system of equations-complex number
$begingroup$
I have a task:
Prove that if for each complex number u,w,z we have:
$uwz=1$ and $u+w+z=u^{-1}+w^{-1}+z^{-1}$
then at least one of them is equal 1.
I tried substituting $u=(uw)^{-1}$ to the second equality and prove that by denying, but I can't solve it.
Thanks in advance.
algebra-precalculus polynomials complex-numbers
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
asked Oct 7 '18 at 18:39
John1357John1357
285
$endgroup$
add a comment |
$begingroup$
I have a task:
Prove that if for each complex number u,w,z we have:
$uwz=1$ and $u+w+z=u^{-1}+w^{-1}+z^{-1}$
then at least one of them is equal 1.
I tried substituting $u=(uw)^{-1}$ to the second equality and prove that by denying, but I can't solve it.
Thanks in advance.
algebra-precalculus polynomials complex-numbers
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
asked Oct 7 '18 at 18:39
John1357John1357
285
$endgroup$
add a comment |
$begingroup$
I have a task:
Prove that if for each complex number u,w,z we have:
$uwz=1$ and $u+w+z=u^{-1}+w^{-1}+z^{-1}$
then at least one of them is equal 1.
I tried substituting $u=(uw)^{-1}$ to the second equality and prove that by denying, but I can't solve it.
Thanks in advance.
algebra-precalculus polynomials complex-numbers
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
asked Oct 7 '18 at 18:39
John1357John1357
285
$endgroup$
I have a task:
Prove that if for each complex number u,w,z we have:
$uwz=1$ and $u+w+z=u^{-1}+w^{-1}+z^{-1}$
then at least one of them is equal 1.
I tried substituting $u=(uw)^{-1}$ to the second equality and prove that by denying, but I can't solve it.
Thanks in advance.
algebra-precalculus polynomials complex-numbers
algebra-precalculus polynomials complex-numbers
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
asked Oct 7 '18 at 18:39
John1357John1357
285
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
asked Oct 7 '18 at 18:39
John1357John1357
285
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
edited Dec 19 '18 at 20:59
Eric Wofsey
187k14215344
187k14215344
asked Oct 7 '18 at 18:39
John1357John1357
285
asked Oct 7 '18 at 18:39
John1357John1357
285
asked Oct 7 '18 at 18:39
John1357John1357
285
285
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$ w = {1over zu} implies z+u+{1over zu} = {1over z} + {1over u} +zu$$
so $$ z^2u+u^2z+1= u+z+z^2u^2$$
so $$z^2u(u-1)-z(u^2-1)+u-1=0$$
so $$(u-1)(z^2u-z(u+1)+1)=0$$
so if $u =1$ we are done else:
$$z^2u-zu-z+1=0implies zu(z-1)-(z-1)=0$$
so if $z=1$ we are done else $zu = 1$ so $w=1$.
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
$endgroup$
add a comment |
$begingroup$
If $uwz=1$ then $u^{-1}+w^{-1}+z^{-1}=uw+uz+wz$, and
$$(X-u)(X-w)(X-z)=X^3-(u+w+z)X^2+(uw+uz+wz)X-uwz=cdots$$
etc.
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
add a comment |
$begingroup$
Let $S=u+w+z$As observed in another answer, the equalities $uvw=1$ and $S=u^{+1}+w^{-1}+w^{-1}$ imply $uw+wz+zu=S$.
Now let's use Vieta's relations: $u,w$ and $z$ are the roots of the cubic polynomial
$$Z^3-SZ^2+SZ-1=Z^3-1-SZ(Z-1)=(Z-1)(Z^2+Z+1-SZ),$$
which has $Z=1$ as a root.
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
$endgroup$
add a comment |
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3 Answers
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3 Answers
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active
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$begingroup$
$$ w = {1over zu} implies z+u+{1over zu} = {1over z} + {1over u} +zu$$
so $$ z^2u+u^2z+1= u+z+z^2u^2$$
so $$z^2u(u-1)-z(u^2-1)+u-1=0$$
so $$(u-1)(z^2u-z(u+1)+1)=0$$
so if $u =1$ we are done else:
$$z^2u-zu-z+1=0implies zu(z-1)-(z-1)=0$$
so if $z=1$ we are done else $zu = 1$ so $w=1$.
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
$endgroup$
add a comment |
$begingroup$
$$ w = {1over zu} implies z+u+{1over zu} = {1over z} + {1over u} +zu$$
so $$ z^2u+u^2z+1= u+z+z^2u^2$$
so $$z^2u(u-1)-z(u^2-1)+u-1=0$$
so $$(u-1)(z^2u-z(u+1)+1)=0$$
so if $u =1$ we are done else:
$$z^2u-zu-z+1=0implies zu(z-1)-(z-1)=0$$
so if $z=1$ we are done else $zu = 1$ so $w=1$.
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
$endgroup$
add a comment |
$begingroup$
$$ w = {1over zu} implies z+u+{1over zu} = {1over z} + {1over u} +zu$$
so $$ z^2u+u^2z+1= u+z+z^2u^2$$
so $$z^2u(u-1)-z(u^2-1)+u-1=0$$
so $$(u-1)(z^2u-z(u+1)+1)=0$$
so if $u =1$ we are done else:
$$z^2u-zu-z+1=0implies zu(z-1)-(z-1)=0$$
so if $z=1$ we are done else $zu = 1$ so $w=1$.
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
$endgroup$
$$ w = {1over zu} implies z+u+{1over zu} = {1over z} + {1over u} +zu$$
so $$ z^2u+u^2z+1= u+z+z^2u^2$$
so $$z^2u(u-1)-z(u^2-1)+u-1=0$$
so $$(u-1)(z^2u-z(u+1)+1)=0$$
so if $u =1$ we are done else:
$$z^2u-zu-z+1=0implies zu(z-1)-(z-1)=0$$
so if $z=1$ we are done else $zu = 1$ so $w=1$.
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
answered Oct 7 '18 at 18:55
greedoidgreedoid
44.4k1156110
44.4k1156110
add a comment |
add a comment |
$begingroup$
If $uwz=1$ then $u^{-1}+w^{-1}+z^{-1}=uw+uz+wz$, and
$$(X-u)(X-w)(X-z)=X^3-(u+w+z)X^2+(uw+uz+wz)X-uwz=cdots$$
etc.
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
add a comment |
$begingroup$
If $uwz=1$ then $u^{-1}+w^{-1}+z^{-1}=uw+uz+wz$, and
$$(X-u)(X-w)(X-z)=X^3-(u+w+z)X^2+(uw+uz+wz)X-uwz=cdots$$
etc.
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
add a comment |
$begingroup$
If $uwz=1$ then $u^{-1}+w^{-1}+z^{-1}=uw+uz+wz$, and
$$(X-u)(X-w)(X-z)=X^3-(u+w+z)X^2+(uw+uz+wz)X-uwz=cdots$$
etc.
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
$endgroup$
If $uwz=1$ then $u^{-1}+w^{-1}+z^{-1}=uw+uz+wz$, and
$$(X-u)(X-w)(X-z)=X^3-(u+w+z)X^2+(uw+uz+wz)X-uwz=cdots$$
etc.
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
answered Oct 7 '18 at 18:48
Lord Shark the UnknownLord Shark the Unknown
105k1160132
105k1160132
add a comment |
add a comment |
$begingroup$
Let $S=u+w+z$As observed in another answer, the equalities $uvw=1$ and $S=u^{+1}+w^{-1}+w^{-1}$ imply $uw+wz+zu=S$.
Now let's use Vieta's relations: $u,w$ and $z$ are the roots of the cubic polynomial
$$Z^3-SZ^2+SZ-1=Z^3-1-SZ(Z-1)=(Z-1)(Z^2+Z+1-SZ),$$
which has $Z=1$ as a root.
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Let $S=u+w+z$As observed in another answer, the equalities $uvw=1$ and $S=u^{+1}+w^{-1}+w^{-1}$ imply $uw+wz+zu=S$.
Now let's use Vieta's relations: $u,w$ and $z$ are the roots of the cubic polynomial
$$Z^3-SZ^2+SZ-1=Z^3-1-SZ(Z-1)=(Z-1)(Z^2+Z+1-SZ),$$
which has $Z=1$ as a root.
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Let $S=u+w+z$As observed in another answer, the equalities $uvw=1$ and $S=u^{+1}+w^{-1}+w^{-1}$ imply $uw+wz+zu=S$.
Now let's use Vieta's relations: $u,w$ and $z$ are the roots of the cubic polynomial
$$Z^3-SZ^2+SZ-1=Z^3-1-SZ(Z-1)=(Z-1)(Z^2+Z+1-SZ),$$
which has $Z=1$ as a root.
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
$endgroup$
Let $S=u+w+z$As observed in another answer, the equalities $uvw=1$ and $S=u^{+1}+w^{-1}+w^{-1}$ imply $uw+wz+zu=S$.
Now let's use Vieta's relations: $u,w$ and $z$ are the roots of the cubic polynomial
$$Z^3-SZ^2+SZ-1=Z^3-1-SZ(Z-1)=(Z-1)(Z^2+Z+1-SZ),$$
which has $Z=1$ as a root.
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
answered Oct 7 '18 at 19:02
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
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