yet another

[Sarahisme]

ok is this now a geomettric series with the comman factor being e^{2pi/n}

hmm...well thats where my thinking is at the moment

??

[Sarahisme]

hmmm that doesnt seem to be getting me anywhere , damn,

[DQW]

It should ! Write down the result you would get from summing a geometric series. What is the first term, and what's the common ratio ?

[DQW]

ok is this now a geomettric series with the comman factor being e^{2pi/n}

hmm...well thats where my thinking is at the moment

??

Shouldn't that be e^{2i pi/n} ?

[Sarahisme]

right right so i was on the right track? yay!

 

ok be back in an hour, sorry again

[Sarahisme]

hey i am back again!

[Sarahisme]

okay this is where i gotten to now:

[DQW]

Nearly there...isn't there a term in that last expression that you can actually evaluate and plug in for ?

 

PS : You have no idea how much nicer it looks to see all the steps in there, following logically from each other !

[DQW]

Just one tiny error in line 2 : check the last term in that expansion (the term for m = n).

[Sarahisme]

you mean sub in z = a^{1/n}?

 

that doesnt seem to simplify it much

 

"PS : You have no idea how much nicer it looks to see all the steps in there, following logically from each other ! "

 

thanks

[Sarahisme]

Just one tiny error in line 2 : check the last term in that expansion (the term for m = n).

 

ok hold on....: here we go i think i fixed it...

[DQW]

you mean sub in z = a^{1/n}?

 

that doesnt seem to simplify it much

Say' date=' what's [imath']e^{2 \pi i} = [/imath] ?

[Sarahisme]

ahh lol i keep thinking in terms of normal exponential functions!

 

wait a sec i'll just do somem more working

[Sarahisme]

one more time now! :

[DQW]

Good !

 

And now that you're actually done with the problem, comes the important part : The Geometry !

[Sarahisme]

lol,, oh oooo...

[DQW]

Of course, this is beyond what the question wants you to do, but by spending another 15 minutes on the geometrical interpretation is valuable. If you think you'd want to do this, ask me and I'll guide you.

[Sarahisme]

ok is this to do with the roots being on a circle or something like that?

 

also the question says

 

"be sure to prove your answer" , i gather i have, but it didnt feel very 'proof like', if you know what i mean

[DQW]

ok is this to do with the roots being on a circle or something like that?

Yes it is.

 

also the question says

 

"be sure to prove your answer" , i gather i have, but it didnt feel very 'proof like', if you know what i mean

No, I don't ! It was very much 'proof like' !

[Sarahisme]

lol ok i see, thanks for all your help DQW! it is really really appreciated!

[Sarahisme]

[imath]

e^{900}-3\Pi\mathcal{12213\varphi}

[/imath]

[Sarahisme]

i think i've got this latex stuff working (i say i think, because it is not showing up on my computer as usual )

 

oh and yeah i would be happy to learn about the geometry of this problem if you are not yet sick of me

[DQW]

Recall how the general complex number [imath]z =r \cdot e^{i \phi} [/imath] is plotted on the argand plane. It is a point that is a distance 'r' from the origin, at an angle [imath]\phi [/imath] from the positive real (x) axis.

 

Now look at your roots. They all have the same modulus = |a|^{1/n}. So they must all lie on a circle of radius r = |a|^{1/n}. In addition, note that their arguments divide [imath]2 \pi [/imath] into n equal parts.

 

The first root has argument [imath] \theta /n [/imath]. The next root's argument is [imath] \theta /n + (2 \pi/n) [/imath]. The one following this has an argument [imath]\theta /n + 2 \cdot (2 \pi/n) [/imath] and so on... So these points form a regular polygon of n sides, centered on the origin. You can visualize the points as the ends of position vectors sticking out radially from the origin.The symmetry of the arragement of these vectors requires that their resultant be a zero vector. So, the sum of the roots is also zero.

 

In the case of even values of n there's a "more direct" argument that can be given. For every point at angle [imath] \phi [/imath] there is a diametrically opposite (antipodal) point at angle [imath] \pi + \phi [/imath]. These points are of the form x + iy and -x -iy, so naturally, their sum = 0. Similarly, each of the roots can be paired up with its antipode and added to give zero. The total sum must hence be zero.

[Sarahisme]

ahh i understand what you are saying, but the latex stuff is not showing up, give me a day or so to try and fix it, sorry....thanks for explaining this though

[Sarahisme]

Recall how the general complex number [imath]z =r \cdot e^{i \phi} [/imath] is plotted on the argand plane. It is a point that is a distance 'r' from the origin' date=' at an angle [imath']\phi [/imath] from the positive real (x) axis.

 

Now look at your roots. They all have the same modulus = |a|^{1/n}. So they must all lie on a circle of radius r = |a|^{1/n}. In addition, note that their arguments divide [imath]2 \pi [/imath] into n equal parts.

 

The first root has argument [imath] \theta /n [/imath]. The next root's argument is [imath] \theta /n + (2 \pi/n) [/imath]. The one following this has an argument [imath]\theta /n + 2 \cdot (2 \pi/n) [/imath] and so on... So these points form a regular polygon of n sides, centered on the origin. You can visualize the points as the ends of position vectors sticking out radially from the origin.The symmetry of the arragement of these vectors requires that their resultant be a zero vector. So, the sum of the roots is also zero.

 

In the case of even values of n there's a "more direct" argument that can be given. For every point at angle [imath] \phi [/imath] there is a diametrically opposite (antipodal) point at angle [imath] \pi + \phi [/imath]. These points are of the form x + iy and -x -iy, so naturally, their sum = 0. Similarly, each of the roots can be paired up with its antipode and added to give zero. The total sum must hence be zero.

 

 

hmm i see, yeah that makes sense geometrically as well now i think, thanks DQW