Proof that $1^alpha$ is dense on the unit circle












0












$begingroup$


We take an irrational, real $alpha$.



I have got as far to show that



$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.



After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.



This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.



Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.



Please help. Thanks :)






complex-analysis number-theory analysis complex-numbers proof-explanation







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  • $begingroup$
    Look at this answer to a very closely related question.
    $endgroup$
    – Greg Martin
    Jan 23 at 4:42
















0












$begingroup$


We take an irrational, real $alpha$.



I have got as far to show that



$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.



After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.



This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.



Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.



Please help. Thanks :)






complex-analysis number-theory analysis complex-numbers proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Look at this answer to a very closely related question.
    $endgroup$
    – Greg Martin
    Jan 23 at 4:42














0












0








0


1



$begingroup$


We take an irrational, real $alpha$.



I have got as far to show that



$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.



After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.



This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.



Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.



Please help. Thanks :)






complex-analysis number-theory analysis complex-numbers proof-explanation







share|cite|improve this question











$endgroup$




We take an irrational, real $alpha$.



I have got as far to show that



$1^{alpha}=e^{log(1^{alpha})}=e^{alpha ln(1)+ialpha arg(1)}=e^{ialpha arg(1)}=e^{ialpha 2pi k},kinmathbb{Z}$.



After that I show that $e^{ialpha 2pi k_1+2pi k_2}neq e^{ialpha 2pi k_1k_3}$ for any $k_1,k_2,k_3inmathbb{Z}$ by noting that $alpha2pi k_1k_3equiv 0modalpha$ but $alpha 2pi k_1+2pi k_2notequiv 0modalpha$ because $alpha$ is irrational.



This entails that no element of $1^{alpha}=e^{ialpha 2pi k}$ is coterminal with any other.



Then I assume you use a pigeonhole argument of some kind, but I am entirely new to this and unsure how to do this or make it rigorous. Additionally, as per instructions, I would like to solve this problem by dividing the unit circle into $ninmathbb{N}$ segments and proving that for any segment, some $1^{alpha}$ is contained in it.



Please help. Thanks :)






complex-analysis number-theory analysis complex-numbers proof-explanation




complex-analysis number-theory analysis complex-numbers proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 4:39







Tejas Rao

















asked Jan 23 at 3:50









Tejas RaoTejas Rao

33711




33711












  • $begingroup$
    Look at this answer to a very closely related question.
    $endgroup$
    – Greg Martin
    Jan 23 at 4:42


















  • $begingroup$
    Look at this answer to a very closely related question.
    $endgroup$
    – Greg Martin
    Jan 23 at 4:42
















$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42




$begingroup$
Look at this answer to a very closely related question.
$endgroup$
– Greg Martin
Jan 23 at 4:42










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