If $z = cis(2kpi/5)$, $z neq 1$, then what is $(z+1/z)^2+(z^2 + 1/z^2)^2=$?
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question 20, part c in the picture:
I substituted the first time as $4 cos^2(2k pi/5)$ and the second term as $4 cos^2(4k pi/5)$, and then tried writing one term in terms of the other using the identity $cos 2a = 2 cos^2 a- 1$. I even tried bringing in $sin$ but I didn't get anywhere. The answer is supposed to be $3$. Can someone solve it?
trigonometry complex-numbers
edited Nov 17 at 11:10
Brahadeesh
5,78341957
asked Nov 17 at 10:45
Vanessa
606
add a comment |
up vote
1
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question 20, part c in the picture:
I substituted the first time as $4 cos^2(2k pi/5)$ and the second term as $4 cos^2(4k pi/5)$, and then tried writing one term in terms of the other using the identity $cos 2a = 2 cos^2 a- 1$. I even tried bringing in $sin$ but I didn't get anywhere. The answer is supposed to be $3$. Can someone solve it?
trigonometry complex-numbers
edited Nov 17 at 11:10
Brahadeesh
5,78341957
asked Nov 17 at 10:45
Vanessa
606
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
question 20, part c in the picture:
I substituted the first time as $4 cos^2(2k pi/5)$ and the second term as $4 cos^2(4k pi/5)$, and then tried writing one term in terms of the other using the identity $cos 2a = 2 cos^2 a- 1$. I even tried bringing in $sin$ but I didn't get anywhere. The answer is supposed to be $3$. Can someone solve it?
trigonometry complex-numbers
edited Nov 17 at 11:10
Brahadeesh
5,78341957
asked Nov 17 at 10:45
Vanessa
606
question 20, part c in the picture:
I substituted the first time as $4 cos^2(2k pi/5)$ and the second term as $4 cos^2(4k pi/5)$, and then tried writing one term in terms of the other using the identity $cos 2a = 2 cos^2 a- 1$. I even tried bringing in $sin$ but I didn't get anywhere. The answer is supposed to be $3$. Can someone solve it?
trigonometry complex-numbers
trigonometry complex-numbers
edited Nov 17 at 11:10
Brahadeesh
5,78341957
asked Nov 17 at 10:45
Vanessa
606
edited Nov 17 at 11:10
Brahadeesh
5,78341957
asked Nov 17 at 10:45
Vanessa
606
edited Nov 17 at 11:10
Brahadeesh
5,78341957
edited Nov 17 at 11:10
Brahadeesh
5,78341957
edited Nov 17 at 11:10
Brahadeesh
5,78341957
5,78341957
asked Nov 17 at 10:45
Vanessa
606
asked Nov 17 at 10:45
Vanessa
606
asked Nov 17 at 10:45
Vanessa
606
606
add a comment |
add a comment |
2 Answers
2
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0
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accepted
The sum is $4+z^2+z^{-2}+z^4+z^{-4}$.
Show that it equals $4+z+z^2+z^3+z^4$.
answered Nov 17 at 11:19
Empy2
33k12159
oh my god it was that simple... thank you!!!
– Vanessa
Nov 17 at 12:05
@Empy2, what is $4+z+z^2+z^3+z^4$.How to use here DeMoivre's theorem?
– Dhamnekar Winod
Nov 17 at 12:30
From part (b), it equals 3+0.
– Empy2
Nov 17 at 12:32
add a comment |
up vote
0
down vote
Hint:
If $5t=2kpi,5nmid k,cos tne1$
$cos3t=cdots==cos2t$
The roots of
$0=dfrac{4cos^3t-2cos^2t-3cos t+1}{cos t-1}=4cos^2t+2cos t-1=0$ will be $$t=2kpi,kequivpm1,pm2pmod5$$
Now $z+dfrac1z=2cosdfrac{2kpi}5, left(z+dfrac1zright)^2=cdots=2left(1+cosdfrac{4kpi}5right)$
$z^2+dfrac1{z^2}=2cosdfrac{4kpi}5,left(z^2+dfrac1{z^2}right)^2=?$
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
$(z^2+frac{1}{z^2})^2$=$2(1+cos{frac{8kpi}{5}})$. Now, how to proceed further?
– Dhamnekar Winod
Nov 17 at 13:06
@Dhanekar, maintain $$cosdfrac{4kpi}5$$
– lab bhattacharjee
Nov 17 at 13:38
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The sum is $4+z^2+z^{-2}+z^4+z^{-4}$.
Show that it equals $4+z+z^2+z^3+z^4$.
answered Nov 17 at 11:19
Empy2
33k12159
oh my god it was that simple... thank you!!!
– Vanessa
Nov 17 at 12:05
@Empy2, what is $4+z+z^2+z^3+z^4$.How to use here DeMoivre's theorem?
– Dhamnekar Winod
Nov 17 at 12:30
From part (b), it equals 3+0.
– Empy2
Nov 17 at 12:32
add a comment |
up vote
0
down vote
accepted
The sum is $4+z^2+z^{-2}+z^4+z^{-4}$.
Show that it equals $4+z+z^2+z^3+z^4$.
answered Nov 17 at 11:19
Empy2
33k12159
oh my god it was that simple... thank you!!!
– Vanessa
Nov 17 at 12:05
@Empy2, what is $4+z+z^2+z^3+z^4$.How to use here DeMoivre's theorem?
– Dhamnekar Winod
Nov 17 at 12:30
From part (b), it equals 3+0.
– Empy2
Nov 17 at 12:32
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The sum is $4+z^2+z^{-2}+z^4+z^{-4}$.
Show that it equals $4+z+z^2+z^3+z^4$.
answered Nov 17 at 11:19
Empy2
33k12159
The sum is $4+z^2+z^{-2}+z^4+z^{-4}$.
Show that it equals $4+z+z^2+z^3+z^4$.
answered Nov 17 at 11:19
Empy2
33k12159
answered Nov 17 at 11:19
Empy2
33k12159
answered Nov 17 at 11:19
Empy2
33k12159
answered Nov 17 at 11:19
Empy2
33k12159
33k12159
oh my god it was that simple... thank you!!!
– Vanessa
Nov 17 at 12:05
@Empy2, what is $4+z+z^2+z^3+z^4$.How to use here DeMoivre's theorem?
– Dhamnekar Winod
Nov 17 at 12:30
From part (b), it equals 3+0.
– Empy2
Nov 17 at 12:32
add a comment |
oh my god it was that simple... thank you!!!
– Vanessa
Nov 17 at 12:05
@Empy2, what is $4+z+z^2+z^3+z^4$.How to use here DeMoivre's theorem?
– Dhamnekar Winod
Nov 17 at 12:30
From part (b), it equals 3+0.
– Empy2
Nov 17 at 12:32
oh my god it was that simple... thank you!!!
– Vanessa
Nov 17 at 12:05
oh my god it was that simple... thank you!!!
– Vanessa
Nov 17 at 12:05
@Empy2, what is $4+z+z^2+z^3+z^4$.How to use here DeMoivre's theorem?
– Dhamnekar Winod
Nov 17 at 12:30
@Empy2, what is $4+z+z^2+z^3+z^4$.How to use here DeMoivre's theorem?
– Dhamnekar Winod
Nov 17 at 12:30
From part (b), it equals 3+0.
– Empy2
Nov 17 at 12:32
From part (b), it equals 3+0.
– Empy2
Nov 17 at 12:32
add a comment |
up vote
0
down vote
Hint:
If $5t=2kpi,5nmid k,cos tne1$
$cos3t=cdots==cos2t$
The roots of
$0=dfrac{4cos^3t-2cos^2t-3cos t+1}{cos t-1}=4cos^2t+2cos t-1=0$ will be $$t=2kpi,kequivpm1,pm2pmod5$$
Now $z+dfrac1z=2cosdfrac{2kpi}5, left(z+dfrac1zright)^2=cdots=2left(1+cosdfrac{4kpi}5right)$
$z^2+dfrac1{z^2}=2cosdfrac{4kpi}5,left(z^2+dfrac1{z^2}right)^2=?$
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
$(z^2+frac{1}{z^2})^2$=$2(1+cos{frac{8kpi}{5}})$. Now, how to proceed further?
– Dhamnekar Winod
Nov 17 at 13:06
@Dhanekar, maintain $$cosdfrac{4kpi}5$$
– lab bhattacharjee
Nov 17 at 13:38
add a comment |
up vote
0
down vote
Hint:
If $5t=2kpi,5nmid k,cos tne1$
$cos3t=cdots==cos2t$
The roots of
$0=dfrac{4cos^3t-2cos^2t-3cos t+1}{cos t-1}=4cos^2t+2cos t-1=0$ will be $$t=2kpi,kequivpm1,pm2pmod5$$
Now $z+dfrac1z=2cosdfrac{2kpi}5, left(z+dfrac1zright)^2=cdots=2left(1+cosdfrac{4kpi}5right)$
$z^2+dfrac1{z^2}=2cosdfrac{4kpi}5,left(z^2+dfrac1{z^2}right)^2=?$
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
$(z^2+frac{1}{z^2})^2$=$2(1+cos{frac{8kpi}{5}})$. Now, how to proceed further?
– Dhamnekar Winod
Nov 17 at 13:06
@Dhanekar, maintain $$cosdfrac{4kpi}5$$
– lab bhattacharjee
Nov 17 at 13:38
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint:
If $5t=2kpi,5nmid k,cos tne1$
$cos3t=cdots==cos2t$
The roots of
$0=dfrac{4cos^3t-2cos^2t-3cos t+1}{cos t-1}=4cos^2t+2cos t-1=0$ will be $$t=2kpi,kequivpm1,pm2pmod5$$
Now $z+dfrac1z=2cosdfrac{2kpi}5, left(z+dfrac1zright)^2=cdots=2left(1+cosdfrac{4kpi}5right)$
$z^2+dfrac1{z^2}=2cosdfrac{4kpi}5,left(z^2+dfrac1{z^2}right)^2=?$
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
Hint:
If $5t=2kpi,5nmid k,cos tne1$
$cos3t=cdots==cos2t$
The roots of
$0=dfrac{4cos^3t-2cos^2t-3cos t+1}{cos t-1}=4cos^2t+2cos t-1=0$ will be $$t=2kpi,kequivpm1,pm2pmod5$$
Now $z+dfrac1z=2cosdfrac{2kpi}5, left(z+dfrac1zright)^2=cdots=2left(1+cosdfrac{4kpi}5right)$
$z^2+dfrac1{z^2}=2cosdfrac{4kpi}5,left(z^2+dfrac1{z^2}right)^2=?$
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
answered Nov 17 at 10:57
lab bhattacharjee
220k15154271
220k15154271
$(z^2+frac{1}{z^2})^2$=$2(1+cos{frac{8kpi}{5}})$. Now, how to proceed further?
– Dhamnekar Winod
Nov 17 at 13:06
@Dhanekar, maintain $$cosdfrac{4kpi}5$$
– lab bhattacharjee
Nov 17 at 13:38
add a comment |
$(z^2+frac{1}{z^2})^2$=$2(1+cos{frac{8kpi}{5}})$. Now, how to proceed further?
– Dhamnekar Winod
Nov 17 at 13:06
@Dhanekar, maintain $$cosdfrac{4kpi}5$$
– lab bhattacharjee
Nov 17 at 13:38
$(z^2+frac{1}{z^2})^2$=$2(1+cos{frac{8kpi}{5}})$. Now, how to proceed further?
– Dhamnekar Winod
Nov 17 at 13:06
$(z^2+frac{1}{z^2})^2$=$2(1+cos{frac{8kpi}{5}})$. Now, how to proceed further?
– Dhamnekar Winod
Nov 17 at 13:06
@Dhanekar, maintain $$cosdfrac{4kpi}5$$
– lab bhattacharjee
Nov 17 at 13:38
@Dhanekar, maintain $$cosdfrac{4kpi}5$$
– lab bhattacharjee
Nov 17 at 13:38
add a comment |
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