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Maneesh Narayanan
Maneesh Narayanan
1:39 PM
@ManeeshNarayanan I added a calculation, have a look. — Zarathustra 6 hours ago
do you understand, how he added exponential term?
 
Maneesh Narayanan
Maneesh Narayanan
1:59 PM
Hey @BAYMAX
 
BAYMAX
BAYMAX
2:22 PM
From en.wikipedia.org/wiki/Tridecagon it is clear that net force experienced by the charge at the center is zero
You can think of such odd sized gons
By comparing to a smaller odd gon
like a triangle
the same concept applies/ the geometry in case of odd n gons
@ManeeshNarayanan
is it clear?
 
Maneesh Narayanan
Maneesh Narayanan
How exponential term coming?
in his answer
@BAYMAX
 
BAYMAX
BAYMAX
3:03 PM
@ManeeshNarayanan i guess roots of unity!
now do u get it ?
 
Maneesh Narayanan
Maneesh Narayanan
How he plug the root of unity?
$\vec{F}=\frac{1}{4\pi \epsilon_0}$ $\frac{qQ}{r^2}$ right?
 
BAYMAX
BAYMAX
The coordinates of the charges on the 13-gon are?
from roots of unity
that is $z^{13} = 1$
$z^{13} = e^{i2\pi}$
 
Maneesh Narayanan
Maneesh Narayanan
yes. that I know
 
BAYMAX
BAYMAX
$z = e^{\frac{2\pi n i }{13}}$
 
Maneesh Narayanan
Maneesh Narayanan
where is $r^2$ term?
ok
unit vector right?
 
BAYMAX
BAYMAX
3:14 PM
yup
 
Maneesh Narayanan
Maneesh Narayanan
yes i got it
Thank you @BAYMAX
 
BAYMAX
BAYMAX
@ManeeshNarayanan $F$ is vector here but on the RHS you have a scalar
 
Maneesh Narayanan
Maneesh Narayanan
sorry
 
BAYMAX
BAYMAX
now the final form of the force is ?
 
Maneesh Narayanan
Maneesh Narayanan
@BAYMAX I had enlightment just above this message :)
 
BAYMAX
BAYMAX
3:18 PM
If I think then I write - $F = K \frac{qQ}{r^3} \hat{r}$
$F$ a vector
 
Maneesh Narayanan
Maneesh Narayanan
yes
r cap are root of unity.
 
BAYMAX
BAYMAX
so whats $\hat{r}$?
 
Maneesh Narayanan
Maneesh Narayanan
here in complex plane
 
BAYMAX
BAYMAX
$\hat{r} = Re(z) \hat{i} + Im(z) \hat{j}$
in vector form
 
Maneesh Narayanan
Maneesh Narayanan
yes
but how would he use the summation in plane?
 
BAYMAX
BAYMAX
3:21 PM
Better if you keep it in expo form
and sum all the forces
 
Maneesh Narayanan
Maneesh Narayanan
but we are actually dealing in plane right?
not complex plane
 
BAYMAX
BAYMAX
that is $\sum e^{\frac{2 n \pi }{13}}$
 
Maneesh Narayanan
Maneesh Narayanan
I know C is isomorphic to R^2
but here in computation will it obey the same?
 
BAYMAX
BAYMAX
@ManeeshNarayanan in your question please do the edit by writing upto this point
 
Maneesh Narayanan
Maneesh Narayanan
suppose $T$ is an isomorphism
 
BAYMAX
BAYMAX
3:25 PM
this is a valid qn
ask this in the edit and i hope u will get the soln
i will be back aftrs m time
 
Maneesh Narayanan
Maneesh Narayanan
yes I got the solution
@BAYMAX
consider the identity map
 
BAYMAX
BAYMAX
post it here
 
Maneesh Narayanan
Maneesh Narayanan
0 maps to 0
vector sign is there
complex number zero maps to (0,0) right?
hence resultant force must be zero.
Am I correct?
@BAYMAX
 
BAYMAX
BAYMAX
4:10 PM
? i did not get it, sorry
 
Maneesh Narayanan
Maneesh Narayanan
which part?
@BAYMAX
 
BAYMAX
BAYMAX
all :'(
 
Maneesh Narayanan
Maneesh Narayanan
($\mathbb C, +$(complex addition)) and ($\mathbb R^2, +$(vector addition)) are isomorphic.
 
BAYMAX
BAYMAX
ok
 
Maneesh Narayanan
Maneesh Narayanan
consider the operation done by him. consider the identity transformation between these structures
 
BAYMAX
BAYMAX
4:19 PM
ok
 
Maneesh Narayanan
Maneesh Narayanan
identity maps to identity under isomorphism
o complex maps to zero vector.
Now I think it is clearer to you.
T($\sum$)=$\sum T()$
If T is an isomorphism.
right?
@BAYMAX
 
BAYMAX
BAYMAX
May be you should ask in main about this.
 
Maneesh Narayanan
Maneesh Narayanan
ok.
I have added in the comment.
please see.
is that enough?
@BAYMAX
 
BAYMAX
BAYMAX
ok, better if you can include in the edit what your actual problem is?
like why we consider complex plane
while we r wrkng on real plane
also about isomorphisms
simply collect the points you asked me in the cht and add them to the edit
 
Maneesh Narayanan
Maneesh Narayanan
ok.
 
BAYMAX
BAYMAX
4:33 PM
and this will attract potential users to see and help
 
Maneesh Narayanan
Maneesh Narayanan
5:02 PM
@BAYMAX I have asked.
are you able to understand the question?
 
BAYMAX
BAYMAX
yup
It is good now
but i see it is on hold now ?? :'(
 
Maneesh Narayanan
Maneesh Narayanan
yes
 
Maneesh Narayanan
Maneesh Narayanan
5:26 PM
they are giving negative votes :'(
 
BAYMAX
BAYMAX
Hmm... let it close.. you can ask it as fresh in MSE I hope... with better Mathematical clarity
 

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