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 (zi)^{n}  (z+i)^{n} = 0
The roots are z = cot(rPi/n) r = 1, 2, ... (n1)
I've tried a few different approaches, and had conflicting results  but no solution.
My current attempt is to say:
(zi)^{n}(z+i)^{n} = 1 = e^{iPi(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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 22032012 12:05

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 22032012 12:49
I think the easiest approach is:
First show what Im(z) must be (use a modulus argument).
Then, consider what arg(zi) and arg(z+i) are. 
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 22032012 13:46
(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(zi) and arg(z+i) are.
I'm still looking at the arguments, but I just thought I'd check that was the result needed from the modulus argument.Last edited by 99wattr89; 22032012 at 14:25. 
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 22032012 14:00
Yes.

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 22032012 14:30
(Original post by DFranklin)
Yes.
I get:
arg(z+i) = 1/z
arg(zi) = 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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 22032012 15:17
I think you're confused. You probably mean tan(arg(z+i)) = 1/z...

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 23032012 10:40
(Original post by DFranklin)
I think you're confused. You probably mean tan(arg(z+i)) = 1/z...
I don't know what I can say about the arguments themselves, since I don't know the value of z.
arg(zi) + arg(z+i) = 2Pi ? 
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 23032012 11:20
That isn't (generally) true.
Suppose arg(zi) = 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(zi)+arg(z+i) is (in terms of w). And basically that's all you need. 
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 23032012 12:25
(Original post by DFranklin)
That isn't (generally) true.
Suppose arg(zi) = 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(zi)+arg(z+i) is (in terms of w). And basically that's all you need.
w + 2arg(z+i) = 2Pi ?
So arg(zi) + 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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 23032012 13:49
(Original post by DFranklin)
That isn't (generally) true.
Suppose arg(zi) = 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(zi)+arg(z+i) is (in terms of w). And basically that's all you need.(Original post by 99watt89)
..
arg((z+i)^n/(zi)^n) = n arg((z+i)/(zi)) = n (arg(z+i)  arg(zi)).
So it is arg(z+i)  arg(zi) you should be looking at. 
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 23032012 14:04
(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/(zi)^n) = n arg((z+i)/(zi)) = n (arg(z+i)  arg(zi)).
So it is arg(z+i)  arg(zi) you should be looking at.
So, given that arg(zi) = arg(z+i), that would mean that adding them gives zero!
So then arg((z+i)^{n}/(zi)^{n}) = 0
But for an argument to be zero, the input has to be a positive real number. So (z+i)/(zi) is a real number.
Rationalizing the fraction, I get [z^{2} +2iz 1]/[z^{2} 1]
So 2iz must be zero. But... that only works if z is zero. This seems like a contradiction. 
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 23032012 14:07
(Original post by 99wattr89)
Don't worry about it.
So, given that arg(zi) = arg(z+i) that would mean that adding them gives zero! 
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 23032012 14:24
(Original post by DFranklin)
Why are you *adding* them, when I've just said you should be looking at arg(z+i)  arg(zi)?
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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 23032012 14:30
(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.
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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 23032012 14:55
(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.
Thanks again! 
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 24032012 10:30
(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.
If arg(z+i) is theta, then arg(zi) is 2Pi  theta, as they're symmetrical  that is what I was meant to take from an argand diagram, right?
So arg(zi)  arg(z+i) = 2Pi  theta  theta = 2Pi  2theta
arg[(zi)^{n}/(z+i)^{n}] = n(2Pi  2theta)
But I don't know what that should be telling me. 
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 24032012 12:17
(Original post by 99wattr89)
arg[(zi)^{n}/(z+i)^{n}] = n(2Pi  2theta)
But I don't know what that should be telling me.
You're wanting (zi)^n = (z+i)^n, which means (zi)^{n}/(z+i)^{n} = 1, which means arg[(zi)^{n}/(z+i)^{n}] must be a multiple of 2pi.
So theta must be... 
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 24032012 14:48
(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 (zi)^n = (z+i)^n, which means (zi)^{n}/(z+i)^{n} = 1, which means arg[(zi)^{n}/(z+i)^{n}] must be a multiple of 2pi.
So theta must be...
But, if n is an integer, then 2n(Pitheta) = 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 (zi)^{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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 24032012 18:15
(Original post by 99wattr89)
So you have to assume that n is an integer?
But, if n is an integer, then 2n(Pitheta) = 2nPi
If so, what you've written is wrong. 2n(pitheta) 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 is an integer.
So is an integer.
So since n is also an integer, what can we say about ? 
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 25032012 13:29
(Original post by DFranklin)
Yes, n is an integer. It's clear from the 1, ..., n1 bit in the given answer.
Are you saying this because we decided that 2n(pitheta) had to be a multiple of pi?
If so, what you've written is wrong. 2n(pitheta) 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 is an integer.
So is an integer.
So since n is also an integer, what can we say about ?
Since n is an integer, has to be too. So 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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Updated: April 3, 2012
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