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Solve (z-i)^n = (z+i)^n

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    I'm doing a multiple part question, and I'm stuck at the second step:

    http://i.imgur.com/0yA8Z.png

    I've shown that e^iθ = cosθ + isinθ

    Now I need to show for (z-i)n - (z+i)n = 0

    The roots are z = cot(rPi/n) r = 1, 2, ... (n-1)

    I've tried a few different approaches, and had conflicting results - but no solution.

    My current attempt is to say:

    (z-i)n(z+i)-n = 1 = eiPi(2n)

    Is my current method a dead end or the right approach?

    Is it correct to say that n is an integer and is positive, or can you not assume that?

    Thanks for reading.
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    I think the easiest approach is:

    First show what Im(z) must be (use a modulus argument).

    Then, consider what arg(z-i) and arg(z+i) are.
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    (Original post by DFranklin)
    I think the easiest approach is:

    First show what Im(z) must be (use a modulus argument).

    Then, consider what arg(z-i) and arg(z+i) are.
    Ah! So the moduli being equal can be used to show that z is a real number!

    I'm still looking at the arguments, but I just thought I'd check that was the result needed from the modulus argument.
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    Yes.
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    (Original post by DFranklin)
    Yes.
    Ok, I've been thinking about what the arguments mean, but I don't think I'm correct in my interpretation, because it seems to me that the answers are at z = 0, Pi/2 and Pi, which is not like the expected answers.

    I get:

    arg(z+i) = 1/z
    arg(z-i) = -1/z

    So, that would suggest to me that for even integer values of n the formula only holds true when the arguments are equal, and for odd integer values of n it also holds true when the arguments are opposite, which gives z = 0 and Pi for n even, and z = 0, Pi/2 and Pi for n odd.

    Is the problem that I can't assume that n is an integer?
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    I think you're confused. You probably mean tan(arg(z+i)) = 1/z...
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    (Original post by DFranklin)
    I think you're confused. You probably mean tan(arg(z+i)) = 1/z...
    I'm sorry, that is what I mean, yes.

    I don't know what I can say about the arguments themselves, since I don't know the value of z.

    arg(z-i) + arg(z+i) = 2Pi ?
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    That isn't (generally) true.

    Suppose arg(z-i) = w. What can you say about arg(z+i)? (Only using the fact that z is real, don't worry about anything else).

    It should be fairly obvious - draw a diagram if necessary.

    From there it's easy to see what arg(z-i)+arg(z+i) is (in terms of w). And basically that's all you need.
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    (Original post by DFranklin)
    That isn't (generally) true.

    Suppose arg(z-i) = w. What can you say about arg(z+i)? (Only using the fact that z is real, don't worry about anything else).

    It should be fairly obvious - draw a diagram if necessary.

    From there it's easy to see what arg(z-i)+arg(z+i) is (in terms of w). And basically that's all you need.
    Well, the two arguments are always symmetrical in the real axis. So if w was the argument of (z-i) then you could say that the argument of (z+i) would be -w. I'm not actually sure how z being real comes into it though.

    w + 2arg(z+i) = 2Pi ?

    So arg(z-i) + arg(z+i) = w/2 +Pi ?


    I'm sorry, I must be being really stupid, because I'm not finding the answer.
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    (Original post by DFranklin)
    That isn't (generally) true.

    Suppose arg(z-i) = w. What can you say about arg(z+i)? (Only using the fact that z is real, don't worry about anything else).

    It should be fairly obvious - draw a diagram if necessary.

    From there it's easy to see what arg(z-i)+arg(z+i) is (in terms of w). And basically that's all you need.
    (Original post by 99watt89)
    ..
    Sorry, the bit in bold is wrong (I didn't reread the thread and was following what you had posted).

    arg((z+i)^n/(z-i)^n) = n arg((z+i)/(z-i)) = n (arg(z+i) - arg(z-i)).

    So it is arg(z+i) - arg(z-i) you should be looking at.
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    (Original post by DFranklin)
    Sorry, the bit in bold is wrong (I didn't reread the thread and was following what you had posted).

    arg((z+i)^n/(z-i)^n) = n arg((z+i)/(z-i)) = n (arg(z+i) - arg(z-i)).

    So it is arg(z+i) - arg(z-i) you should be looking at.
    Don't worry about it.

    So, given that arg(z-i) = -arg(z+i), that would mean that adding them gives zero!

    So then arg((z+i)n/(z-i)n) = 0

    But for an argument to be zero, the input has to be a positive real number. So (z+i)/(z-i) is a real number.

    Rationalizing the fraction, I get [z2 +2iz -1]/[z2 -1]

    So 2iz must be zero. But... that only works if z is zero. This seems like a contradiction.
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    (Original post by 99wattr89)
    Don't worry about it.

    So, given that arg(z-i) = -arg(z+i) that would mean that adding them gives zero!
    Why are you *adding* them, when I've just said you should be looking at arg(z+i) - arg(z-i)?
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    (Original post by DFranklin)
    Why are you *adding* them, when I've just said you should be looking at arg(z+i) - arg(z-i)?
    I'm sorry. I'm being terrible today.

    I'll stop wasting your time, and come back to this tomorrow instead and try again. Thanks for all the help and patience.
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    (Original post by 99wattr89)
    I'm sorry. I'm being terrible today.

    I'll stop wasting your time, and come back to this tomorrow instead and try again. Thanks for all the help and patience.
    Up to you, but don't let me get you down.

    I do think that once you think about subtracting them rather than adding, you'll not be far from a solution.
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    (Original post by DFranklin)
    Up to you, but don't let me get you down.

    I do think that once you think about subtracting them rather than adding, you'll not be far from a solution.
    Oh no, it's not you, you've been great. I'm just really off today - I have a lot of other stuff on my mind and I keep cocking this up. I'll start afresh looking at the subtracting case tomorrow and let you know how it goes.

    Thanks again!
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    (Original post by DFranklin)
    I do think that once you think about subtracting them rather than adding, you'll not be far from a solution.
    I'm starting to get really worried now, because I have absolutely no idea what I'm doing.

    If arg(z+i) is theta, then arg(z-i) is 2Pi - theta, as they're symmetrical - that is what I was meant to take from an argand diagram, right?

    So arg(z-i) - arg(z+i) = 2Pi - theta - theta = 2Pi - 2theta

    arg[(z-i)n/(z+i)n] = n(2Pi - 2theta)

    But I don't know what that should be telling me.
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    (Original post by 99wattr89)
    arg[(z-i)n/(z+i)n] = n(2Pi - 2theta)

    But I don't know what that should be telling me.
    OK, so this is the point where you need to look at what you're trying to solve.

    You're wanting (z-i)^n = (z+i)^n, which means (z-i)n/(z+i)n = 1, which means arg[(z-i)n/(z+i)n] must be a multiple of 2pi.

    So theta must be...
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    (Original post by DFranklin)
    OK, so this is the point where you need to look at what you're trying to solve.

    You're wanting (z-i)^n = (z+i)^n, which means (z-i)n/(z+i)n = 1, which means arg[(z-i)n/(z+i)n] must be a multiple of 2pi.

    So theta must be...
    So you have to assume that n is an integer? Because you can't solve it if n can be non-integers. :confused:

    But, if n is an integer, then 2n(Pi-theta) = 2nPi

    So theta is Pi/2.

    So arg(z+i) = Pi/2

    So since z is real, z must be 1.

    But that doesn't make sense, because if you expand (z-i)n=(z+i)n binomially, you get cancelling terms for the even powers, and opposite terms for the odd powers.

    So you get i -i/3! +i/5! ... = -i +i/3! -i/5! ...

    So 1 -1/3! + 1/5! ... = -(1 -1/3! + 1/5! ...)

    Which can only work if the terms are zero. Which would mean n was zero.
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    (Original post by 99wattr89)
    So you have to assume that n is an integer?
    Yes, n is an integer. It's clear from the 1, ..., n-1 bit in the given answer.

    But, if n is an integer, then 2n(Pi-theta) = 2nPi
    Are you saying this because we decided that 2n(pi-theta) had to be a multiple of pi?

    If so, what you've written is wrong. 2n(pi-theta) is going to be 2mPi for some integer m, but there's no reason to think that m will be the same as n (and in general it won't).

    Instead, I suggest:

    We know 2n pi - 2n \theta is a multiple of 2pi.

    So \dfrac{2n\pi - 2n \theta}{2\pi} is an integer.

    So  n - \dfrac{n\theta}{\pi} is an integer.

    So since n is also an integer, what can we say about \dfrac{n\theta}{\pi}?
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    (Original post by DFranklin)
    Yes, n is an integer. It's clear from the 1, ..., n-1 bit in the given answer.

    Are you saying this because we decided that 2n(pi-theta) had to be a multiple of pi?

    If so, what you've written is wrong. 2n(pi-theta) is going to be 2mPi for some integer m, but there's no reason to think that m will be the same as n (and in general it won't).

    Instead, I suggest:

    We know 2n pi - 2n \theta is a multiple of 2pi.

    So \dfrac{2n\pi - 2n \theta}{2\pi} is an integer.

    So  n - \dfrac{n\theta}{\pi} is an integer.

    So since n is also an integer, what can we say about \dfrac{n\theta}{\pi}?
    Ah, I think I understand what I'm doing a little more now.

    Since n is an integer, \dfrac{n\theta}{\pi} has to be too. So \dfrac{\theta}{\pi} has to be an integer as well.

    So theta has to be an integer multiple of pi. (Yet another independently varying integer?)

    But... that can't be right, because integer multiple of pi give angles that have no imaginary component.

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