Roots of unity and large expression
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $omega$ be a complex number such that $omega^5 = 1$ and $omega neq 1$. Find
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3.$$
I have tried combining the first and third terms & first and last terms. Here is what I have so far:
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^41 + omega^3 + fracomega^21 + omega^4 + fracomega^31 + omega \
&= dfracomega(1+omega^3) + omega^4(1+omega^2)(1+omega^2)(1+omega^3) + dfracomega^2(1+omega) + omega^3(1+omega^4)(1+omega^4)(1+omega) \
&= dfracomega + 2omega^4 +omega^61+omega^2 + omega^3 + omega^5 + dfracomega^2 + 2omega^3 + omega^71+omega + omega^4 + omega^5 \
&= dfrac2omega + 2omega^42+omega^2 + omega^3 +
dfrac2omega^2 + 2omega^32+omega+omega^4
endalign*
OR
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^31 + omega + fracomega^41 + omega^3 + fracomega^21 + omega^4 \
&= dfracomega(1+omega) + omega^3(1+omega^2)(1+omega)(1+omega^2) + dfracomega^2(1+omega^3) + omega^4(1+omega^4)(1+omega^3)(1+omega^4) \
&= dfracomega + omega^2 + omega^3 + omega^51+omega + omega^2 + omega^3 + dfracomega^2 + omega^4 + omega^5 + omega^81 + omega^3 + omega^4 + omega^7 \
&= dfrac2omega+omega^2+omega^31+omega+omega^2+omega^4 + dfrac1+omega+omega^2+omega^41+2omega^3+omega^4
endalign*
algebra-precalculus complex-numbers roots-of-unity
edited Aug 18 at 12:43
greedoid
27k93575
asked Aug 18 at 6:07
Ojasw Upadhyay
119111
add a comment |Â
up vote
2
down vote
favorite
Let $omega$ be a complex number such that $omega^5 = 1$ and $omega neq 1$. Find
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3.$$
I have tried combining the first and third terms & first and last terms. Here is what I have so far:
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^41 + omega^3 + fracomega^21 + omega^4 + fracomega^31 + omega \
&= dfracomega(1+omega^3) + omega^4(1+omega^2)(1+omega^2)(1+omega^3) + dfracomega^2(1+omega) + omega^3(1+omega^4)(1+omega^4)(1+omega) \
&= dfracomega + 2omega^4 +omega^61+omega^2 + omega^3 + omega^5 + dfracomega^2 + 2omega^3 + omega^71+omega + omega^4 + omega^5 \
&= dfrac2omega + 2omega^42+omega^2 + omega^3 +
dfrac2omega^2 + 2omega^32+omega+omega^4
endalign*
OR
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^31 + omega + fracomega^41 + omega^3 + fracomega^21 + omega^4 \
&= dfracomega(1+omega) + omega^3(1+omega^2)(1+omega)(1+omega^2) + dfracomega^2(1+omega^3) + omega^4(1+omega^4)(1+omega^3)(1+omega^4) \
&= dfracomega + omega^2 + omega^3 + omega^51+omega + omega^2 + omega^3 + dfracomega^2 + omega^4 + omega^5 + omega^81 + omega^3 + omega^4 + omega^7 \
&= dfrac2omega+omega^2+omega^31+omega+omega^2+omega^4 + dfrac1+omega+omega^2+omega^41+2omega^3+omega^4
endalign*
algebra-precalculus complex-numbers roots-of-unity
edited Aug 18 at 12:43
greedoid
27k93575
asked Aug 18 at 6:07
Ojasw Upadhyay
119111
Please note that this is different from previous similar questions because of the denominators of the terms in the original question. Their signs are flipped. Thanks!
– Ojasw Upadhyay
Aug 18 at 6:08
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $omega$ be a complex number such that $omega^5 = 1$ and $omega neq 1$. Find
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3.$$
I have tried combining the first and third terms & first and last terms. Here is what I have so far:
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^41 + omega^3 + fracomega^21 + omega^4 + fracomega^31 + omega \
&= dfracomega(1+omega^3) + omega^4(1+omega^2)(1+omega^2)(1+omega^3) + dfracomega^2(1+omega) + omega^3(1+omega^4)(1+omega^4)(1+omega) \
&= dfracomega + 2omega^4 +omega^61+omega^2 + omega^3 + omega^5 + dfracomega^2 + 2omega^3 + omega^71+omega + omega^4 + omega^5 \
&= dfrac2omega + 2omega^42+omega^2 + omega^3 +
dfrac2omega^2 + 2omega^32+omega+omega^4
endalign*
OR
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^31 + omega + fracomega^41 + omega^3 + fracomega^21 + omega^4 \
&= dfracomega(1+omega) + omega^3(1+omega^2)(1+omega)(1+omega^2) + dfracomega^2(1+omega^3) + omega^4(1+omega^4)(1+omega^3)(1+omega^4) \
&= dfracomega + omega^2 + omega^3 + omega^51+omega + omega^2 + omega^3 + dfracomega^2 + omega^4 + omega^5 + omega^81 + omega^3 + omega^4 + omega^7 \
&= dfrac2omega+omega^2+omega^31+omega+omega^2+omega^4 + dfrac1+omega+omega^2+omega^41+2omega^3+omega^4
endalign*
algebra-precalculus complex-numbers roots-of-unity
edited Aug 18 at 12:43
greedoid
27k93575
asked Aug 18 at 6:07
Ojasw Upadhyay
119111
Let $omega$ be a complex number such that $omega^5 = 1$ and $omega neq 1$. Find
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3.$$
I have tried combining the first and third terms & first and last terms. Here is what I have so far:
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^41 + omega^3 + fracomega^21 + omega^4 + fracomega^31 + omega \
&= dfracomega(1+omega^3) + omega^4(1+omega^2)(1+omega^2)(1+omega^3) + dfracomega^2(1+omega) + omega^3(1+omega^4)(1+omega^4)(1+omega) \
&= dfracomega + 2omega^4 +omega^61+omega^2 + omega^3 + omega^5 + dfracomega^2 + 2omega^3 + omega^71+omega + omega^4 + omega^5 \
&= dfrac2omega + 2omega^42+omega^2 + omega^3 +
dfrac2omega^2 + 2omega^32+omega+omega^4
endalign*
OR
beginalign*
fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3
&= fracomega1 + omega^2 + fracomega^31 + omega + fracomega^41 + omega^3 + fracomega^21 + omega^4 \
&= dfracomega(1+omega) + omega^3(1+omega^2)(1+omega)(1+omega^2) + dfracomega^2(1+omega^3) + omega^4(1+omega^4)(1+omega^3)(1+omega^4) \
&= dfracomega + omega^2 + omega^3 + omega^51+omega + omega^2 + omega^3 + dfracomega^2 + omega^4 + omega^5 + omega^81 + omega^3 + omega^4 + omega^7 \
&= dfrac2omega+omega^2+omega^31+omega+omega^2+omega^4 + dfrac1+omega+omega^2+omega^41+2omega^3+omega^4
endalign*
algebra-precalculus complex-numbers roots-of-unity
edited Aug 18 at 12:43
greedoid
27k93575
asked Aug 18 at 6:07
Ojasw Upadhyay
119111
edited Aug 18 at 12:43
greedoid
27k93575
edited Aug 18 at 12:43
greedoid
27k93575
edited Aug 18 at 12:43
greedoid
27k93575
27k93575
asked Aug 18 at 6:07
Ojasw Upadhyay
119111
asked Aug 18 at 6:07
Ojasw Upadhyay
119111
asked Aug 18 at 6:07
Ojasw Upadhyay
119111
119111
Please note that this is different from previous similar questions because of the denominators of the terms in the original question. Their signs are flipped. Thanks!
– Ojasw Upadhyay
Aug 18 at 6:08
add a comment |Â
Please note that this is different from previous similar questions because of the denominators of the terms in the original question. Their signs are flipped. Thanks!
– Ojasw Upadhyay
Aug 18 at 6:08
Please note that this is different from previous similar questions because of the denominators of the terms in the original question. Their signs are flipped. Thanks!
– Ojasw Upadhyay
Aug 18 at 6:08
Please note that this is different from previous similar questions because of the denominators of the terms in the original question. Their signs are flipped. Thanks!
– Ojasw Upadhyay
Aug 18 at 6:08
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Expand 2nd and 4th fraction with $omega $ and $omega ^2$ respectively: $$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3=fracomega1 + omega^2 + fracomega^3omega+ 1 + fracomega^31 + omega + fracomegaomega^2+1$$
$$=2fracomega1 + omega^2 + 2fracomega^3omega+ 1 $$
$$=2fracomega^2+omega + omega^3+1(omega^2+1)(omega+ 1)=2 $$
edited Aug 18 at 21:45
Henning Makholm
229k16294525
answered Aug 18 at 6:12
greedoid
27k93575
@henningmakholm nice
– user2661923
Aug 18 at 22:31
add a comment |Â
up vote
1
down vote
Alt. hint: Â let $,z=omega+dfrac1omega,$ so that $,z^2=omega^2+dfrac1omega^2+2,$, then use that $,omega^4=baromega,$ and $,omega^3=baromega^2,$ so the sum is:
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracbaromega^21 + baromega^4 + fracbaromega1 + baromega^2 = 2 operatornameReleft(fracomega1 + omega^2 + fracomega^21 + omega^4 right) = 2 operatornameReleft(frac1z + frac1z^2-2right)$$
But $,0=omega^5-1=(omega-1)left(omega^4+omega^3+omega^2+omega+1right)=omega^2(omega-1)left(z^2 + z - 1right),$, so $,z^2+z-1=0,$ and:
$$requirecancel
frac1z + frac1z^2-2 = frac1z+frac1-z-1 = fraccancel-z-1+cancelz-z^2-z = frac-1-1
$$
answered Aug 18 at 21:40
dxiv
55.2k64798
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Expand 2nd and 4th fraction with $omega $ and $omega ^2$ respectively: $$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3=fracomega1 + omega^2 + fracomega^3omega+ 1 + fracomega^31 + omega + fracomegaomega^2+1$$
$$=2fracomega1 + omega^2 + 2fracomega^3omega+ 1 $$
$$=2fracomega^2+omega + omega^3+1(omega^2+1)(omega+ 1)=2 $$
edited Aug 18 at 21:45
Henning Makholm
229k16294525
answered Aug 18 at 6:12
greedoid
27k93575
@henningmakholm nice
– user2661923
Aug 18 at 22:31
add a comment |Â
up vote
4
down vote
accepted
Expand 2nd and 4th fraction with $omega $ and $omega ^2$ respectively: $$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3=fracomega1 + omega^2 + fracomega^3omega+ 1 + fracomega^31 + omega + fracomegaomega^2+1$$
$$=2fracomega1 + omega^2 + 2fracomega^3omega+ 1 $$
$$=2fracomega^2+omega + omega^3+1(omega^2+1)(omega+ 1)=2 $$
edited Aug 18 at 21:45
Henning Makholm
229k16294525
answered Aug 18 at 6:12
greedoid
27k93575
@henningmakholm nice
– user2661923
Aug 18 at 22:31
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Expand 2nd and 4th fraction with $omega $ and $omega ^2$ respectively: $$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3=fracomega1 + omega^2 + fracomega^3omega+ 1 + fracomega^31 + omega + fracomegaomega^2+1$$
$$=2fracomega1 + omega^2 + 2fracomega^3omega+ 1 $$
$$=2fracomega^2+omega + omega^3+1(omega^2+1)(omega+ 1)=2 $$
edited Aug 18 at 21:45
Henning Makholm
229k16294525
answered Aug 18 at 6:12
greedoid
27k93575
Expand 2nd and 4th fraction with $omega $ and $omega ^2$ respectively: $$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracomega^31 + omega + fracomega^41 + omega^3=fracomega1 + omega^2 + fracomega^3omega+ 1 + fracomega^31 + omega + fracomegaomega^2+1$$
$$=2fracomega1 + omega^2 + 2fracomega^3omega+ 1 $$
$$=2fracomega^2+omega + omega^3+1(omega^2+1)(omega+ 1)=2 $$
edited Aug 18 at 21:45
Henning Makholm
229k16294525
answered Aug 18 at 6:12
greedoid
27k93575
edited Aug 18 at 21:45
Henning Makholm
229k16294525
edited Aug 18 at 21:45
Henning Makholm
229k16294525
edited Aug 18 at 21:45
Henning Makholm
229k16294525
229k16294525
answered Aug 18 at 6:12
greedoid
27k93575
answered Aug 18 at 6:12
greedoid
27k93575
answered Aug 18 at 6:12
greedoid
27k93575
27k93575
@henningmakholm nice
– user2661923
Aug 18 at 22:31
add a comment |Â
@henningmakholm nice
– user2661923
Aug 18 at 22:31
@henningmakholm nice
– user2661923
Aug 18 at 22:31
@henningmakholm nice
– user2661923
Aug 18 at 22:31
add a comment |Â
up vote
1
down vote
Alt. hint: Â let $,z=omega+dfrac1omega,$ so that $,z^2=omega^2+dfrac1omega^2+2,$, then use that $,omega^4=baromega,$ and $,omega^3=baromega^2,$ so the sum is:
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracbaromega^21 + baromega^4 + fracbaromega1 + baromega^2 = 2 operatornameReleft(fracomega1 + omega^2 + fracomega^21 + omega^4 right) = 2 operatornameReleft(frac1z + frac1z^2-2right)$$
But $,0=omega^5-1=(omega-1)left(omega^4+omega^3+omega^2+omega+1right)=omega^2(omega-1)left(z^2 + z - 1right),$, so $,z^2+z-1=0,$ and:
$$requirecancel
frac1z + frac1z^2-2 = frac1z+frac1-z-1 = fraccancel-z-1+cancelz-z^2-z = frac-1-1
$$
answered Aug 18 at 21:40
dxiv
55.2k64798
add a comment |Â
up vote
1
down vote
Alt. hint: Â let $,z=omega+dfrac1omega,$ so that $,z^2=omega^2+dfrac1omega^2+2,$, then use that $,omega^4=baromega,$ and $,omega^3=baromega^2,$ so the sum is:
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracbaromega^21 + baromega^4 + fracbaromega1 + baromega^2 = 2 operatornameReleft(fracomega1 + omega^2 + fracomega^21 + omega^4 right) = 2 operatornameReleft(frac1z + frac1z^2-2right)$$
But $,0=omega^5-1=(omega-1)left(omega^4+omega^3+omega^2+omega+1right)=omega^2(omega-1)left(z^2 + z - 1right),$, so $,z^2+z-1=0,$ and:
$$requirecancel
frac1z + frac1z^2-2 = frac1z+frac1-z-1 = fraccancel-z-1+cancelz-z^2-z = frac-1-1
$$
answered Aug 18 at 21:40
dxiv
55.2k64798
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Alt. hint: Â let $,z=omega+dfrac1omega,$ so that $,z^2=omega^2+dfrac1omega^2+2,$, then use that $,omega^4=baromega,$ and $,omega^3=baromega^2,$ so the sum is:
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracbaromega^21 + baromega^4 + fracbaromega1 + baromega^2 = 2 operatornameReleft(fracomega1 + omega^2 + fracomega^21 + omega^4 right) = 2 operatornameReleft(frac1z + frac1z^2-2right)$$
But $,0=omega^5-1=(omega-1)left(omega^4+omega^3+omega^2+omega+1right)=omega^2(omega-1)left(z^2 + z - 1right),$, so $,z^2+z-1=0,$ and:
$$requirecancel
frac1z + frac1z^2-2 = frac1z+frac1-z-1 = fraccancel-z-1+cancelz-z^2-z = frac-1-1
$$
answered Aug 18 at 21:40
dxiv
55.2k64798
Alt. hint: Â let $,z=omega+dfrac1omega,$ so that $,z^2=omega^2+dfrac1omega^2+2,$, then use that $,omega^4=baromega,$ and $,omega^3=baromega^2,$ so the sum is:
$$fracomega1 + omega^2 + fracomega^21 + omega^4 + fracbaromega^21 + baromega^4 + fracbaromega1 + baromega^2 = 2 operatornameReleft(fracomega1 + omega^2 + fracomega^21 + omega^4 right) = 2 operatornameReleft(frac1z + frac1z^2-2right)$$
But $,0=omega^5-1=(omega-1)left(omega^4+omega^3+omega^2+omega+1right)=omega^2(omega-1)left(z^2 + z - 1right),$, so $,z^2+z-1=0,$ and:
$$requirecancel
frac1z + frac1z^2-2 = frac1z+frac1-z-1 = fraccancel-z-1+cancelz-z^2-z = frac-1-1
$$
answered Aug 18 at 21:40
dxiv
55.2k64798
answered Aug 18 at 21:40
dxiv
55.2k64798
answered Aug 18 at 21:40
dxiv
55.2k64798
answered Aug 18 at 21:40
dxiv
55.2k64798
55.2k64798
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2886460%2froots-of-unity-and-large-expression%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Please note that this is different from previous similar questions because of the denominators of the terms in the original question. Their signs are flipped. Thanks!
– Ojasw Upadhyay
Aug 18 at 6:08