Complex complex number problem

!nval!d_us3rnam3 · 2019-04-19 03:22:54

Hello there! I haven't been here in a long time.

Find all possible integer values of $n$ such that the following system of equations has a solution for z:

z^n = 1,
(z + 1/z)^n = 1.

I'm not sure at all where to start; can you explain how to go about the harder bits?
Thanks!

Last edited by !nval!d_us3rnam3 (2019-04-19 03:23:57)

!nval!d_us3rnam3 · 2019-04-20 03:00:19

Bump this thread; can anyone help?

Bob · 2019-04-20 04:15:56

Hi,

I've been hoping someone else would make a suggestion here. The first equation alone gives the nth roots of 1. In an Argand diagram, draw a unit circle around the origin and mark off points every 2π/n radians

Obviously you need more ... I'll keep thinking.

Bob

!nval!d_us3rnam3 · 2019-04-20 04:50:35

Should I just use roots of unity on the first equation? We don't have a definitive value for n, though...
People are saying to try proving that 1/z is equal to the conjugate of z. Is that possible? Can you show me how?

Bob · 2019-04-20 06:15:45

Oh, that's interesting.

Let's say z = a + ib.

If z lies on that unit circle then a^2 + b^2 = 1

1/z = 1/(a + ib) = (a - ib)/((a + ib)(a - ib)) = (a - ib)/(a^2 + b^2) = a - ib

So, yes, once equation 1 is given, 1/z is the conjugate of z.

Bob

!nval!d_us3rnam3 · 2019-04-20 06:21:26

That's great!
Can you help with the rest of the problem? I'm kind of still stuck.

Bob · 2019-04-20 07:36:11

I've realised that I've been assuming that n > 0.

Let's stick with that for the moment.

The first equation fixes z as any of the nth roots of 1 and that will work for any n > 0.

The second equation becomes ( as a + ib + a - ib = 2a ) .... (2a)^n = 1 which implies a = 1/2 and therefore b = √3/2 . This equation also has these solutions for all n > 0.

Now to consider other values of n.

If n = 0, and as anything to the power zero is 1, we still have solutions.

If n < o. What ?

The first equation can be re written as (1/z)^positive integer = 1 let's say (1/z)^m = 1 where m >0

Have to go back to pencil and paper to explore some more. Continued in next post.

B

Bob · 2019-04-20 07:49:55

Let's say w^m = 1 where w = 1/z

By the same argument w is a root of 1, so w = a + ib where a^2 + b^2 = 1, and z + 1/z = 2a again.

Seems to me n can be any integer.

Is any of this correct? Now I'm worried my whole reasoning is faulty. I guess that's why I originally hoped someone else would do this question for us.

Still thinking ?

Bob

!nval!d_us3rnam3 · 2019-04-20 08:06:13

I got the same thing. That means either we're both right or we're both wrong.
I used the fact that 1/z = conjugate(z) on the second equation, and got a=1/2. Not sure what I'd do with that; people are talking about that too, but haven't said much else.

Bob · 2019-04-20 08:10:15

Ok. I think I have it.

If z = 1/2 + √3/2 i then 'n' isn't any value. Specially it is 6 because only this complex number raised to the power 6 (or any higher multiple of 6) will actually give 1. Use the angle made with the X axis ... It's π/3 so needs six of these to make 2π.

So, best answer I can come up with: n must be a multiple of 6.

I think I can sleep on that.

Bob

Math Is Fun Forum

#1 2019-04-19 03:22:54

Complex complex number problem

#2 2019-04-20 03:00:19

Re: Complex complex number problem

#3 2019-04-20 04:15:56

Re: Complex complex number problem

#4 2019-04-20 04:50:35

Re: Complex complex number problem

#5 2019-04-20 06:15:45

Re: Complex complex number problem

#6 2019-04-20 06:21:26

Re: Complex complex number problem

#7 2019-04-20 07:36:11

Re: Complex complex number problem

#8 2019-04-20 07:49:55

Re: Complex complex number problem

#9 2019-04-20 08:06:13

Re: Complex complex number problem

#10 2019-04-20 08:10:15

Re: Complex complex number problem

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