Mixing
Solve $(x+1)^n=(x-1)^n$, assuming $x$ is a complex number and $n>0$.
$begingroup$
How do I solve $(x+1)^n=(x-1)^n$?
I assumed $x=a+bi$, getting the equation $((a+1)+bi)^n=((a-1)+bi)^n$.
How do I solve it using Moivre's n-th root theorem?
algebra-precalculus
edited Jul 2 '15 at 18:01
Sloan
2,1301718
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
$endgroup$
add a comment |
$begingroup$
How do I solve $(x+1)^n=(x-1)^n$?
I assumed $x=a+bi$, getting the equation $((a+1)+bi)^n=((a-1)+bi)^n$.
How do I solve it using Moivre's n-th root theorem?
algebra-precalculus
edited Jul 2 '15 at 18:01
Sloan
2,1301718
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
$endgroup$
add a comment |
$begingroup$
How do I solve $(x+1)^n=(x-1)^n$?
I assumed $x=a+bi$, getting the equation $((a+1)+bi)^n=((a-1)+bi)^n$.
How do I solve it using Moivre's n-th root theorem?
algebra-precalculus
edited Jul 2 '15 at 18:01
Sloan
2,1301718
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
$endgroup$
How do I solve $(x+1)^n=(x-1)^n$?
I assumed $x=a+bi$, getting the equation $((a+1)+bi)^n=((a-1)+bi)^n$.
How do I solve it using Moivre's n-th root theorem?
algebra-precalculus
algebra-precalculus
edited Jul 2 '15 at 18:01
Sloan
2,1301718
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
edited Jul 2 '15 at 18:01
Sloan
2,1301718
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
edited Jul 2 '15 at 18:01
Sloan
2,1301718
edited Jul 2 '15 at 18:01
Sloan
2,1301718
edited Jul 2 '15 at 18:01
Sloan
2,1301718
2,1301718
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
asked Jul 2 '15 at 17:56
Franciele DaltoéFranciele Daltoé
271210
271210
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your equation is equivalent to
$$ left(frac{x+1}{x-1}right)^n = 1 tag{1}$$
and since the inverse function of $f(x)=frac{x+1}{x-1}$ is given by $frac{y-1}{y+1}$, we have:
$$ x = frac{expfrac{2pi k i}{n}+1}{expfrac{2pi k i}{n}-1}=-icotfrac{pi k}{n},qquad kinleft[1,ldots,n-1right].tag{2}$$
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
$endgroup$
$begingroup$
There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=itan(pi/3)$. Then, $frac{x+1}{x-1}=frac{itan(pi/3)+1}{itan(pi/3)-1}=frac{cos(pi/3)+isin(pi/3)}{-cos(pi/3)+isin(pi/3)}=-e^{2pi/3}$ and $left(-e^{2pi/3}right)^3=-1ne 1$.
$endgroup$
– Mark Viola
Jul 2 '15 at 20:00
$begingroup$
@Dr.MV: you are right, now fixed, thanks.
$endgroup$
– Jack D'Aurizio
Jul 2 '15 at 20:02
$begingroup$
Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-))
$endgroup$
– Mark Viola
Jul 2 '15 at 20:03
$begingroup$
@FrancieleDaltoé: so you have to solve $y^n=1$, then $frac{z+1}{z-1}=y$, that is exactly what I did.
$endgroup$
– Jack D'Aurizio
Jul 3 '15 at 14:53
$begingroup$
Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n leftrightarrow frac{(z+1)^n}{(z-1)^n}=frac{(z-1)^n}{(z-1)^n}=1 leftrightarrow left (frac{z+1}{z-1}right)^n=1 leftrightarrow sqrt[n]{left (frac{z+1}{z-1}right)^n}=sqrt[n]{1}$ What do I do from there? I thought I'd write $frac{z+1}{z-1}=x$ and then solve $x^n=1=(cos(2kpi)+isin(2kpi))$, but I'm not sure how to do that.
$endgroup$
– Franciele Daltoé
Jul 3 '15 at 14:58
|
show 2 more comments
$begingroup$
Hint: $x = 1$ is not a solution, so we might assume $x neq 1$ and re-write this as:
$$left(frac{x + 1}{x - 1} right)^n = 1$$
It boils down to just listing the $n$th roots of $1$ and computing $x$.
$endgroup$
$begingroup$
Why can I equal this equation to 1?
$endgroup$
– Franciele Daltoé
Jul 2 '15 at 18:03
1
$begingroup$
@FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$.
$endgroup$
– user230734
Jul 2 '15 at 18:03
add a comment |
$begingroup$
For $z^n=w$, one can write $w=we^{i2 pi ell}$ for integer $ell$. Then, solving for $z$ reveals that
$$z=w^{1/n}e^{i2 pi ell/n}$$
for $ell =0, 1, 2,cdots n-1.$
Now for $z=x+1$ and $w=(x-1)^n$ we have that
$$x+1=(x-1)e^{i2 pi ell/n} implies x = frac{e^{i2 pi ell/n}+1}{e^{i2 pi ell/n}-1}=-icot{piell/n},,text{for},,ell =1,2,cdots n-1$$
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
$endgroup$
$begingroup$
@FrancieleDaltoe Please let me know how I can improve my answer, which was the first correct answer posted. I really want to give the best answer I can.
$endgroup$
– Mark Viola
Jul 11 '15 at 4:16
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your equation is equivalent to
$$ left(frac{x+1}{x-1}right)^n = 1 tag{1}$$
and since the inverse function of $f(x)=frac{x+1}{x-1}$ is given by $frac{y-1}{y+1}$, we have:
$$ x = frac{expfrac{2pi k i}{n}+1}{expfrac{2pi k i}{n}-1}=-icotfrac{pi k}{n},qquad kinleft[1,ldots,n-1right].tag{2}$$
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
$endgroup$
$begingroup$
There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=itan(pi/3)$. Then, $frac{x+1}{x-1}=frac{itan(pi/3)+1}{itan(pi/3)-1}=frac{cos(pi/3)+isin(pi/3)}{-cos(pi/3)+isin(pi/3)}=-e^{2pi/3}$ and $left(-e^{2pi/3}right)^3=-1ne 1$.
$endgroup$
– Mark Viola
Jul 2 '15 at 20:00
$begingroup$
@Dr.MV: you are right, now fixed, thanks.
$endgroup$
– Jack D'Aurizio
Jul 2 '15 at 20:02
$begingroup$
Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-))
$endgroup$
– Mark Viola
Jul 2 '15 at 20:03
$begingroup$
@FrancieleDaltoé: so you have to solve $y^n=1$, then $frac{z+1}{z-1}=y$, that is exactly what I did.
$endgroup$
– Jack D'Aurizio
Jul 3 '15 at 14:53
$begingroup$
Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n leftrightarrow frac{(z+1)^n}{(z-1)^n}=frac{(z-1)^n}{(z-1)^n}=1 leftrightarrow left (frac{z+1}{z-1}right)^n=1 leftrightarrow sqrt[n]{left (frac{z+1}{z-1}right)^n}=sqrt[n]{1}$ What do I do from there? I thought I'd write $frac{z+1}{z-1}=x$ and then solve $x^n=1=(cos(2kpi)+isin(2kpi))$, but I'm not sure how to do that.
$endgroup$
– Franciele Daltoé
Jul 3 '15 at 14:58
|
show 2 more comments
$begingroup$
Your equation is equivalent to
$$ left(frac{x+1}{x-1}right)^n = 1 tag{1}$$
and since the inverse function of $f(x)=frac{x+1}{x-1}$ is given by $frac{y-1}{y+1}$, we have:
$$ x = frac{expfrac{2pi k i}{n}+1}{expfrac{2pi k i}{n}-1}=-icotfrac{pi k}{n},qquad kinleft[1,ldots,n-1right].tag{2}$$
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
$endgroup$
$begingroup$
There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=itan(pi/3)$. Then, $frac{x+1}{x-1}=frac{itan(pi/3)+1}{itan(pi/3)-1}=frac{cos(pi/3)+isin(pi/3)}{-cos(pi/3)+isin(pi/3)}=-e^{2pi/3}$ and $left(-e^{2pi/3}right)^3=-1ne 1$.
$endgroup$
– Mark Viola
Jul 2 '15 at 20:00
$begingroup$
@Dr.MV: you are right, now fixed, thanks.
$endgroup$
– Jack D'Aurizio
Jul 2 '15 at 20:02
$begingroup$
Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-))
$endgroup$
– Mark Viola
Jul 2 '15 at 20:03
$begingroup$
@FrancieleDaltoé: so you have to solve $y^n=1$, then $frac{z+1}{z-1}=y$, that is exactly what I did.
$endgroup$
– Jack D'Aurizio
Jul 3 '15 at 14:53
$begingroup$
Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n leftrightarrow frac{(z+1)^n}{(z-1)^n}=frac{(z-1)^n}{(z-1)^n}=1 leftrightarrow left (frac{z+1}{z-1}right)^n=1 leftrightarrow sqrt[n]{left (frac{z+1}{z-1}right)^n}=sqrt[n]{1}$ What do I do from there? I thought I'd write $frac{z+1}{z-1}=x$ and then solve $x^n=1=(cos(2kpi)+isin(2kpi))$, but I'm not sure how to do that.
$endgroup$
– Franciele Daltoé
Jul 3 '15 at 14:58
|
show 2 more comments
$begingroup$
Your equation is equivalent to
$$ left(frac{x+1}{x-1}right)^n = 1 tag{1}$$
and since the inverse function of $f(x)=frac{x+1}{x-1}$ is given by $frac{y-1}{y+1}$, we have:
$$ x = frac{expfrac{2pi k i}{n}+1}{expfrac{2pi k i}{n}-1}=-icotfrac{pi k}{n},qquad kinleft[1,ldots,n-1right].tag{2}$$
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
$endgroup$
Your equation is equivalent to
$$ left(frac{x+1}{x-1}right)^n = 1 tag{1}$$
and since the inverse function of $f(x)=frac{x+1}{x-1}$ is given by $frac{y-1}{y+1}$, we have:
$$ x = frac{expfrac{2pi k i}{n}+1}{expfrac{2pi k i}{n}-1}=-icotfrac{pi k}{n},qquad kinleft[1,ldots,n-1right].tag{2}$$
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
edited Jul 2 '15 at 20:02
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
answered Jul 2 '15 at 18:02
Jack D'AurizioJack D'Aurizio
290k33282662
290k33282662
$begingroup$
There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=itan(pi/3)$. Then, $frac{x+1}{x-1}=frac{itan(pi/3)+1}{itan(pi/3)-1}=frac{cos(pi/3)+isin(pi/3)}{-cos(pi/3)+isin(pi/3)}=-e^{2pi/3}$ and $left(-e^{2pi/3}right)^3=-1ne 1$.
$endgroup$
– Mark Viola
Jul 2 '15 at 20:00
$begingroup$
@Dr.MV: you are right, now fixed, thanks.
$endgroup$
– Jack D'Aurizio
Jul 2 '15 at 20:02
$begingroup$
Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-))
$endgroup$
– Mark Viola
Jul 2 '15 at 20:03
$begingroup$
@FrancieleDaltoé: so you have to solve $y^n=1$, then $frac{z+1}{z-1}=y$, that is exactly what I did.
$endgroup$
– Jack D'Aurizio
Jul 3 '15 at 14:53
$begingroup$
Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n leftrightarrow frac{(z+1)^n}{(z-1)^n}=frac{(z-1)^n}{(z-1)^n}=1 leftrightarrow left (frac{z+1}{z-1}right)^n=1 leftrightarrow sqrt[n]{left (frac{z+1}{z-1}right)^n}=sqrt[n]{1}$ What do I do from there? I thought I'd write $frac{z+1}{z-1}=x$ and then solve $x^n=1=(cos(2kpi)+isin(2kpi))$, but I'm not sure how to do that.
$endgroup$
– Franciele Daltoé
Jul 3 '15 at 14:58
|
show 2 more comments
$begingroup$
There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=itan(pi/3)$. Then, $frac{x+1}{x-1}=frac{itan(pi/3)+1}{itan(pi/3)-1}=frac{cos(pi/3)+isin(pi/3)}{-cos(pi/3)+isin(pi/3)}=-e^{2pi/3}$ and $left(-e^{2pi/3}right)^3=-1ne 1$.
$endgroup$
– Mark Viola
Jul 2 '15 at 20:00
$begingroup$
@Dr.MV: you are right, now fixed, thanks.
$endgroup$
– Jack D'Aurizio
Jul 2 '15 at 20:02
$begingroup$
Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-))
$endgroup$
– Mark Viola
Jul 2 '15 at 20:03
$begingroup$
@FrancieleDaltoé: so you have to solve $y^n=1$, then $frac{z+1}{z-1}=y$, that is exactly what I did.
$endgroup$
– Jack D'Aurizio
Jul 3 '15 at 14:53
$begingroup$
Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n leftrightarrow frac{(z+1)^n}{(z-1)^n}=frac{(z-1)^n}{(z-1)^n}=1 leftrightarrow left (frac{z+1}{z-1}right)^n=1 leftrightarrow sqrt[n]{left (frac{z+1}{z-1}right)^n}=sqrt[n]{1}$ What do I do from there? I thought I'd write $frac{z+1}{z-1}=x$ and then solve $x^n=1=(cos(2kpi)+isin(2kpi))$, but I'm not sure how to do that.
$endgroup$
– Franciele Daltoé
Jul 3 '15 at 14:58
$begingroup$
There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=itan(pi/3)$. Then, $frac{x+1}{x-1}=frac{itan(pi/3)+1}{itan(pi/3)-1}=frac{cos(pi/3)+isin(pi/3)}{-cos(pi/3)+isin(pi/3)}=-e^{2pi/3}$ and $left(-e^{2pi/3}right)^3=-1ne 1$.
$endgroup$
– Mark Viola
Jul 2 '15 at 20:00
$begingroup$
There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=itan(pi/3)$. Then, $frac{x+1}{x-1}=frac{itan(pi/3)+1}{itan(pi/3)-1}=frac{cos(pi/3)+isin(pi/3)}{-cos(pi/3)+isin(pi/3)}=-e^{2pi/3}$ and $left(-e^{2pi/3}right)^3=-1ne 1$.
$endgroup$
– Mark Viola
Jul 2 '15 at 20:00
$begingroup$
@Dr.MV: you are right, now fixed, thanks.
$endgroup$
– Jack D'Aurizio
Jul 2 '15 at 20:02
$begingroup$
@Dr.MV: you are right, now fixed, thanks.
$endgroup$
– Jack D'Aurizio
Jul 2 '15 at 20:02
$begingroup$
Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-))
$endgroup$
– Mark Viola
Jul 2 '15 at 20:03
$begingroup$
Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-))
$endgroup$
– Mark Viola
Jul 2 '15 at 20:03
$begingroup$
@FrancieleDaltoé: so you have to solve $y^n=1$, then $frac{z+1}{z-1}=y$, that is exactly what I did.
$endgroup$
– Jack D'Aurizio
Jul 3 '15 at 14:53
$begingroup$
@FrancieleDaltoé: so you have to solve $y^n=1$, then $frac{z+1}{z-1}=y$, that is exactly what I did.
$endgroup$
– Jack D'Aurizio
Jul 3 '15 at 14:53
$begingroup$
Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n leftrightarrow frac{(z+1)^n}{(z-1)^n}=frac{(z-1)^n}{(z-1)^n}=1 leftrightarrow left (frac{z+1}{z-1}right)^n=1 leftrightarrow sqrt[n]{left (frac{z+1}{z-1}right)^n}=sqrt[n]{1}$ What do I do from there? I thought I'd write $frac{z+1}{z-1}=x$ and then solve $x^n=1=(cos(2kpi)+isin(2kpi))$, but I'm not sure how to do that.
$endgroup$
– Franciele Daltoé
Jul 3 '15 at 14:58
$begingroup$
Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n leftrightarrow frac{(z+1)^n}{(z-1)^n}=frac{(z-1)^n}{(z-1)^n}=1 leftrightarrow left (frac{z+1}{z-1}right)^n=1 leftrightarrow sqrt[n]{left (frac{z+1}{z-1}right)^n}=sqrt[n]{1}$ What do I do from there? I thought I'd write $frac{z+1}{z-1}=x$ and then solve $x^n=1=(cos(2kpi)+isin(2kpi))$, but I'm not sure how to do that.
$endgroup$
– Franciele Daltoé
Jul 3 '15 at 14:58
|
show 2 more comments
$begingroup$
Hint: $x = 1$ is not a solution, so we might assume $x neq 1$ and re-write this as:
$$left(frac{x + 1}{x - 1} right)^n = 1$$
It boils down to just listing the $n$th roots of $1$ and computing $x$.
$endgroup$
$begingroup$
Why can I equal this equation to 1?
$endgroup$
– Franciele Daltoé
Jul 2 '15 at 18:03
1
$begingroup$
@FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$.
$endgroup$
– user230734
Jul 2 '15 at 18:03
add a comment |
$begingroup$
Hint: $x = 1$ is not a solution, so we might assume $x neq 1$ and re-write this as:
$$left(frac{x + 1}{x - 1} right)^n = 1$$
It boils down to just listing the $n$th roots of $1$ and computing $x$.
$endgroup$
$begingroup$
Why can I equal this equation to 1?
$endgroup$
– Franciele Daltoé
Jul 2 '15 at 18:03
1
$begingroup$
@FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$.
$endgroup$
– user230734
Jul 2 '15 at 18:03
add a comment |
$begingroup$
Hint: $x = 1$ is not a solution, so we might assume $x neq 1$ and re-write this as:
$$left(frac{x + 1}{x - 1} right)^n = 1$$
It boils down to just listing the $n$th roots of $1$ and computing $x$.
$endgroup$
Hint: $x = 1$ is not a solution, so we might assume $x neq 1$ and re-write this as:
$$left(frac{x + 1}{x - 1} right)^n = 1$$
It boils down to just listing the $n$th roots of $1$ and computing $x$.
answered Jul 2 '15 at 17:59
user230734
$begingroup$
Why can I equal this equation to 1?
$endgroup$
– Franciele Daltoé
Jul 2 '15 at 18:03
1
$begingroup$
@FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$.
$endgroup$
– user230734
Jul 2 '15 at 18:03
add a comment |
$begingroup$
Why can I equal this equation to 1?
$endgroup$
– Franciele Daltoé
Jul 2 '15 at 18:03
1
$begingroup$
@FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$.
$endgroup$
– user230734
Jul 2 '15 at 18:03
$begingroup$
Why can I equal this equation to 1?
$endgroup$
– Franciele Daltoé
Jul 2 '15 at 18:03
$begingroup$
Why can I equal this equation to 1?
$endgroup$
– Franciele Daltoé
Jul 2 '15 at 18:03
1
1
$begingroup$
@FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$.
$endgroup$
– user230734
Jul 2 '15 at 18:03
$begingroup$
@FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$.
$endgroup$
– user230734
Jul 2 '15 at 18:03
add a comment |
$begingroup$
For $z^n=w$, one can write $w=we^{i2 pi ell}$ for integer $ell$. Then, solving for $z$ reveals that
$$z=w^{1/n}e^{i2 pi ell/n}$$
for $ell =0, 1, 2,cdots n-1.$
Now for $z=x+1$ and $w=(x-1)^n$ we have that
$$x+1=(x-1)e^{i2 pi ell/n} implies x = frac{e^{i2 pi ell/n}+1}{e^{i2 pi ell/n}-1}=-icot{piell/n},,text{for},,ell =1,2,cdots n-1$$
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
$endgroup$
$begingroup$
@FrancieleDaltoe Please let me know how I can improve my answer, which was the first correct answer posted. I really want to give the best answer I can.
$endgroup$
– Mark Viola
Jul 11 '15 at 4:16
add a comment |
$begingroup$
For $z^n=w$, one can write $w=we^{i2 pi ell}$ for integer $ell$. Then, solving for $z$ reveals that
$$z=w^{1/n}e^{i2 pi ell/n}$$
for $ell =0, 1, 2,cdots n-1.$
Now for $z=x+1$ and $w=(x-1)^n$ we have that
$$x+1=(x-1)e^{i2 pi ell/n} implies x = frac{e^{i2 pi ell/n}+1}{e^{i2 pi ell/n}-1}=-icot{piell/n},,text{for},,ell =1,2,cdots n-1$$
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
$endgroup$
$begingroup$
@FrancieleDaltoe Please let me know how I can improve my answer, which was the first correct answer posted. I really want to give the best answer I can.
$endgroup$
– Mark Viola
Jul 11 '15 at 4:16
add a comment |
$begingroup$
For $z^n=w$, one can write $w=we^{i2 pi ell}$ for integer $ell$. Then, solving for $z$ reveals that
$$z=w^{1/n}e^{i2 pi ell/n}$$
for $ell =0, 1, 2,cdots n-1.$
Now for $z=x+1$ and $w=(x-1)^n$ we have that
$$x+1=(x-1)e^{i2 pi ell/n} implies x = frac{e^{i2 pi ell/n}+1}{e^{i2 pi ell/n}-1}=-icot{piell/n},,text{for},,ell =1,2,cdots n-1$$
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
$endgroup$
For $z^n=w$, one can write $w=we^{i2 pi ell}$ for integer $ell$. Then, solving for $z$ reveals that
$$z=w^{1/n}e^{i2 pi ell/n}$$
for $ell =0, 1, 2,cdots n-1.$
Now for $z=x+1$ and $w=(x-1)^n$ we have that
$$x+1=(x-1)e^{i2 pi ell/n} implies x = frac{e^{i2 pi ell/n}+1}{e^{i2 pi ell/n}-1}=-icot{piell/n},,text{for},,ell =1,2,cdots n-1$$
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
answered Jul 2 '15 at 18:48
Mark ViolaMark Viola
132k1276174
132k1276174
$begingroup$
@FrancieleDaltoe Please let me know how I can improve my answer, which was the first correct answer posted. I really want to give the best answer I can.
$endgroup$
– Mark Viola
Jul 11 '15 at 4:16
add a comment |
$begingroup$
@FrancieleDaltoe Please let me know how I can improve my answer, which was the first correct answer posted. I really want to give the best answer I can.
$endgroup$
– Mark Viola
Jul 11 '15 at 4:16
$begingroup$
@FrancieleDaltoe Please let me know how I can improve my answer, which was the first correct answer posted. I really want to give the best answer I can.
$endgroup$
– Mark Viola
Jul 11 '15 at 4:16
$begingroup$
@FrancieleDaltoe Please let me know how I can improve my answer, which was the first correct answer posted. I really want to give the best answer I can.
$endgroup$
– Mark Viola
Jul 11 '15 at 4:16
add a comment |
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