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    So, I understand that a\sin{x} + b\cos{x} can usually be written in the form R\cos{(x-a)}, where R = \sqrt{a^2 + b^2} and a is equal to blah blah blah..
    But since  e^{i\theta} = \cos\theta +i \sin\theta, R = \sqrt{1 + i^2} = 0

    Obviously algebraically, I understand why it's  0 , but intuitively, why can't it be written in this form? I've tried it for other coefficients of i and it seems that \cos(a) must be imaginary [is that possible?]. I just found it interesting and would kind of like an explanation as to why it only works for real numbers, thanks
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    (Original post by Callum Scott)
    So, I understand that a\sin{x} + b\cos{x} can usually be written in the form R\cos{(x-a)}, where R = \sqrt{a^2 + b^2} and a is equal to blah blah blah..
    But since  e^{i\pi} = \cos\theta +i \sin\theta, R = \sqrt{1 + i^2} = 0

    Obviously algebraically, I understand why it's , but intuitively, why can't it be written in this form? I've tried it for other coefficients of i and it seems that \cos(a) must be imaginary [is that possible?]. I just found it interesting and would kind of like an explanation as to why it only works for real numbers, thanks
    e^(ipi) = cos pi + isin(pi) = cos pi = cos (pi - 0) so it's already in that form with R = 1 and a = 0
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    (Original post by davros)
    e^(ipi) = cos pi + isin(pi) = cos pi = cos (pi - 0) so it's already in that form with R = 1 and a = 0
    SORRY! e^{ix}, I have Euler's formula as my wallpaper and it distracted me ok :argh:, I'll try to change it now lol

    Edit: Turns out I can't change the title, dammit
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    ... I see ...
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    I actually thought you could probably do this if you allowed R and a to be complex, but turns out it doesn't work:

    We want to find R, a such that

    e^iz = cos z + i sin z = R (cos z cos a + sin z sin a)

    setting z = 0 we have R cos a = 1
    setting z = pi/2 we have R sin a = i
    dividing we find tan a = i, that is, sin a /cos a = i and so i sin a / cos a = -1.

    Since e^ia - e^-ia = 2i sin a and e^ia+e-ia = 2 cos a, we must have \dfrac{e^{ia} - e^{-ia}}{e^{ia} + e^{-ia}} = -1. Writing s = ia, this is:

    \dfrac{e^s - e^{-s}}{e^s+e^{-s}} = -1 from which we see we must have -s = \infty and so a=i\infty. Which isn't terribly helpful (I'm not sure if you can make it work if you're prepared to deal with this quantity, but I can't see it being fruitful).

    Not sure if this sheds any light on anything to be honest...
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    (Original post by DFranklin)
    I actually thought you could probably do this if you allowed R and a to be complex, but turns out it doesn't work:

    We want to find R, a such that

    e^iz = cos z + i sin z = R (cos z cos a + sin z sin a)

    setting z = 0 we have R cos a = 1
    setting z = pi/2 we have R sin a = i
    dividing we find tan a = i, that is, sin a /cos a = i and so i sin a / cos a = -1.

    Since e^ia - e^-ia = 2i sin a and e^ia+e-ia = 2 cos a, we must have \dfrac{e^{ia} - e^{-ia}}{e^{ia} + e^{-ia}} = -1. Writing s = ia, this is:

    \dfrac{e^s - e^{-s}}{e^s+e^{-s}} = -1 from which we see we must have -s = \infty and so a=i\infty. Which isn't terribly helpful (I'm not sure if you can make it work if you're prepared to deal with this quantity, but I can't see it being fruitful).

    Not sure if this sheds any light on anything to be honest...
    I don't have a uni level education, so I have no idea how to correctly manipulate infinities without limits, but:

    Substituting a = i\infty into R = \frac{i}{\sin{a}} \equiv \frac{1}{\cos{a}}

    Then,  \cos a = \frac{e^{-\infty} + e^\infty}{2} = \frac{e^\infty}{2} \therefore R = \frac{2}{e^\infty} = 0 <- what I got before anyway

    So in essence, does \cos{x} + i\sin{x} = \frac{2}{e^\infty} \cos(x-i\infty) = \frac{2}{e^\infty}\left(\cosh \infty \cos x + \sinh\infty \sin x\right), which doesn't really mean anything, but if it does mean that, I'll take it, haha.
    I was just perplexed as to the conceptual, non-mathematical reasoning behind why complex numbers can't be written in this manner. It all seems a bit weird.

    Could it be that when you plot 2 graphs of a\sin x and b\cos x that adding them sort of represents their interference with each other, but purely real and purely imaginary functions have a sort of inherent, natural perpedicularity to them. I've gotten to the point where I have no idea what I'm even saying anymore, I just think it's awesome how they can't be merged together; and if it does mean whatever I just mumbled on about, that's pretty awesome. Nonetheless, it's still awesome.
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    DFranklin has put his finger on it, but maybe I could elaborate just a little more...

    If you take everthing in sight as being defined in the complex plane, then you still come out with these equations:

     a = R \cos(\alpha)
     b = R \sin(\alpha)

    and therefore

     \alpha = \arctan(b/a) and  R^2 = a^2 + b^2

    The point is that the arctan function is well behaved when you restrict attention to the real line - there's always a solution. When you extend attention to the complex plane, this is no longer the case. For example \tan(\alpha) = i has no finite solutions. Similarly, the equation in R^2 can end up giving an answer zero.

    As an extra credit exercise: do the values of (a, b) that fail to give an answer for \alpha (the "poles" of \arctan(z)) corresond to the values that fail to give a non-zero answer for R?
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    I should mention that the formulae in the complex domain do work as long as you keep away from these "tricky" points. For example, messing around with Mathematica will give you

    2 \sqrt 6 \cos(z - \arctan(i/5)) = 5 \cos(z) + i \sin(z)
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    (Original post by atsruser)
    What are your a and b here?
    If you go back to the start, we are trying to solve for R and \alpha in

    a \cos(z) + b \sin(z) = R \cos(z - \alpha)

    for fixed constants a, b and for a complex variable z. Crunch through the equations and you get to the simultaneous equations I stated above. What I am saying is that there are solutions to these equations except for certain circumstances involving the values of a and b

    So the whole attempt fails for *any* complex number, no?
    No. There are values of a and b that mess things up (such as  a = 1, b = i ) but otherwise you can solve the equations. Above, I gave an example where the procedure works:

    2 \sqrt 6 \cos(z - \arctan(i/5)) = 5 \cos(z) + i \sin(z)

    It's a very nice question though.
    Definitely.
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    (Original post by Gregorius)
    DFranklin has put his finger on it, but maybe I could elaborate just a little more...

    If you take everthing in sight as being defined in the complex plane, then you still come out with these equations:

     a = R \cos(\alpha)
     b = R \sin(\alpha)

    and therefore

     \alpha = \arctan(b/a) and  R^2 = a^2 + b^2
    Thanks for providing/verifying these equations - I was mentally blanking on the approach for deriving them and so basically bailed once I got to \arctan i = i\infty.
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    (Original post by Gregorius)
    If you go back to the start, we are trying to solve for R and \alpha in

    a \cos(z) + b \sin(z) = R \cos(z - \alpha)
    You ended up replying to a deleted post, as I didn't get rid of it quickly enough. I realised that I hadn't read your post properly before I submitted it.

    However, I think that you have changed the original question posed the OP, who is an A level student, I think. As originally posed, I think that it is correct to state that there are no solutions, as he was explicitly looking for a solution based on re^{ix} = r(\cos x +i \sin x) i.e. with a=1,b=i
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    (Original post by atsruser)
    However, I think that you have changed the original question posed the OP, who is an A level student, I think. As originally posed, I think that it is correct to state that there are no solutions, as he was explicitly looking for a solution based on re^{ix} = r(\cos x +i \sin x) i.e. with a=1,b=i
    Oh indeed, yes. But one component of OP's original question was an inquiry about why this was happening; asking for intuition. OP had pointed out that he understood the algebraic manipulation and how this arrived at the conclusion of no solution for this particular case.

    In order to answer this more general question, one has to generalize the original observation; what we mathematicians do all the time!
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    (Original post by Gregorius)
    [user=94857]

     a = R \cos(\alpha)
     b = R \sin(\alpha)
     \alpha = \arctan(b/a) and  R^2 = a^2 + b^2

    As an extra credit exercise: do the values of (a, b) that fail to give an answer for \alpha (the "poles" of \arctan(z)) corresond to the values that fail to give a non-zero answer for R?
    Well, if we have a^2+b^2=0 \Rightarrow a=\pm ib then \tan \alpha = b/a = \pm i which have no solutions. So I'd say that the answer to your question is yes.
 
 
 
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