How to write $f(z)=sqrt{z}$ as a complex series around the origin












3












$begingroup$


I'm confused about the way to approach this exercises, I've been introduced complex analysis a few weeks ago.



If I have, for example, $f(z)=sqrt z$ what I'm tempted to do is to find some derivarives $f,f',f'',f''',...$ to see if I can guess the n-th term and write it's taylor polynomial around the origin.



Another way is to find some way to write $f(z)$ with terms whose series expansion are used frequently and well known.



My question is: what is necessary to be able to expand $f(z)$ as a power series around $z_0$? Is there a way am I missing? how can we do this kind of exercises?



Thanks for your time.






calculus complex-analysis taylor-expansion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Are you aware that there is no single-valued and continuous $sqrt{z}$ around the origin?
    $endgroup$
    – Adayah
    Jun 20 '18 at 15:33












  • $begingroup$
    Beginning complex analysis, you perhaps learned the definition of "analytic function". Check that $sqrt{z}$ is not analytic at $z=0$.
    $endgroup$
    – GEdgar
    Jun 20 '18 at 15:35










  • $begingroup$
    If you write down a definition of $sqrt z,$ the impossibility of doing this will become evident.
    $endgroup$
    – zhw.
    Jun 20 '18 at 15:59
















3












$begingroup$


I'm confused about the way to approach this exercises, I've been introduced complex analysis a few weeks ago.



If I have, for example, $f(z)=sqrt z$ what I'm tempted to do is to find some derivarives $f,f',f'',f''',...$ to see if I can guess the n-th term and write it's taylor polynomial around the origin.



Another way is to find some way to write $f(z)$ with terms whose series expansion are used frequently and well known.



My question is: what is necessary to be able to expand $f(z)$ as a power series around $z_0$? Is there a way am I missing? how can we do this kind of exercises?



Thanks for your time.






calculus complex-analysis taylor-expansion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Are you aware that there is no single-valued and continuous $sqrt{z}$ around the origin?
    $endgroup$
    – Adayah
    Jun 20 '18 at 15:33












  • $begingroup$
    Beginning complex analysis, you perhaps learned the definition of "analytic function". Check that $sqrt{z}$ is not analytic at $z=0$.
    $endgroup$
    – GEdgar
    Jun 20 '18 at 15:35










  • $begingroup$
    If you write down a definition of $sqrt z,$ the impossibility of doing this will become evident.
    $endgroup$
    – zhw.
    Jun 20 '18 at 15:59














3












3








3





$begingroup$


I'm confused about the way to approach this exercises, I've been introduced complex analysis a few weeks ago.



If I have, for example, $f(z)=sqrt z$ what I'm tempted to do is to find some derivarives $f,f',f'',f''',...$ to see if I can guess the n-th term and write it's taylor polynomial around the origin.



Another way is to find some way to write $f(z)$ with terms whose series expansion are used frequently and well known.



My question is: what is necessary to be able to expand $f(z)$ as a power series around $z_0$? Is there a way am I missing? how can we do this kind of exercises?



Thanks for your time.






calculus complex-analysis taylor-expansion







share|cite|improve this question











$endgroup$




I'm confused about the way to approach this exercises, I've been introduced complex analysis a few weeks ago.



If I have, for example, $f(z)=sqrt z$ what I'm tempted to do is to find some derivarives $f,f',f'',f''',...$ to see if I can guess the n-th term and write it's taylor polynomial around the origin.



Another way is to find some way to write $f(z)$ with terms whose series expansion are used frequently and well known.



My question is: what is necessary to be able to expand $f(z)$ as a power series around $z_0$? Is there a way am I missing? how can we do this kind of exercises?



Thanks for your time.






calculus complex-analysis taylor-expansion




calculus complex-analysis taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 3:34









Eric Wofsey

185k13213339




185k13213339










asked Jun 20 '18 at 15:30









RelureRelure

2,1671035




2,1671035








  • 4




    $begingroup$
    Are you aware that there is no single-valued and continuous $sqrt{z}$ around the origin?
    $endgroup$
    – Adayah
    Jun 20 '18 at 15:33












  • $begingroup$
    Beginning complex analysis, you perhaps learned the definition of "analytic function". Check that $sqrt{z}$ is not analytic at $z=0$.
    $endgroup$
    – GEdgar
    Jun 20 '18 at 15:35










  • $begingroup$
    If you write down a definition of $sqrt z,$ the impossibility of doing this will become evident.
    $endgroup$
    – zhw.
    Jun 20 '18 at 15:59














  • 4




    $begingroup$
    Are you aware that there is no single-valued and continuous $sqrt{z}$ around the origin?
    $endgroup$
    – Adayah
    Jun 20 '18 at 15:33












  • $begingroup$
    Beginning complex analysis, you perhaps learned the definition of "analytic function". Check that $sqrt{z}$ is not analytic at $z=0$.
    $endgroup$
    – GEdgar
    Jun 20 '18 at 15:35










  • $begingroup$
    If you write down a definition of $sqrt z,$ the impossibility of doing this will become evident.
    $endgroup$
    – zhw.
    Jun 20 '18 at 15:59








4




4




$begingroup$
Are you aware that there is no single-valued and continuous $sqrt{z}$ around the origin?
$endgroup$
– Adayah
Jun 20 '18 at 15:33






$begingroup$
Are you aware that there is no single-valued and continuous $sqrt{z}$ around the origin?
$endgroup$
– Adayah
Jun 20 '18 at 15:33














$begingroup$
Beginning complex analysis, you perhaps learned the definition of "analytic function". Check that $sqrt{z}$ is not analytic at $z=0$.
$endgroup$
– GEdgar
Jun 20 '18 at 15:35




$begingroup$
Beginning complex analysis, you perhaps learned the definition of "analytic function". Check that $sqrt{z}$ is not analytic at $z=0$.
$endgroup$
– GEdgar
Jun 20 '18 at 15:35












$begingroup$
If you write down a definition of $sqrt z,$ the impossibility of doing this will become evident.
$endgroup$
– zhw.
Jun 20 '18 at 15:59




$begingroup$
If you write down a definition of $sqrt z,$ the impossibility of doing this will become evident.
$endgroup$
– zhw.
Jun 20 '18 at 15:59










1 Answer
1






active

oldest

votes


















7












$begingroup$

A Taylor series expansion is possible for a function $fcolonmathbb Ctomathbb C$ is possible (converges) at $z_0$ if $f$ is complex analytic in a neighborhood of the point $z_0$.



The square root function has two issues.
First, at the origin it is not even analytic: it does not satisfy the Cauchy–Riemann equations because it is not even differentiable.
In addition, outside the origin it is not single-valued.
If you choose a branch (see below) of the square root, then it is analytic in a neighborhood of any $z_0neq0$.



There is also a possibility to have a series expansion around singular points.
This is known as a Laurent series, but at this stage you should focus on analytic functions and Taylor series.
But even the Laurent series does not exist for all functions.
It requires that the singularity is of a suitable nice kind, and the one of the square root at the origin does not qualify.



Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign.
Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued.
This concept should appear at some later point in your studies in more detail, but this is the idea.
Some other functions have more than two options to choose from.
For example, the number $1+i$ has three cubic roots and infinitely many logarithms.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi @Joonas, what do you mean by a branch?
    $endgroup$
    – Relure
    Jun 20 '18 at 15:45






  • 1




    $begingroup$
    @Relure Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign. Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued. This concept should appear at some later point in your studies in more detail, but this is the idea. Some other functions have more than two options to choose from.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 15:50






  • 2




    $begingroup$
    I like most of this answer (+1) but I worry that the third paragraph might give the OP or other readers the idea that there will be a Laurent series for $sqrt z$ around $0$. The best one can actually get is (as far as I know) a Puiseux series (which in this case would be trivial: $z^{1/2}$).
    $endgroup$
    – Andreas Blass
    Jun 20 '18 at 16:55










  • $begingroup$
    @AndreasBlass Fair enough, that was indeed misleading. I edited the paragraph a bit.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 18:59











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

A Taylor series expansion is possible for a function $fcolonmathbb Ctomathbb C$ is possible (converges) at $z_0$ if $f$ is complex analytic in a neighborhood of the point $z_0$.



The square root function has two issues.
First, at the origin it is not even analytic: it does not satisfy the Cauchy–Riemann equations because it is not even differentiable.
In addition, outside the origin it is not single-valued.
If you choose a branch (see below) of the square root, then it is analytic in a neighborhood of any $z_0neq0$.



There is also a possibility to have a series expansion around singular points.
This is known as a Laurent series, but at this stage you should focus on analytic functions and Taylor series.
But even the Laurent series does not exist for all functions.
It requires that the singularity is of a suitable nice kind, and the one of the square root at the origin does not qualify.



Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign.
Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued.
This concept should appear at some later point in your studies in more detail, but this is the idea.
Some other functions have more than two options to choose from.
For example, the number $1+i$ has three cubic roots and infinitely many logarithms.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi @Joonas, what do you mean by a branch?
    $endgroup$
    – Relure
    Jun 20 '18 at 15:45






  • 1




    $begingroup$
    @Relure Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign. Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued. This concept should appear at some later point in your studies in more detail, but this is the idea. Some other functions have more than two options to choose from.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 15:50






  • 2




    $begingroup$
    I like most of this answer (+1) but I worry that the third paragraph might give the OP or other readers the idea that there will be a Laurent series for $sqrt z$ around $0$. The best one can actually get is (as far as I know) a Puiseux series (which in this case would be trivial: $z^{1/2}$).
    $endgroup$
    – Andreas Blass
    Jun 20 '18 at 16:55










  • $begingroup$
    @AndreasBlass Fair enough, that was indeed misleading. I edited the paragraph a bit.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 18:59
















7












$begingroup$

A Taylor series expansion is possible for a function $fcolonmathbb Ctomathbb C$ is possible (converges) at $z_0$ if $f$ is complex analytic in a neighborhood of the point $z_0$.



The square root function has two issues.
First, at the origin it is not even analytic: it does not satisfy the Cauchy–Riemann equations because it is not even differentiable.
In addition, outside the origin it is not single-valued.
If you choose a branch (see below) of the square root, then it is analytic in a neighborhood of any $z_0neq0$.



There is also a possibility to have a series expansion around singular points.
This is known as a Laurent series, but at this stage you should focus on analytic functions and Taylor series.
But even the Laurent series does not exist for all functions.
It requires that the singularity is of a suitable nice kind, and the one of the square root at the origin does not qualify.



Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign.
Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued.
This concept should appear at some later point in your studies in more detail, but this is the idea.
Some other functions have more than two options to choose from.
For example, the number $1+i$ has three cubic roots and infinitely many logarithms.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi @Joonas, what do you mean by a branch?
    $endgroup$
    – Relure
    Jun 20 '18 at 15:45






  • 1




    $begingroup$
    @Relure Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign. Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued. This concept should appear at some later point in your studies in more detail, but this is the idea. Some other functions have more than two options to choose from.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 15:50






  • 2




    $begingroup$
    I like most of this answer (+1) but I worry that the third paragraph might give the OP or other readers the idea that there will be a Laurent series for $sqrt z$ around $0$. The best one can actually get is (as far as I know) a Puiseux series (which in this case would be trivial: $z^{1/2}$).
    $endgroup$
    – Andreas Blass
    Jun 20 '18 at 16:55










  • $begingroup$
    @AndreasBlass Fair enough, that was indeed misleading. I edited the paragraph a bit.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 18:59














7












7








7





$begingroup$

A Taylor series expansion is possible for a function $fcolonmathbb Ctomathbb C$ is possible (converges) at $z_0$ if $f$ is complex analytic in a neighborhood of the point $z_0$.



The square root function has two issues.
First, at the origin it is not even analytic: it does not satisfy the Cauchy–Riemann equations because it is not even differentiable.
In addition, outside the origin it is not single-valued.
If you choose a branch (see below) of the square root, then it is analytic in a neighborhood of any $z_0neq0$.



There is also a possibility to have a series expansion around singular points.
This is known as a Laurent series, but at this stage you should focus on analytic functions and Taylor series.
But even the Laurent series does not exist for all functions.
It requires that the singularity is of a suitable nice kind, and the one of the square root at the origin does not qualify.



Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign.
Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued.
This concept should appear at some later point in your studies in more detail, but this is the idea.
Some other functions have more than two options to choose from.
For example, the number $1+i$ has three cubic roots and infinitely many logarithms.






share|cite|improve this answer











$endgroup$



A Taylor series expansion is possible for a function $fcolonmathbb Ctomathbb C$ is possible (converges) at $z_0$ if $f$ is complex analytic in a neighborhood of the point $z_0$.



The square root function has two issues.
First, at the origin it is not even analytic: it does not satisfy the Cauchy–Riemann equations because it is not even differentiable.
In addition, outside the origin it is not single-valued.
If you choose a branch (see below) of the square root, then it is analytic in a neighborhood of any $z_0neq0$.



There is also a possibility to have a series expansion around singular points.
This is known as a Laurent series, but at this stage you should focus on analytic functions and Taylor series.
But even the Laurent series does not exist for all functions.
It requires that the singularity is of a suitable nice kind, and the one of the square root at the origin does not qualify.



Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign.
Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued.
This concept should appear at some later point in your studies in more detail, but this is the idea.
Some other functions have more than two options to choose from.
For example, the number $1+i$ has three cubic roots and infinitely many logarithms.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 20 '18 at 18:59

























answered Jun 20 '18 at 15:41









Joonas IlmavirtaJoonas Ilmavirta

20.6k94282




20.6k94282












  • $begingroup$
    Hi @Joonas, what do you mean by a branch?
    $endgroup$
    – Relure
    Jun 20 '18 at 15:45






  • 1




    $begingroup$
    @Relure Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign. Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued. This concept should appear at some later point in your studies in more detail, but this is the idea. Some other functions have more than two options to choose from.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 15:50






  • 2




    $begingroup$
    I like most of this answer (+1) but I worry that the third paragraph might give the OP or other readers the idea that there will be a Laurent series for $sqrt z$ around $0$. The best one can actually get is (as far as I know) a Puiseux series (which in this case would be trivial: $z^{1/2}$).
    $endgroup$
    – Andreas Blass
    Jun 20 '18 at 16:55










  • $begingroup$
    @AndreasBlass Fair enough, that was indeed misleading. I edited the paragraph a bit.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 18:59


















  • $begingroup$
    Hi @Joonas, what do you mean by a branch?
    $endgroup$
    – Relure
    Jun 20 '18 at 15:45






  • 1




    $begingroup$
    @Relure Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign. Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued. This concept should appear at some later point in your studies in more detail, but this is the idea. Some other functions have more than two options to choose from.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 15:50






  • 2




    $begingroup$
    I like most of this answer (+1) but I worry that the third paragraph might give the OP or other readers the idea that there will be a Laurent series for $sqrt z$ around $0$. The best one can actually get is (as far as I know) a Puiseux series (which in this case would be trivial: $z^{1/2}$).
    $endgroup$
    – Andreas Blass
    Jun 20 '18 at 16:55










  • $begingroup$
    @AndreasBlass Fair enough, that was indeed misleading. I edited the paragraph a bit.
    $endgroup$
    – Joonas Ilmavirta
    Jun 20 '18 at 18:59
















$begingroup$
Hi @Joonas, what do you mean by a branch?
$endgroup$
– Relure
Jun 20 '18 at 15:45




$begingroup$
Hi @Joonas, what do you mean by a branch?
$endgroup$
– Relure
Jun 20 '18 at 15:45




1




1




$begingroup$
@Relure Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign. Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued. This concept should appear at some later point in your studies in more detail, but this is the idea. Some other functions have more than two options to choose from.
$endgroup$
– Joonas Ilmavirta
Jun 20 '18 at 15:50




$begingroup$
@Relure Every non-zero complex number has two square roots, just like a number $x>0$ has two square roots which differ by sign. Taking a branch means choosing one of the two in a consistent way so that the square root function becomes continuous and single-valued. This concept should appear at some later point in your studies in more detail, but this is the idea. Some other functions have more than two options to choose from.
$endgroup$
– Joonas Ilmavirta
Jun 20 '18 at 15:50




2




2




$begingroup$
I like most of this answer (+1) but I worry that the third paragraph might give the OP or other readers the idea that there will be a Laurent series for $sqrt z$ around $0$. The best one can actually get is (as far as I know) a Puiseux series (which in this case would be trivial: $z^{1/2}$).
$endgroup$
– Andreas Blass
Jun 20 '18 at 16:55




$begingroup$
I like most of this answer (+1) but I worry that the third paragraph might give the OP or other readers the idea that there will be a Laurent series for $sqrt z$ around $0$. The best one can actually get is (as far as I know) a Puiseux series (which in this case would be trivial: $z^{1/2}$).
$endgroup$
– Andreas Blass
Jun 20 '18 at 16:55












$begingroup$
@AndreasBlass Fair enough, that was indeed misleading. I edited the paragraph a bit.
$endgroup$
– Joonas Ilmavirta
Jun 20 '18 at 18:59




$begingroup$
@AndreasBlass Fair enough, that was indeed misleading. I edited the paragraph a bit.
$endgroup$
– Joonas Ilmavirta
Jun 20 '18 at 18:59


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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Gyllenstierna

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