Show that if $z^n$ + $z^{n-1}$ + … + $z$ + 1 = 0 then $z$ $neq$ 1 and $z^{n+1}$ =1.
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Show that if $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 0 then $z$ $neq$ 1 and $z^{n+1}$ =1.
I thought I had managed to show that $z$ $neq$ 1 since if $z$ = 1 then
$z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 1 + 1 ... 1 + 1 = n $neq$ 0.
But then when I tried to prove the second part of the question I got a contradiction as follows:
Factorising $z^{n+1}$ - 1 = 0 gives ($z$ - 1)($z^n$ + $z^{n-1}$ + ... + $z$ + 1) = 0.
But since $z$ $neq$ 1 how does this work?
complex-analysis proof-verification complex-numbers proof-explanation
asked Nov 17 at 23:54
sam
498
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up vote
0
down vote
favorite
Show that if $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 0 then $z$ $neq$ 1 and $z^{n+1}$ =1.
I thought I had managed to show that $z$ $neq$ 1 since if $z$ = 1 then
$z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 1 + 1 ... 1 + 1 = n $neq$ 0.
But then when I tried to prove the second part of the question I got a contradiction as follows:
Factorising $z^{n+1}$ - 1 = 0 gives ($z$ - 1)($z^n$ + $z^{n-1}$ + ... + $z$ + 1) = 0.
But since $z$ $neq$ 1 how does this work?
complex-analysis proof-verification complex-numbers proof-explanation
asked Nov 17 at 23:54
sam
498
2
I don't get your confusion; you have already shown that $zneq1$. It is $z^{n+1}-1=(z-1)(z^n+dots+1)=0$ (the second factor is $0$). Therefore $z^{n+1}=1$
– JustDroppedIn
Nov 18 at 0:00
If $z=1$ then $z^n + z^{n-1} +dots+ z + 1 = n+1.$
– Somos
Nov 18 at 0:07
@JustDroppedIn ohhhhhh ohmnygod i dont know what i was thinking i think i forgot about the original part and dividing by $z$ - 1 lmfao but thankyou
– sam
Nov 18 at 0:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that if $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 0 then $z$ $neq$ 1 and $z^{n+1}$ =1.
I thought I had managed to show that $z$ $neq$ 1 since if $z$ = 1 then
$z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 1 + 1 ... 1 + 1 = n $neq$ 0.
But then when I tried to prove the second part of the question I got a contradiction as follows:
Factorising $z^{n+1}$ - 1 = 0 gives ($z$ - 1)($z^n$ + $z^{n-1}$ + ... + $z$ + 1) = 0.
But since $z$ $neq$ 1 how does this work?
complex-analysis proof-verification complex-numbers proof-explanation
asked Nov 17 at 23:54
sam
498
Show that if $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 0 then $z$ $neq$ 1 and $z^{n+1}$ =1.
I thought I had managed to show that $z$ $neq$ 1 since if $z$ = 1 then
$z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 1 + 1 ... 1 + 1 = n $neq$ 0.
But then when I tried to prove the second part of the question I got a contradiction as follows:
Factorising $z^{n+1}$ - 1 = 0 gives ($z$ - 1)($z^n$ + $z^{n-1}$ + ... + $z$ + 1) = 0.
But since $z$ $neq$ 1 how does this work?
complex-analysis proof-verification complex-numbers proof-explanation
complex-analysis proof-verification complex-numbers proof-explanation
asked Nov 17 at 23:54
sam
498
asked Nov 17 at 23:54
sam
498
asked Nov 17 at 23:54
sam
498
asked Nov 17 at 23:54
sam
498
asked Nov 17 at 23:54
sam
498
498
2
I don't get your confusion; you have already shown that $zneq1$. It is $z^{n+1}-1=(z-1)(z^n+dots+1)=0$ (the second factor is $0$). Therefore $z^{n+1}=1$
– JustDroppedIn
Nov 18 at 0:00
If $z=1$ then $z^n + z^{n-1} +dots+ z + 1 = n+1.$
– Somos
Nov 18 at 0:07
@JustDroppedIn ohhhhhh ohmnygod i dont know what i was thinking i think i forgot about the original part and dividing by $z$ - 1 lmfao but thankyou
– sam
Nov 18 at 0:13
add a comment |
2
I don't get your confusion; you have already shown that $zneq1$. It is $z^{n+1}-1=(z-1)(z^n+dots+1)=0$ (the second factor is $0$). Therefore $z^{n+1}=1$
– JustDroppedIn
Nov 18 at 0:00
If $z=1$ then $z^n + z^{n-1} +dots+ z + 1 = n+1.$
– Somos
Nov 18 at 0:07
@JustDroppedIn ohhhhhh ohmnygod i dont know what i was thinking i think i forgot about the original part and dividing by $z$ - 1 lmfao but thankyou
– sam
Nov 18 at 0:13
2
2
I don't get your confusion; you have already shown that $zneq1$. It is $z^{n+1}-1=(z-1)(z^n+dots+1)=0$ (the second factor is $0$). Therefore $z^{n+1}=1$
– JustDroppedIn
Nov 18 at 0:00
I don't get your confusion; you have already shown that $zneq1$. It is $z^{n+1}-1=(z-1)(z^n+dots+1)=0$ (the second factor is $0$). Therefore $z^{n+1}=1$
– JustDroppedIn
Nov 18 at 0:00
If $z=1$ then $z^n + z^{n-1} +dots+ z + 1 = n+1.$
– Somos
Nov 18 at 0:07
If $z=1$ then $z^n + z^{n-1} +dots+ z + 1 = n+1.$
– Somos
Nov 18 at 0:07
@JustDroppedIn ohhhhhh ohmnygod i dont know what i was thinking i think i forgot about the original part and dividing by $z$ - 1 lmfao but thankyou
– sam
Nov 18 at 0:13
@JustDroppedIn ohhhhhh ohmnygod i dont know what i was thinking i think i forgot about the original part and dividing by $z$ - 1 lmfao but thankyou
– sam
Nov 18 at 0:13
add a comment |
2 Answers
2
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0
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Write $(z - 1)(z^n + z^{n-1} + ... + z + 1) = 0$, using the null product law and since $z ne 1$, $(z^n + z^{n-1} + ... + z + 1) = 0$.
Using the formula for $(z^n + z^{n-1} + ... + z + 1)$, we get $$frac{1-z^{n+1}}{1-z} = 0$$
Hence $z^{n+1} = 1$.
answered Nov 18 at 0:08
Euler Pythagoras
4369
add a comment |
up vote
-1
down vote
It comes down to the high-school factorisation formula, used to compute the sum of the first $n$ terms of a geometric progression:
$$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$
answered Nov 18 at 0:06
Bernard
116k637108
The maniac downvoter struck again!
– Bernard
Nov 18 at 11:15
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
Write $(z - 1)(z^n + z^{n-1} + ... + z + 1) = 0$, using the null product law and since $z ne 1$, $(z^n + z^{n-1} + ... + z + 1) = 0$.
Using the formula for $(z^n + z^{n-1} + ... + z + 1)$, we get $$frac{1-z^{n+1}}{1-z} = 0$$
Hence $z^{n+1} = 1$.
answered Nov 18 at 0:08
Euler Pythagoras
4369
add a comment |
up vote
0
down vote
Write $(z - 1)(z^n + z^{n-1} + ... + z + 1) = 0$, using the null product law and since $z ne 1$, $(z^n + z^{n-1} + ... + z + 1) = 0$.
Using the formula for $(z^n + z^{n-1} + ... + z + 1)$, we get $$frac{1-z^{n+1}}{1-z} = 0$$
Hence $z^{n+1} = 1$.
answered Nov 18 at 0:08
Euler Pythagoras
4369
add a comment |
up vote
0
down vote
up vote
0
down vote
Write $(z - 1)(z^n + z^{n-1} + ... + z + 1) = 0$, using the null product law and since $z ne 1$, $(z^n + z^{n-1} + ... + z + 1) = 0$.
Using the formula for $(z^n + z^{n-1} + ... + z + 1)$, we get $$frac{1-z^{n+1}}{1-z} = 0$$
Hence $z^{n+1} = 1$.
answered Nov 18 at 0:08
Euler Pythagoras
4369
Write $(z - 1)(z^n + z^{n-1} + ... + z + 1) = 0$, using the null product law and since $z ne 1$, $(z^n + z^{n-1} + ... + z + 1) = 0$.
Using the formula for $(z^n + z^{n-1} + ... + z + 1)$, we get $$frac{1-z^{n+1}}{1-z} = 0$$
Hence $z^{n+1} = 1$.
answered Nov 18 at 0:08
Euler Pythagoras
4369
answered Nov 18 at 0:08
Euler Pythagoras
4369
answered Nov 18 at 0:08
Euler Pythagoras
4369
answered Nov 18 at 0:08
Euler Pythagoras
4369
4369
add a comment |
add a comment |
up vote
-1
down vote
It comes down to the high-school factorisation formula, used to compute the sum of the first $n$ terms of a geometric progression:
$$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$
answered Nov 18 at 0:06
Bernard
116k637108
The maniac downvoter struck again!
– Bernard
Nov 18 at 11:15
add a comment |
up vote
-1
down vote
It comes down to the high-school factorisation formula, used to compute the sum of the first $n$ terms of a geometric progression:
$$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$
answered Nov 18 at 0:06
Bernard
116k637108
The maniac downvoter struck again!
– Bernard
Nov 18 at 11:15
add a comment |
up vote
-1
down vote
up vote
-1
down vote
It comes down to the high-school factorisation formula, used to compute the sum of the first $n$ terms of a geometric progression:
$$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$
answered Nov 18 at 0:06
Bernard
116k637108
It comes down to the high-school factorisation formula, used to compute the sum of the first $n$ terms of a geometric progression:
$$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$
answered Nov 18 at 0:06
Bernard
116k637108
answered Nov 18 at 0:06
Bernard
116k637108
answered Nov 18 at 0:06
Bernard
116k637108
answered Nov 18 at 0:06
Bernard
116k637108
116k637108
The maniac downvoter struck again!
– Bernard
Nov 18 at 11:15
add a comment |
The maniac downvoter struck again!
– Bernard
Nov 18 at 11:15
The maniac downvoter struck again!
– Bernard
Nov 18 at 11:15
The maniac downvoter struck again!
– Bernard
Nov 18 at 11:15
add a comment |
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I don't get your confusion; you have already shown that $zneq1$. It is $z^{n+1}-1=(z-1)(z^n+dots+1)=0$ (the second factor is $0$). Therefore $z^{n+1}=1$
– JustDroppedIn
Nov 18 at 0:00
If $z=1$ then $z^n + z^{n-1} +dots+ z + 1 = n+1.$
– Somos
Nov 18 at 0:07
@JustDroppedIn ohhhhhh ohmnygod i dont know what i was thinking i think i forgot about the original part and dividing by $z$ - 1 lmfao but thankyou
– sam
Nov 18 at 0:13