| Marcos |
When we write eiax do we mean (cos x+i sin x)a or cosax+i sin ax? Also please explain if it doesn't matter and why because the way I see it, one is single-valued whereas the other can be multi-valued... Marcos |
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| Anuj
Shah |
You are obviously not familiar with De Moivre's Theorem, which states that (cos(q)+i sin(q))n=cos(nq)+i sin(nq) This can be proved by induction 1) Is the theorem true for n=1 (cos(q)+i sin(q))1=cos(q)+i sin (q) 2) Assume it is true for n=k (cos(q)+i sin(q))k=cos(kq)+i sin(kq) 3) Consider n=k+1 (cos(q)+i sin(q))k+1 =(cos(q)+i sin(q))k (cos(q)+i sin(q))1 =(cos(kq)+i sin(kq))(cos(q)+i sin(q)) =cos(kq)cos(q)+i2 sin(kq)sin (q)+i sin(kq)cos(q)+i cos(kq) sin(q) =cos(kq)cos(q)-sin(kq)sin(q) +i(sin(kq)cos(q)+cos(kq)sin(q)) (using the trig formulae sin(A+B)=sin(A)cos(B)+ cos(A)sin(B) and cos(A+B)=cos(A)cos(B)-sin(A) sin(B)) =cos(kq+q)+i sin(kq+q) =cos((k+1)q)+i sin((k+1)q) 4) if the theorem is true for n=k then it is true for n=k+1, but it is true for n=1 so it is true for n=2 and n=3 and so on. So theorem is true for all positive integer values of n. There are similar proofs for negative and fraction n. |
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| Chris
Tynan |
I always thought, by De Moivre, that your two latter expressions were equal. This link seems to confirm this: http://mathworld.wolfram.com/deMoivresIdentity.html Chris |
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| Chris
Tynan |
Well, Anuj beat me to it. |
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| Paul
Smith |
Actually, Chris and Anuj, this is not correct: Marcos is right, (cos x+i sin x)a is in general infinitely many-valued (unless a is real and rational), whereas cosax+i sinax is single-valued. Back to Marcos' question, I suspect the answer is probably the latter of the two equations (simply due to the symmetry in x and a), but I think you might like to consult the final chapter of Hardy's Pure Mathematics where he gives different definitions to exp z and ez!!! :) Perhaps someone else can help out here (or maybe I'm being stupid as well...) Paul |
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| Paul
Smith |
...Hmmm, now I'm pretty sure it is the latter of the two equations! :) The reason is that in general, the relation azz ' =(az) z ' doesn't hold (every value of the LHS is a value of the RHS, but the converse is not true). This strangely seems to make sense ... :) Paul |
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| Marcos |
Anuj + Chris, see Paul's post... Paul... I remember seeing 2 separate definitions for ez and exp(z) in the downloadable booklet on Complex Analysis that Ben Tormey suggested to Yatir in a past thread (I forgot the discussion/site of booklet)! It's quite confusing in certain cases, I think! I also think it's probably the latter expression in my original post but I just wanted to make sure/get other people's feedback on the subject... Marcos |
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| Dan
Goodman |
My take on this is that unless the context indicates otherwise, ez always means exp(z). I can't really imagine any situation where you would use ez to be something other than exp(z) although it might crop up in a detailed bit of complex analysis. However, if anything was meant by ez other than exp(z) then it would be normal practice to say so explicitly. One other point, ez can be considered multivalued itself, and it would be very strange to define "eiax "=(exp(ix))a rather than simply eiax (for a suitable branch of the function ez ) or exp(iax). If you don't know the term "branch" applied to a complex valued function post again. I don't think you can make much sense of this sort of question without understanding branching. |
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| Colin
Prue |
Would it be excessive to explain what you mean by 'branching'? If not, could you... thanks |
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| Dan
Goodman |
Well, a full explanation might be excessive, but I think I can probably give you some idea. Basically, the problem is that lots of things which should be functions aren't because they're not single valued. The definition of a function implies it must be single valued. So, for example, if you want to define the square root function sqrt(x) for the positive real numbers, and you want it to satisfy sqrt(x)2 =x then you have two choices for every x , either the positive or negative square root. So, for example, my sqrt(x) function might give the positive square root whenever x> 1 or the negative square root for 0 < x < = 1. The problem with this function is that it has jumps. So, ideally we want functions that are continuous (no jumps) and single valued. We can do that with the sqrt(x) function for the positive reals because we can either always choose the positive square root or always choose the negative one. You could even extend this definition to all the reals. Simply define sqrt(x) to be the positive square root of x if x> =0 or sqrt(x)=i.sqrt(-x) for x < 0. This function is continuous from the real line to the complex plane and is single valued. We could equally we have defined sqrt(x)=-i.sqrt(-x) for x < 0. Having solved the problem for the reals, it's natural to see if you can do the same for the complex numbers and define sqrt(x) as a continuous function for the whole complex plane. Unfortunately, it can't be done. Here is a way of proving this: draw a circular arc about 0 from 0 degrees to almost 360 degrees. Suppose we had defined sqrt(x) on the complex plane and it is continuous. This means, as x travels around the unit circle at 0 the value of sqrt(x) shouldn't jump. However, suppose sqrt(1)=1 (rather than the other possibility, -1). Move x anticlockwise around the circle. By the time we get to -1, sqrt(x) will be i. As we keep going around the circle approaching 1 from the other side of the circle, sqrt(x) will get close to -1. In other words, there has to be a jump in sqrt(x) at x=1. One way of solving this problem is to use "branch cuts". Rather than defining sqrt(x) for all x in the complex plane, we only define it for a subset. For example, I can define sqrt(x) for all complex numbers except positive real numbers or 0. I can do this, for example, by taking the positive square root of the modulus of a complex number and half the angle (where the angle must be between 0 and 360). The positive real numbers together with 0 are a "branch cut" (because we've "cut" them out of the complex plane). In fact, any reasonably nice path from 0 all the way out to infinity can be used to define a branch cut. A "branch" of a multivalued function is a choice of domain for the function, in this case we chose to make a branch cut from 0 to +infinity, and a choice of the function on that domain (we could have chosen -sqrt(x) as our sqrt(x) function). Now, back to exp(z). We can define exp(z) everywhere on the complex plane, no worries there. Suppose we want to define log(z). We know we can't do it at 0, but can we do it everywhere else? Again, the answer is no for much the same reason as last time. However, if we make a cut from 0 to +infinity we can solve the problem. We can then define log(z)=log|z|+i.arg(z) where |z| is the modulus of z, log|z| is the log function defined for the positive reals, and arg(z) gives the argument of z (the angle between 0 and 2pi). You can check that this works if you like. However, we could have chosen arg(z) to be the angle between 2pi and 4pi . Or between 124pi and 126pi . In general, we could have chosen the angle to be between 2npi and (2n+1)pi. Each choice of n will give us a different "branch" of log. Now think about defining, say, f(z)=iz . A suggestive possibility would be to reason by analogy to the real numbers and say iz =exp(z.log(i)). However, we know that we can't define log everywhere, so to define iz we need to choose a branch of log. Similarly, rather than defining ez =exp(z), we could have defined ez =exp(z.log(e)). There is one branch of log which will give this new definition of ez =exp(z). However, different branches of log will give different values of exp(z.log(e)). For example, suppose we chose a branch of log that gave log(e)=1+2i pi . Now, ei pi =exp(i pi (1+2i pi ))=exp(i pi )exp(-2pi 2 )=-exp(-2p 2 ) rather than -1pi Now, that's quite a lot to take in so if you have any questions, fire away... |
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| Colin
Prue |
So a 'branch cut' is what you remove to ensure that the definition of function is preserved? i still cannot see though, why (cos x+i sin x)a is many-valued, other than due to the nature of ez that you pointed out in your previous post...obviously i did not understand as well as i had thought... |
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| Dan
Goodman |
Yes, I think that's right. Think about how you would define (cos x+i sin x)a. As above, I would define it to mean exp(alog(cos x+i sin x)). However, that depends on a choice of branch of log again. A different branch of log would give a different value of (cos x + i sin x)a. As an exercise, try and calculate all possible values of ii (interestingly, one of them is a real number). |