Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

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

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 46

Complex complex number problem

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)


"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Offline

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

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 46

Re: Complex complex number problem

Bump this thread; can anyone help?


"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Offline

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

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,541

Re: Complex complex number problem

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Bundy smile

Offline

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

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 46

Re: Complex complex number problem

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?


"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Offline

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

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,541

Re: Complex complex number problem

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Bundy smile

Offline

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

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 46

Re: Complex complex number problem

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


"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Offline

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

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,541

Re: Complex complex number problem

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Bundy smile

Offline

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

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,541

Re: Complex complex number problem

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Bundy smile

Offline

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

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 46

Re: Complex complex number problem

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.


"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

Offline

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

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,541

Re: Complex complex number problem

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Bundy smile

Offline

Board footer

Powered by FluxBB