurius]

string theory flashes in my mind looking at that even though they are different.

[ZetaX]

Nevermind I just looked it up on wikipedia >_>

Spoiler: Shit still is weird. (click to show/hide)

Like a roller coaster. Whoooo!

That's just a picture, not an explaination.

[Darvi]

I know, but it's fun to look at

[Christes]

I just let my students use a 3x5 inch index card on the exams.  Memorizing all of the formulas is supremely pointless IMHO.  (Of course, then I can write problems to test if they really understand the formulas)

Eh, it's easy. Unless you'd also want me to prove that cos(2x)=cos^2(x)-sin^2(x),' cuz that would take a while and perhaps a graph too.

The real slight of hand here is proving it, but only doing so for acute angles.  That seems pretty standard.

The one best way of proving and memorizing stupid trig. stuff is via e^(ix) = cos(x)+i·sin(x) and e^(x+y) = e^x · e^y. The trig. formulas are then just properties if real and imaginary parts.

We do complex numbers right afterwards, but this formula is not mentioned.  Honestly, it would confuse all but a couple students.  Even moreso, since you can't prove it until you have at least of differential calc.  I mentioned it offhand last quarter on the day before thanksgiving when only a handful were there, since we had nothing better to do.  (On the same day, a fellow grad student of mine talked about the Cantor Set (!!!) in his trig class)

[ZetaX]

I doubt it would confuse most people if explained carefully. You just explain that you CAN actually have complex exponents, then explain what I did in my second last post to show that it actually is on the unit circle. And if one uses exponentiation, one wants e^(x+y) = e^x·e^y as this is the most basic rule of exponentiation. Then suggest that one may DEFINE it for all complex numbers via e^(x+iy) = e^x·e^(iy) = e^x·(cos(x)+i·sin(x)) as this definition then works (fulfills e^(x+y) = e^x·e^y) by those trig. theorems. From that, just conclude e^(i·pi)+1=0 for extra fun^^

[Darvi]

... and now I'm picturing a pie filled with eyes...

[Virex]

I just let my students use a 3x5 inch index card on the exams.  Memorizing all of the formulas is supremely pointless IMHO.  (Of course, then I can write problems to test if they really understand the formulas)

Eh, it's easy. Unless you'd also want me to prove that cos(2x)=cos^2(x)-sin^2(x),' cuz that would take a while and perhaps a graph too.

The real slight of hand here is proving it, but only doing so for acute angles.  That seems pretty standard.

The one best way of proving and memorizing stupid trig. stuff is via e^(ix) = cos(x)+i·sin(x) and e^(x+y) = e^x · e^y. The trig. formulas are then just properties if real and imaginary parts.

We do complex numbers right afterwards, but this formula is not mentioned.  Honestly, it would confuse all but a couple students.  Even moreso, since you can't prove it until you have at least of differential calc.  I mentioned it offhand last quarter on the day before thanksgiving when only a handful were there, since we had nothing better to do.  (On the same day, a fellow grad student of mine talked about the Cantor Set (!!!) in his trig class)

I thought that formula was like half the point of imaginary numbers? I can't really remember many cases in which I used i for a different reason.

[ZetaX]

The point goes much further, like complex analysis, algebraic closedness, etc.

[Christes]

I doubt it would confuse most people if explained carefully. You just explain that you CAN actually have complex exponents, then explain what I did in my second last post to show that it actually is on the unit circle. And if one uses exponentiation, one wants e^(x+y) = e^x·e^y as this is the most basic rule of exponentiation. Then suggest that one may DEFINE it for all complex numbers via e^(x+iy) = e^x·e^(iy) = e^x·(cos(x)+i·sin(x)) as this definition then works (fulfills e^(x+y) = e^x·e^y) by those trig. theorems. From that, just conclude e^(i·pi)+1=0 for extra fun^^

We're talking about students who often get confused by rational exponents, let alone real ones.  I agree with what you are saying and will likely mention it offhand for those who are genuinely interested.

The major problem with math at this level is that there is little justification given to studying anything.  There are so many awesome things that go on in the complex numbers.  I really wish I could share that with my students, but there are three problems:

1) Most are simply not prepared for any "math" that goes much beyond rote memorization.
2) Most don't care and just want a passing grade to satisfy some requirement or another.
3) Even if parts 1 and 2 don't hold, I need to uphold the standard curriculum, which means I don't have too much time to spare.

I usually make a few offhand remarks about more general results, and emphasize that I use a lot of this material regularly.  That helps, but there is only so much that can be done.

[Darvi]

1) Most are simply not prepared for any "math" that goes much beyond rote memorization.

Heh, I'm nott prepared for anything that requires memorization.

Which is why I failed my last year :/