/ ith root of -1

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ablackett - on 21 Sep 2016
I'm a maths teacher, and have been marking some Y10 papers.

Believe it or not, one of them has left me some mathematical doodling in the margin which at one point finds the ith root of -1.

e^i*Pi = -1
e^Pi = (-1)^(1/i)

He then incorrectly writes that
e^Pi = -i

and goes onto write
-e^pi = i
(-e^pi)^2 = -1
(e^pi)^2 = 535.49

So 535.49 = -1

His mistake is thinking that the ith root of -1 is i, Wolfram tells me that the ith root of -1 is -1.

I have a maths degree, so am reasonable comfortable with Argand diagrams roots of unity etc, but have never come across ith roots before.

Can anyone explain it to me in a way that I have a hope in hell of explaining it to an inquisitive y10.
jonny taylor on 21 Sep 2016
In reply to ablackett:

Sounds like a lad who will go far!

Certainly got me thinking. He is totally right, and he’s basically there with his second line.

If I define the nth root of a as satisfying the equation x^n = a, then in this case we are trying to solve x^i = -1.

I happen to remember (as he does) that e^{i pi} = -1, or to write it in a less familiar, but equally correct way, (e^pi}^i = -1. This implies that an ith root of -1 is actually e^pi.

I don’t know if there are other roots (in the same way that there are two square roots of a number, a positive and a negative). That might be one for a mathematican not a physicist!
Robert Durran - on 21 Sep 2016
In reply to jonny taylor:
> I happen to remember (as he does) that e^{i pi} = -1, or to write it in a less familiar, but equally correct way, (e^pi}^i = -1. This implies that an ith root of -1 is actually e^pi.

> I don£t know if there are other roots (in the same way that there are two square roots of a number, a positive and a negative). That might be one for a mathematican not a physicist!

It is fairly easy to see that any number of the form e^kpi where k is an odd number (or the negative of an odd number) is an ith root of -1:

(e^kpi)^i
=e^kipi
=(e^ipi)k
=(-1)^k

Which is -1 when k is odd.
Obviously when k is even we have an ith root of 1

Note that I have not proved that these are the only ith roots of -1 ans 1............. but I suspect they are!

Edit: proved it now I think!
Post edited at 18:05
Robert Durran - on 21 Sep 2016
In reply to ablackett:
Wolfram tells me that the ith root of -1 is -1.

. Can't be right......

(-1)^i
=(e^ipi)^i
=e^(iipi)
=e^(-pi)

Which is not -1
Post edited at 18:06
Robert Durran - on 21 Sep 2016
In reply to ablackett:

> Wolfram tells me that the ith root of -1 is -1.

I've just put it into Wolfram and it agrees with e^kpi (k odd). I think you must have entered it carelessly.
ablackett - on 21 Sep 2016
In reply to Robert Durran:
> I've just put it into Wolfram and it agrees with e^kpi (k odd). I think you must have entered it carelessly.

Ah, yes.

I have now entered it as (-1)^(1/i)

http://m.wolframalpha.com/input/?i=%28-1%29%5E%281%2Fi%29&x=10&y=3

And it gives e^pi

How do I enter it to give the result e^kpi
Post edited at 19:32
Robert Durran - on 21 Sep 2016
In reply to ablackett:

> Ah, yes.

> I have now entered it as (-1)^(1/i)

> And it gives e^pi

> How do I enter it to give the result e^kpi

Enter (-1)^(1/i) as you did, but scroll down the results and it gives them all at the bottom.
crossdressingrodney - on 21 Sep 2016
In reply to ablackett:

To prove that you've found all the solutions to x^i = -1
- zero is not a solution
- any non-zero complex number can be written as x = e^(a + ib) where a is real and b lies in [0, 2pi).
- then x^i = e^(ia-b) so we have e^(ia-b) = -1.
- taking absolute values shows that b = 0.
- solving e^(ia) = -1 gives a an odd multiple of pi.

Very impressive stuff from a year 10!
jonny taylor on 22 Sep 2016
In reply to crossdressingrodney:

Nice!

Any suggestions about where this result is useful? I can't think of anything in physics/engineering, at least in my own areas. Do the properties of complex roots feature in any well-known pure maths problems?
planetmarshall on 22 Sep 2016
In reply to jonny taylor:

> Do the properties of complex roots feature in any well-known pure maths problems?

The Riemann Hypothesis.

planetmarshall on 22 Sep 2016
In reply to jonny taylor:

They are also crucial to Wiles' eventual proof of Fermat's Last Theorem, via the Taniyama–Shimura-Weil conjecture.
LittleRob - on 22 Sep 2016
In reply to planetmarshall:


And if you've ever read "The Humans" by Matt Haig https://www.amazon.co.uk/Humans-Matt-Haig/dp/0857868780 you'll know why you don't want to solve that ;-)
Post edited at 10:57
planetmarshall on 22 Sep 2016
In reply to LittleRob:

> And if you've ever read "The Humans" by Matt Haig...

I haven't, but I need more non-mountaineering literature so thanks for the recommendation.
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jonny taylor on 22 Sep 2016
In reply to planetmarshall:

> The Riemann Hypothesis.

Ah yes, of course.

Looks like this year 10 student should get right on that as his next move!

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