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

519211




519211












  • $begingroup$
    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

105k1160132




105k1160132












  • $begingroup$
    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

45.2k31954




45.2k31954












  • $begingroup$
    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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Wiesbaden

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Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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