Mixing
Removable Singularity with Entire Function
Clash Royale CLAN TAG#URR8PPP
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Let $f$ be an entire function with $sup_zinmathbbC|f(z)/z|<infty$. Show that $z=0$ is a removable singularity of $g(z):=f(z)/z$.
To prove the claim, I need to show that $0=lim_zto 0(z-0)g(z)=lim_zto 0(f(z)$, that seems really obvious to me because if a fraction where the denominator goes to zero is bounded, then the nominator has to go to zero?
I would like to do that more formally. So what can I write for a proof?
complex-analysis singularity entire-functions
edited Aug 3 at 15:24
Clayton
17.6k22781
asked Aug 3 at 14:34
mathstackuser
1497
add a comment |Â
up vote
0
down vote
favorite
Let $f$ be an entire function with $sup_zinmathbbC|f(z)/z|<infty$. Show that $z=0$ is a removable singularity of $g(z):=f(z)/z$.
To prove the claim, I need to show that $0=lim_zto 0(z-0)g(z)=lim_zto 0(f(z)$, that seems really obvious to me because if a fraction where the denominator goes to zero is bounded, then the nominator has to go to zero?
I would like to do that more formally. So what can I write for a proof?
complex-analysis singularity entire-functions
edited Aug 3 at 15:24
Clayton
17.6k22781
asked Aug 3 at 14:34
mathstackuser
1497
1
Hint: you can look at the laurent series of f(z)/z around zero. Use the integral formula for the coefficients and the fact that f(z)/z is bounded to show that all of the coefficients of z^n (where n is negative) are zero. Thus 0 is a removeable singilarity of f(z)/z
– zokomoko
Aug 3 at 14:43
1
Use the Taylor expansion of $f(z)$ at $z_0=0$.
– Riemann
Aug 3 at 14:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f$ be an entire function with $sup_zinmathbbC|f(z)/z|<infty$. Show that $z=0$ is a removable singularity of $g(z):=f(z)/z$.
To prove the claim, I need to show that $0=lim_zto 0(z-0)g(z)=lim_zto 0(f(z)$, that seems really obvious to me because if a fraction where the denominator goes to zero is bounded, then the nominator has to go to zero?
I would like to do that more formally. So what can I write for a proof?
complex-analysis singularity entire-functions
edited Aug 3 at 15:24
Clayton
17.6k22781
asked Aug 3 at 14:34
mathstackuser
1497
Let $f$ be an entire function with $sup_zinmathbbC|f(z)/z|<infty$. Show that $z=0$ is a removable singularity of $g(z):=f(z)/z$.
To prove the claim, I need to show that $0=lim_zto 0(z-0)g(z)=lim_zto 0(f(z)$, that seems really obvious to me because if a fraction where the denominator goes to zero is bounded, then the nominator has to go to zero?
I would like to do that more formally. So what can I write for a proof?
complex-analysis singularity entire-functions
edited Aug 3 at 15:24
Clayton
17.6k22781
asked Aug 3 at 14:34
mathstackuser
1497
edited Aug 3 at 15:24
Clayton
17.6k22781
edited Aug 3 at 15:24
Clayton
17.6k22781
edited Aug 3 at 15:24
Clayton
17.6k22781
17.6k22781
asked Aug 3 at 14:34
mathstackuser
1497
asked Aug 3 at 14:34
mathstackuser
1497
asked Aug 3 at 14:34
mathstackuser
1497
1497
1
Hint: you can look at the laurent series of f(z)/z around zero. Use the integral formula for the coefficients and the fact that f(z)/z is bounded to show that all of the coefficients of z^n (where n is negative) are zero. Thus 0 is a removeable singilarity of f(z)/z
– zokomoko
Aug 3 at 14:43
1
Use the Taylor expansion of $f(z)$ at $z_0=0$.
– Riemann
Aug 3 at 14:48
add a comment |Â
1
Hint: you can look at the laurent series of f(z)/z around zero. Use the integral formula for the coefficients and the fact that f(z)/z is bounded to show that all of the coefficients of z^n (where n is negative) are zero. Thus 0 is a removeable singilarity of f(z)/z
– zokomoko
Aug 3 at 14:43
1
Use the Taylor expansion of $f(z)$ at $z_0=0$.
– Riemann
Aug 3 at 14:48
1
1
Hint: you can look at the laurent series of f(z)/z around zero. Use the integral formula for the coefficients and the fact that f(z)/z is bounded to show that all of the coefficients of z^n (where n is negative) are zero. Thus 0 is a removeable singilarity of f(z)/z
– zokomoko
Aug 3 at 14:43
Hint: you can look at the laurent series of f(z)/z around zero. Use the integral formula for the coefficients and the fact that f(z)/z is bounded to show that all of the coefficients of z^n (where n is negative) are zero. Thus 0 is a removeable singilarity of f(z)/z
– zokomoko
Aug 3 at 14:43
1
1
Use the Taylor expansion of $f(z)$ at $z_0=0$.
– Riemann
Aug 3 at 14:48
Use the Taylor expansion of $f(z)$ at $z_0=0$.
– Riemann
Aug 3 at 14:48
add a comment |Â
2 Answers
2
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up vote
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In order to prove that $lim_zto0zg(z)=0$, take $varepsilon>0$. Let $S=sup_zinmathbbCsetminus0left|fracf(z)zright|$. If $S=0$, then $f$ is the null function and the statment is trivial. Otherwise, let $delta=fracvarepsilon S$. Then$$|z|<deltaiff|z|<fracvarepsilon Simpliesbigl|f(z)bigr|=|z|.left|fracf(z)zright|<fracvarepsilon S.S=varepsilon.$$Therefore, by the definition of limit, $lim_zto0zg(z)=0$.
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
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up vote
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Since $g(z)$ is analytic in $0 < left| z right| < infty $ by laurent's theorem we can write:
$$g(z) = sumlimits_ - infty ^infty a_nz^n $$
for all $0 < left| z right| < infty $,
and $$a_n = 1 over 2pi iointlimits_gamma gleft( z right) over z^n + 1dz $$
for any simple-closed curve $gamma $ that lies in the annulus $0 < left| z right| < infty $ and circles $z=0$.
Notice that for all $R>0$ by the estimation lemma:
$$eqalign cr $$
Let $M equiv mathop sup limits_z in Bbb C left| fracfleft( z right)z right|$.
We now have $left| a_n right| le M over R^n$ for all $R>0$
Since $a_n$ does not depend on $R$ (which was arbitrary) we can take the limit $R to 0^ + $:
Thus, if $n leqslant - 1$ we have
$$left| a_n right| = mathop lim limits_R to 0^ + left| a_n right| leqslant mathop lim limits_R to 0^ + fracMR^n = 0$$
Since $a_n = 0$ for all $n leqslant - 1$ we can conclude that $z=0$ is a removable singularity for $g(z)$.
But we can actually show more than that!
We can also take the limit $R to infty $:
Thus, if $n geqslant 0$ we have
$$left| a_n right| = mathop lim limits_R to infty left| a_n right| leqslant mathop lim limits_R to infty fracMR^n = 0$$
In conclusion, $gleft( z right) = sumlimits_ - infty ^infty a_nz_n = a_0$ (a constant)
answered Aug 3 at 15:51
zokomoko
305213
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In order to prove that $lim_zto0zg(z)=0$, take $varepsilon>0$. Let $S=sup_zinmathbbCsetminus0left|fracf(z)zright|$. If $S=0$, then $f$ is the null function and the statment is trivial. Otherwise, let $delta=fracvarepsilon S$. Then$$|z|<deltaiff|z|<fracvarepsilon Simpliesbigl|f(z)bigr|=|z|.left|fracf(z)zright|<fracvarepsilon S.S=varepsilon.$$Therefore, by the definition of limit, $lim_zto0zg(z)=0$.
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
add a comment |Â
up vote
1
down vote
accepted
In order to prove that $lim_zto0zg(z)=0$, take $varepsilon>0$. Let $S=sup_zinmathbbCsetminus0left|fracf(z)zright|$. If $S=0$, then $f$ is the null function and the statment is trivial. Otherwise, let $delta=fracvarepsilon S$. Then$$|z|<deltaiff|z|<fracvarepsilon Simpliesbigl|f(z)bigr|=|z|.left|fracf(z)zright|<fracvarepsilon S.S=varepsilon.$$Therefore, by the definition of limit, $lim_zto0zg(z)=0$.
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In order to prove that $lim_zto0zg(z)=0$, take $varepsilon>0$. Let $S=sup_zinmathbbCsetminus0left|fracf(z)zright|$. If $S=0$, then $f$ is the null function and the statment is trivial. Otherwise, let $delta=fracvarepsilon S$. Then$$|z|<deltaiff|z|<fracvarepsilon Simpliesbigl|f(z)bigr|=|z|.left|fracf(z)zright|<fracvarepsilon S.S=varepsilon.$$Therefore, by the definition of limit, $lim_zto0zg(z)=0$.
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
In order to prove that $lim_zto0zg(z)=0$, take $varepsilon>0$. Let $S=sup_zinmathbbCsetminus0left|fracf(z)zright|$. If $S=0$, then $f$ is the null function and the statment is trivial. Otherwise, let $delta=fracvarepsilon S$. Then$$|z|<deltaiff|z|<fracvarepsilon Simpliesbigl|f(z)bigr|=|z|.left|fracf(z)zright|<fracvarepsilon S.S=varepsilon.$$Therefore, by the definition of limit, $lim_zto0zg(z)=0$.
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
answered Aug 3 at 15:17
José Carlos Santos
112k1696172
112k1696172
add a comment |Â
add a comment |Â
up vote
0
down vote
Since $g(z)$ is analytic in $0 < left| z right| < infty $ by laurent's theorem we can write:
$$g(z) = sumlimits_ - infty ^infty a_nz^n $$
for all $0 < left| z right| < infty $,
and $$a_n = 1 over 2pi iointlimits_gamma gleft( z right) over z^n + 1dz $$
for any simple-closed curve $gamma $ that lies in the annulus $0 < left| z right| < infty $ and circles $z=0$.
Notice that for all $R>0$ by the estimation lemma:
$$eqalign cr $$
Let $M equiv mathop sup limits_z in Bbb C left| fracfleft( z right)z right|$.
We now have $left| a_n right| le M over R^n$ for all $R>0$
Since $a_n$ does not depend on $R$ (which was arbitrary) we can take the limit $R to 0^ + $:
Thus, if $n leqslant - 1$ we have
$$left| a_n right| = mathop lim limits_R to 0^ + left| a_n right| leqslant mathop lim limits_R to 0^ + fracMR^n = 0$$
Since $a_n = 0$ for all $n leqslant - 1$ we can conclude that $z=0$ is a removable singularity for $g(z)$.
But we can actually show more than that!
We can also take the limit $R to infty $:
Thus, if $n geqslant 0$ we have
$$left| a_n right| = mathop lim limits_R to infty left| a_n right| leqslant mathop lim limits_R to infty fracMR^n = 0$$
In conclusion, $gleft( z right) = sumlimits_ - infty ^infty a_nz_n = a_0$ (a constant)
answered Aug 3 at 15:51
zokomoko
305213
add a comment |Â
up vote
0
down vote
Since $g(z)$ is analytic in $0 < left| z right| < infty $ by laurent's theorem we can write:
$$g(z) = sumlimits_ - infty ^infty a_nz^n $$
for all $0 < left| z right| < infty $,
and $$a_n = 1 over 2pi iointlimits_gamma gleft( z right) over z^n + 1dz $$
for any simple-closed curve $gamma $ that lies in the annulus $0 < left| z right| < infty $ and circles $z=0$.
Notice that for all $R>0$ by the estimation lemma:
$$eqalign cr $$
Let $M equiv mathop sup limits_z in Bbb C left| fracfleft( z right)z right|$.
We now have $left| a_n right| le M over R^n$ for all $R>0$
Since $a_n$ does not depend on $R$ (which was arbitrary) we can take the limit $R to 0^ + $:
Thus, if $n leqslant - 1$ we have
$$left| a_n right| = mathop lim limits_R to 0^ + left| a_n right| leqslant mathop lim limits_R to 0^ + fracMR^n = 0$$
Since $a_n = 0$ for all $n leqslant - 1$ we can conclude that $z=0$ is a removable singularity for $g(z)$.
But we can actually show more than that!
We can also take the limit $R to infty $:
Thus, if $n geqslant 0$ we have
$$left| a_n right| = mathop lim limits_R to infty left| a_n right| leqslant mathop lim limits_R to infty fracMR^n = 0$$
In conclusion, $gleft( z right) = sumlimits_ - infty ^infty a_nz_n = a_0$ (a constant)
answered Aug 3 at 15:51
zokomoko
305213
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $g(z)$ is analytic in $0 < left| z right| < infty $ by laurent's theorem we can write:
$$g(z) = sumlimits_ - infty ^infty a_nz^n $$
for all $0 < left| z right| < infty $,
and $$a_n = 1 over 2pi iointlimits_gamma gleft( z right) over z^n + 1dz $$
for any simple-closed curve $gamma $ that lies in the annulus $0 < left| z right| < infty $ and circles $z=0$.
Notice that for all $R>0$ by the estimation lemma:
$$eqalign cr $$
Let $M equiv mathop sup limits_z in Bbb C left| fracfleft( z right)z right|$.
We now have $left| a_n right| le M over R^n$ for all $R>0$
Since $a_n$ does not depend on $R$ (which was arbitrary) we can take the limit $R to 0^ + $:
Thus, if $n leqslant - 1$ we have
$$left| a_n right| = mathop lim limits_R to 0^ + left| a_n right| leqslant mathop lim limits_R to 0^ + fracMR^n = 0$$
Since $a_n = 0$ for all $n leqslant - 1$ we can conclude that $z=0$ is a removable singularity for $g(z)$.
But we can actually show more than that!
We can also take the limit $R to infty $:
Thus, if $n geqslant 0$ we have
$$left| a_n right| = mathop lim limits_R to infty left| a_n right| leqslant mathop lim limits_R to infty fracMR^n = 0$$
In conclusion, $gleft( z right) = sumlimits_ - infty ^infty a_nz_n = a_0$ (a constant)
answered Aug 3 at 15:51
zokomoko
305213
Since $g(z)$ is analytic in $0 < left| z right| < infty $ by laurent's theorem we can write:
$$g(z) = sumlimits_ - infty ^infty a_nz^n $$
for all $0 < left| z right| < infty $,
and $$a_n = 1 over 2pi iointlimits_gamma gleft( z right) over z^n + 1dz $$
for any simple-closed curve $gamma $ that lies in the annulus $0 < left| z right| < infty $ and circles $z=0$.
Notice that for all $R>0$ by the estimation lemma:
$$eqalign cr $$
Let $M equiv mathop sup limits_z in Bbb C left| fracfleft( z right)z right|$.
We now have $left| a_n right| le M over R^n$ for all $R>0$
Since $a_n$ does not depend on $R$ (which was arbitrary) we can take the limit $R to 0^ + $:
Thus, if $n leqslant - 1$ we have
$$left| a_n right| = mathop lim limits_R to 0^ + left| a_n right| leqslant mathop lim limits_R to 0^ + fracMR^n = 0$$
Since $a_n = 0$ for all $n leqslant - 1$ we can conclude that $z=0$ is a removable singularity for $g(z)$.
But we can actually show more than that!
We can also take the limit $R to infty $:
Thus, if $n geqslant 0$ we have
$$left| a_n right| = mathop lim limits_R to infty left| a_n right| leqslant mathop lim limits_R to infty fracMR^n = 0$$
In conclusion, $gleft( z right) = sumlimits_ - infty ^infty a_nz_n = a_0$ (a constant)
answered Aug 3 at 15:51
zokomoko
305213
answered Aug 3 at 15:51
zokomoko
305213
answered Aug 3 at 15:51
zokomoko
305213
answered Aug 3 at 15:51
zokomoko
305213
305213
add a comment |Â
add a comment |Â
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1
Hint: you can look at the laurent series of f(z)/z around zero. Use the integral formula for the coefficients and the fact that f(z)/z is bounded to show that all of the coefficients of z^n (where n is negative) are zero. Thus 0 is a removeable singilarity of f(z)/z
– zokomoko
Aug 3 at 14:43
1
Use the Taylor expansion of $f(z)$ at $z_0=0$.
– Riemann
Aug 3 at 14:48