Showing $f(z)=|z|^{1/2}z$ is differentiable at $z=0$ but not holomorphic.












2












$begingroup$


Given $f(z)=|z|^{1/2}z$, it is obviously complex differentiable at $z=0$ because $$f'(0)=lim_{zrightarrow 0}frac{|z|^{1/2}z}{z}=0$$ I have done similar examples with $|z|^2$, but the $|z|^{1/2}$ is throwing me off because you cannot simply expand it like you would with $|z|^2$. How can I show that it is not holomorphic at $z=0$?



Also, what would be the set of all $z$ such that $f$ is complex differentiable? I assume I have to use Cauchy-Riemann equation, but again, I'm having trouble working with the $|z|^{1/2}$






complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you tried looking at the partial derivatives and determining where they obey the Cauchy Riemann equations?
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 15:33










  • $begingroup$
    One approach might be (not sure if it works): since $g(z)=z^2$ is holomorphic, consider $gcirc f(z)$. I believe this fails to be holomorphic, so $f(z)$ is not holomorphic. I’m not $100%$ sure this is valid (though I am reasonably sure), but it at least provides a nice heuristic.
    $endgroup$
    – Clayton
    Dec 19 '18 at 15:36








  • 1




    $begingroup$
    @Kevin What for? And surely you are aware there exists no continuous square root in a neighborhood of the origin?
    $endgroup$
    – Did
    Dec 19 '18 at 19:26










  • $begingroup$
    @Did Mind blank. Yes I was aware, I had a bad day :-(
    $endgroup$
    – Kevin
    Dec 20 '18 at 10:24










  • $begingroup$
    @Kevin No problem. Next day surely will be better... :-)
    $endgroup$
    – Did
    Dec 20 '18 at 11:29
















2












$begingroup$


Given $f(z)=|z|^{1/2}z$, it is obviously complex differentiable at $z=0$ because $$f'(0)=lim_{zrightarrow 0}frac{|z|^{1/2}z}{z}=0$$ I have done similar examples with $|z|^2$, but the $|z|^{1/2}$ is throwing me off because you cannot simply expand it like you would with $|z|^2$. How can I show that it is not holomorphic at $z=0$?



Also, what would be the set of all $z$ such that $f$ is complex differentiable? I assume I have to use Cauchy-Riemann equation, but again, I'm having trouble working with the $|z|^{1/2}$






complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you tried looking at the partial derivatives and determining where they obey the Cauchy Riemann equations?
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 15:33










  • $begingroup$
    One approach might be (not sure if it works): since $g(z)=z^2$ is holomorphic, consider $gcirc f(z)$. I believe this fails to be holomorphic, so $f(z)$ is not holomorphic. I’m not $100%$ sure this is valid (though I am reasonably sure), but it at least provides a nice heuristic.
    $endgroup$
    – Clayton
    Dec 19 '18 at 15:36








  • 1




    $begingroup$
    @Kevin What for? And surely you are aware there exists no continuous square root in a neighborhood of the origin?
    $endgroup$
    – Did
    Dec 19 '18 at 19:26










  • $begingroup$
    @Did Mind blank. Yes I was aware, I had a bad day :-(
    $endgroup$
    – Kevin
    Dec 20 '18 at 10:24










  • $begingroup$
    @Kevin No problem. Next day surely will be better... :-)
    $endgroup$
    – Did
    Dec 20 '18 at 11:29














2












2








2


1



$begingroup$


Given $f(z)=|z|^{1/2}z$, it is obviously complex differentiable at $z=0$ because $$f'(0)=lim_{zrightarrow 0}frac{|z|^{1/2}z}{z}=0$$ I have done similar examples with $|z|^2$, but the $|z|^{1/2}$ is throwing me off because you cannot simply expand it like you would with $|z|^2$. How can I show that it is not holomorphic at $z=0$?



Also, what would be the set of all $z$ such that $f$ is complex differentiable? I assume I have to use Cauchy-Riemann equation, but again, I'm having trouble working with the $|z|^{1/2}$






complex-analysis







share|cite|improve this question









$endgroup$




Given $f(z)=|z|^{1/2}z$, it is obviously complex differentiable at $z=0$ because $$f'(0)=lim_{zrightarrow 0}frac{|z|^{1/2}z}{z}=0$$ I have done similar examples with $|z|^2$, but the $|z|^{1/2}$ is throwing me off because you cannot simply expand it like you would with $|z|^2$. How can I show that it is not holomorphic at $z=0$?



Also, what would be the set of all $z$ such that $f$ is complex differentiable? I assume I have to use Cauchy-Riemann equation, but again, I'm having trouble working with the $|z|^{1/2}$






complex-analysis




complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 15:25









Ya GYa G

469211




469211












  • $begingroup$
    Have you tried looking at the partial derivatives and determining where they obey the Cauchy Riemann equations?
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 15:33










  • $begingroup$
    One approach might be (not sure if it works): since $g(z)=z^2$ is holomorphic, consider $gcirc f(z)$. I believe this fails to be holomorphic, so $f(z)$ is not holomorphic. I’m not $100%$ sure this is valid (though I am reasonably sure), but it at least provides a nice heuristic.
    $endgroup$
    – Clayton
    Dec 19 '18 at 15:36








  • 1




    $begingroup$
    @Kevin What for? And surely you are aware there exists no continuous square root in a neighborhood of the origin?
    $endgroup$
    – Did
    Dec 19 '18 at 19:26










  • $begingroup$
    @Did Mind blank. Yes I was aware, I had a bad day :-(
    $endgroup$
    – Kevin
    Dec 20 '18 at 10:24










  • $begingroup$
    @Kevin No problem. Next day surely will be better... :-)
    $endgroup$
    – Did
    Dec 20 '18 at 11:29


















  • $begingroup$
    Have you tried looking at the partial derivatives and determining where they obey the Cauchy Riemann equations?
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 15:33










  • $begingroup$
    One approach might be (not sure if it works): since $g(z)=z^2$ is holomorphic, consider $gcirc f(z)$. I believe this fails to be holomorphic, so $f(z)$ is not holomorphic. I’m not $100%$ sure this is valid (though I am reasonably sure), but it at least provides a nice heuristic.
    $endgroup$
    – Clayton
    Dec 19 '18 at 15:36








  • 1




    $begingroup$
    @Kevin What for? And surely you are aware there exists no continuous square root in a neighborhood of the origin?
    $endgroup$
    – Did
    Dec 19 '18 at 19:26










  • $begingroup$
    @Did Mind blank. Yes I was aware, I had a bad day :-(
    $endgroup$
    – Kevin
    Dec 20 '18 at 10:24










  • $begingroup$
    @Kevin No problem. Next day surely will be better... :-)
    $endgroup$
    – Did
    Dec 20 '18 at 11:29
















$begingroup$
Have you tried looking at the partial derivatives and determining where they obey the Cauchy Riemann equations?
$endgroup$
– SmileyCraft
Dec 19 '18 at 15:33




$begingroup$
Have you tried looking at the partial derivatives and determining where they obey the Cauchy Riemann equations?
$endgroup$
– SmileyCraft
Dec 19 '18 at 15:33












$begingroup$
One approach might be (not sure if it works): since $g(z)=z^2$ is holomorphic, consider $gcirc f(z)$. I believe this fails to be holomorphic, so $f(z)$ is not holomorphic. I’m not $100%$ sure this is valid (though I am reasonably sure), but it at least provides a nice heuristic.
$endgroup$
– Clayton
Dec 19 '18 at 15:36






$begingroup$
One approach might be (not sure if it works): since $g(z)=z^2$ is holomorphic, consider $gcirc f(z)$. I believe this fails to be holomorphic, so $f(z)$ is not holomorphic. I’m not $100%$ sure this is valid (though I am reasonably sure), but it at least provides a nice heuristic.
$endgroup$
– Clayton
Dec 19 '18 at 15:36






1




1




$begingroup$
@Kevin What for? And surely you are aware there exists no continuous square root in a neighborhood of the origin?
$endgroup$
– Did
Dec 19 '18 at 19:26




$begingroup$
@Kevin What for? And surely you are aware there exists no continuous square root in a neighborhood of the origin?
$endgroup$
– Did
Dec 19 '18 at 19:26












$begingroup$
@Did Mind blank. Yes I was aware, I had a bad day :-(
$endgroup$
– Kevin
Dec 20 '18 at 10:24




$begingroup$
@Did Mind blank. Yes I was aware, I had a bad day :-(
$endgroup$
– Kevin
Dec 20 '18 at 10:24












$begingroup$
@Kevin No problem. Next day surely will be better... :-)
$endgroup$
– Did
Dec 20 '18 at 11:29




$begingroup$
@Kevin No problem. Next day surely will be better... :-)
$endgroup$
– Did
Dec 20 '18 at 11:29










2 Answers
2






active

oldest

votes


















4












$begingroup$

Holomorphicity at $0$ means that it's complex-differentiable in a neighbourhood
of $0$. But it's not complex-differentiable other than at $0$.



For the C-R equations, one has $f(x+iy)=u+iv$
where
$$u=x(x^2+y^2)^{1/4}$$
and
$$v=y(x^2+y^2)^{1/4}.$$
Then
$$u_x=(x^2+y^2)^{1/4}+frac{x^2}{2(x^2+y^2)^{3/4}}
=frac{3x^2+2y^2}{2(x^2+y^2)^{3/4}}$$
and similarly,
$$v_y=frac{2x^2+3y^2}{2(x^2+y^2)^{3/4}}.$$
These are different at nonzero points on the $x$-axis, so $f$ isn't
complex-differentiable there, and so not in any neighbourhood of $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    shouldn't $u=xleft(x^2+y^2right)^{1/4}$? since $|z|$ itself is $sqrt{x^2+y^2}$
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:37










  • $begingroup$
    @YaG Thanks${}$!
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 15:42



















0












$begingroup$

Substistute $z = x+iy$



Separate your function into real parts and imaginary parts.



$|z|^frac12$ is strictly real



$|z| = (x^2 + y^2)^frac 12\
|z|^{frac 12} = (x^2 + y^2)^frac 14\
|z|^{frac 12}z = (x^2 + y^2)^frac 14(x+iy)$



$f(x+iy) = u+iv\
u = (x^2+y^2)^frac 14x\
v = (x^2+y^2)^frac 14y$



Do these satisfy they Cauchy-Reimann equations?



$frac {partial u}{partial x} = frac {partial v}{partial y}\
frac {partial u}{partial y} = -frac {partial v}{partial x}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Got it! However, your $u$ and $v$ looks wrong.
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:59











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Holomorphicity at $0$ means that it's complex-differentiable in a neighbourhood
of $0$. But it's not complex-differentiable other than at $0$.



For the C-R equations, one has $f(x+iy)=u+iv$
where
$$u=x(x^2+y^2)^{1/4}$$
and
$$v=y(x^2+y^2)^{1/4}.$$
Then
$$u_x=(x^2+y^2)^{1/4}+frac{x^2}{2(x^2+y^2)^{3/4}}
=frac{3x^2+2y^2}{2(x^2+y^2)^{3/4}}$$
and similarly,
$$v_y=frac{2x^2+3y^2}{2(x^2+y^2)^{3/4}}.$$
These are different at nonzero points on the $x$-axis, so $f$ isn't
complex-differentiable there, and so not in any neighbourhood of $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    shouldn't $u=xleft(x^2+y^2right)^{1/4}$? since $|z|$ itself is $sqrt{x^2+y^2}$
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:37










  • $begingroup$
    @YaG Thanks${}$!
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 15:42
















4












$begingroup$

Holomorphicity at $0$ means that it's complex-differentiable in a neighbourhood
of $0$. But it's not complex-differentiable other than at $0$.



For the C-R equations, one has $f(x+iy)=u+iv$
where
$$u=x(x^2+y^2)^{1/4}$$
and
$$v=y(x^2+y^2)^{1/4}.$$
Then
$$u_x=(x^2+y^2)^{1/4}+frac{x^2}{2(x^2+y^2)^{3/4}}
=frac{3x^2+2y^2}{2(x^2+y^2)^{3/4}}$$
and similarly,
$$v_y=frac{2x^2+3y^2}{2(x^2+y^2)^{3/4}}.$$
These are different at nonzero points on the $x$-axis, so $f$ isn't
complex-differentiable there, and so not in any neighbourhood of $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    shouldn't $u=xleft(x^2+y^2right)^{1/4}$? since $|z|$ itself is $sqrt{x^2+y^2}$
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:37










  • $begingroup$
    @YaG Thanks${}$!
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 15:42














4












4








4





$begingroup$

Holomorphicity at $0$ means that it's complex-differentiable in a neighbourhood
of $0$. But it's not complex-differentiable other than at $0$.



For the C-R equations, one has $f(x+iy)=u+iv$
where
$$u=x(x^2+y^2)^{1/4}$$
and
$$v=y(x^2+y^2)^{1/4}.$$
Then
$$u_x=(x^2+y^2)^{1/4}+frac{x^2}{2(x^2+y^2)^{3/4}}
=frac{3x^2+2y^2}{2(x^2+y^2)^{3/4}}$$
and similarly,
$$v_y=frac{2x^2+3y^2}{2(x^2+y^2)^{3/4}}.$$
These are different at nonzero points on the $x$-axis, so $f$ isn't
complex-differentiable there, and so not in any neighbourhood of $0$.






share|cite|improve this answer











$endgroup$



Holomorphicity at $0$ means that it's complex-differentiable in a neighbourhood
of $0$. But it's not complex-differentiable other than at $0$.



For the C-R equations, one has $f(x+iy)=u+iv$
where
$$u=x(x^2+y^2)^{1/4}$$
and
$$v=y(x^2+y^2)^{1/4}.$$
Then
$$u_x=(x^2+y^2)^{1/4}+frac{x^2}{2(x^2+y^2)^{3/4}}
=frac{3x^2+2y^2}{2(x^2+y^2)^{3/4}}$$
and similarly,
$$v_y=frac{2x^2+3y^2}{2(x^2+y^2)^{3/4}}.$$
These are different at nonzero points on the $x$-axis, so $f$ isn't
complex-differentiable there, and so not in any neighbourhood of $0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 15:41

























answered Dec 19 '18 at 15:34









Lord Shark the UnknownLord Shark the Unknown

104k1160132




104k1160132












  • $begingroup$
    shouldn't $u=xleft(x^2+y^2right)^{1/4}$? since $|z|$ itself is $sqrt{x^2+y^2}$
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:37










  • $begingroup$
    @YaG Thanks${}$!
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 15:42


















  • $begingroup$
    shouldn't $u=xleft(x^2+y^2right)^{1/4}$? since $|z|$ itself is $sqrt{x^2+y^2}$
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:37










  • $begingroup$
    @YaG Thanks${}$!
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 15:42
















$begingroup$
shouldn't $u=xleft(x^2+y^2right)^{1/4}$? since $|z|$ itself is $sqrt{x^2+y^2}$
$endgroup$
– Ya G
Dec 19 '18 at 15:37




$begingroup$
shouldn't $u=xleft(x^2+y^2right)^{1/4}$? since $|z|$ itself is $sqrt{x^2+y^2}$
$endgroup$
– Ya G
Dec 19 '18 at 15:37












$begingroup$
@YaG Thanks${}$!
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 15:42




$begingroup$
@YaG Thanks${}$!
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 15:42











0












$begingroup$

Substistute $z = x+iy$



Separate your function into real parts and imaginary parts.



$|z|^frac12$ is strictly real



$|z| = (x^2 + y^2)^frac 12\
|z|^{frac 12} = (x^2 + y^2)^frac 14\
|z|^{frac 12}z = (x^2 + y^2)^frac 14(x+iy)$



$f(x+iy) = u+iv\
u = (x^2+y^2)^frac 14x\
v = (x^2+y^2)^frac 14y$



Do these satisfy they Cauchy-Reimann equations?



$frac {partial u}{partial x} = frac {partial v}{partial y}\
frac {partial u}{partial y} = -frac {partial v}{partial x}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Got it! However, your $u$ and $v$ looks wrong.
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:59
















0












$begingroup$

Substistute $z = x+iy$



Separate your function into real parts and imaginary parts.



$|z|^frac12$ is strictly real



$|z| = (x^2 + y^2)^frac 12\
|z|^{frac 12} = (x^2 + y^2)^frac 14\
|z|^{frac 12}z = (x^2 + y^2)^frac 14(x+iy)$



$f(x+iy) = u+iv\
u = (x^2+y^2)^frac 14x\
v = (x^2+y^2)^frac 14y$



Do these satisfy they Cauchy-Reimann equations?



$frac {partial u}{partial x} = frac {partial v}{partial y}\
frac {partial u}{partial y} = -frac {partial v}{partial x}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Got it! However, your $u$ and $v$ looks wrong.
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:59














0












0








0





$begingroup$

Substistute $z = x+iy$



Separate your function into real parts and imaginary parts.



$|z|^frac12$ is strictly real



$|z| = (x^2 + y^2)^frac 12\
|z|^{frac 12} = (x^2 + y^2)^frac 14\
|z|^{frac 12}z = (x^2 + y^2)^frac 14(x+iy)$



$f(x+iy) = u+iv\
u = (x^2+y^2)^frac 14x\
v = (x^2+y^2)^frac 14y$



Do these satisfy they Cauchy-Reimann equations?



$frac {partial u}{partial x} = frac {partial v}{partial y}\
frac {partial u}{partial y} = -frac {partial v}{partial x}$






share|cite|improve this answer











$endgroup$



Substistute $z = x+iy$



Separate your function into real parts and imaginary parts.



$|z|^frac12$ is strictly real



$|z| = (x^2 + y^2)^frac 12\
|z|^{frac 12} = (x^2 + y^2)^frac 14\
|z|^{frac 12}z = (x^2 + y^2)^frac 14(x+iy)$



$f(x+iy) = u+iv\
u = (x^2+y^2)^frac 14x\
v = (x^2+y^2)^frac 14y$



Do these satisfy they Cauchy-Reimann equations?



$frac {partial u}{partial x} = frac {partial v}{partial y}\
frac {partial u}{partial y} = -frac {partial v}{partial x}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 17:23

























answered Dec 19 '18 at 15:57









Doug MDoug M

45.2k31854




45.2k31854












  • $begingroup$
    Got it! However, your $u$ and $v$ looks wrong.
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:59


















  • $begingroup$
    Got it! However, your $u$ and $v$ looks wrong.
    $endgroup$
    – Ya G
    Dec 19 '18 at 15:59
















$begingroup$
Got it! However, your $u$ and $v$ looks wrong.
$endgroup$
– Ya G
Dec 19 '18 at 15:59




$begingroup$
Got it! However, your $u$ and $v$ looks wrong.
$endgroup$
– Ya G
Dec 19 '18 at 15:59


















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build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

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0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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