To show that the function is not holomorphic at origin but satisfies the Cauchy-Reimann equation












0












$begingroup$


Let $$ f(z) = left{ begin{align}
&e^{-frac{1}{z^4}} &hspace{1mm} mbox{if} hspace{1mm} z neq 0 \
&0 &hspace{1mm} mbox{if} hspace{1mm} z = 0 \
end{align} right. $$



I have to show that this function satisfies the Cauchy-Riemann equation at $z=0$, but is not holomorphic.



I have been able to prove that it satisfies Cauchy-Riemann. But I am struggling with proving that it is not holomorphic at $z=0$.



My attempt:



Let $z=x+iy$ where $x,yinmathbb{R}$. Then
$$frac{1}{z}=frac{x-iy}{x^2+y^2}.$$
So
$$frac{1}{z^4}=frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}+ifrac{4xy^3-4x^3y}{(x^2+y^2)^4}.$$



Then
$$e^{-frac{1}{z^4}}=e^{-frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}}left(cosleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)-i sinleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)right)$$



My idea is the following:




If we can show that $f$ is not continuous at $z=0$, then it will follow automatically that $f$ is not holomorphic.




To show that $f$ is not continuous, I am trying to show that $f$ has different limits as $(x,y)to (0,0)$ along different paths. I have tried the paths $y=x$, $y=-x$, $y=x^2$. But all these paths give the limit to be $0$, which is the value of $f$ at $z=0$.



(1) Is my idea wrong and the function is continuous at $z=0$?



(2) If $f$ is not continuous, which path can be used to prove it?






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If it were holomorphic it would have a power-series expansion around zero. What can you say about the coefficients?
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:47






  • 1




    $begingroup$
    @uniquesolution I suppose that the OP meant to say that it is not differentiable at the origin.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:49






  • 1




    $begingroup$
    @Jose Carlos Santos But it is differentiable at the origin. In fact, it is differentiable infinitely many times at the origin, but it is not holmorphic there. (holomorphic=analytic). I suppose the OP meant to say exactly what (s)he said.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:53








  • 1




    $begingroup$
    @uniquesolution No, it is not differentiable at the origin. It is not even continuous there.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:55








  • 1




    $begingroup$
    Yes you are right -- I assumed - mistakenly - that $z$ is real.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 8:02
















0












$begingroup$


Let $$ f(z) = left{ begin{align}
&e^{-frac{1}{z^4}} &hspace{1mm} mbox{if} hspace{1mm} z neq 0 \
&0 &hspace{1mm} mbox{if} hspace{1mm} z = 0 \
end{align} right. $$



I have to show that this function satisfies the Cauchy-Riemann equation at $z=0$, but is not holomorphic.



I have been able to prove that it satisfies Cauchy-Riemann. But I am struggling with proving that it is not holomorphic at $z=0$.



My attempt:



Let $z=x+iy$ where $x,yinmathbb{R}$. Then
$$frac{1}{z}=frac{x-iy}{x^2+y^2}.$$
So
$$frac{1}{z^4}=frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}+ifrac{4xy^3-4x^3y}{(x^2+y^2)^4}.$$



Then
$$e^{-frac{1}{z^4}}=e^{-frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}}left(cosleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)-i sinleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)right)$$



My idea is the following:




If we can show that $f$ is not continuous at $z=0$, then it will follow automatically that $f$ is not holomorphic.




To show that $f$ is not continuous, I am trying to show that $f$ has different limits as $(x,y)to (0,0)$ along different paths. I have tried the paths $y=x$, $y=-x$, $y=x^2$. But all these paths give the limit to be $0$, which is the value of $f$ at $z=0$.



(1) Is my idea wrong and the function is continuous at $z=0$?



(2) If $f$ is not continuous, which path can be used to prove it?






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If it were holomorphic it would have a power-series expansion around zero. What can you say about the coefficients?
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:47






  • 1




    $begingroup$
    @uniquesolution I suppose that the OP meant to say that it is not differentiable at the origin.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:49






  • 1




    $begingroup$
    @Jose Carlos Santos But it is differentiable at the origin. In fact, it is differentiable infinitely many times at the origin, but it is not holmorphic there. (holomorphic=analytic). I suppose the OP meant to say exactly what (s)he said.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:53








  • 1




    $begingroup$
    @uniquesolution No, it is not differentiable at the origin. It is not even continuous there.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:55








  • 1




    $begingroup$
    Yes you are right -- I assumed - mistakenly - that $z$ is real.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 8:02














0












0








0





$begingroup$


Let $$ f(z) = left{ begin{align}
&e^{-frac{1}{z^4}} &hspace{1mm} mbox{if} hspace{1mm} z neq 0 \
&0 &hspace{1mm} mbox{if} hspace{1mm} z = 0 \
end{align} right. $$



I have to show that this function satisfies the Cauchy-Riemann equation at $z=0$, but is not holomorphic.



I have been able to prove that it satisfies Cauchy-Riemann. But I am struggling with proving that it is not holomorphic at $z=0$.



My attempt:



Let $z=x+iy$ where $x,yinmathbb{R}$. Then
$$frac{1}{z}=frac{x-iy}{x^2+y^2}.$$
So
$$frac{1}{z^4}=frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}+ifrac{4xy^3-4x^3y}{(x^2+y^2)^4}.$$



Then
$$e^{-frac{1}{z^4}}=e^{-frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}}left(cosleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)-i sinleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)right)$$



My idea is the following:




If we can show that $f$ is not continuous at $z=0$, then it will follow automatically that $f$ is not holomorphic.




To show that $f$ is not continuous, I am trying to show that $f$ has different limits as $(x,y)to (0,0)$ along different paths. I have tried the paths $y=x$, $y=-x$, $y=x^2$. But all these paths give the limit to be $0$, which is the value of $f$ at $z=0$.



(1) Is my idea wrong and the function is continuous at $z=0$?



(2) If $f$ is not continuous, which path can be used to prove it?






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$




Let $$ f(z) = left{ begin{align}
&e^{-frac{1}{z^4}} &hspace{1mm} mbox{if} hspace{1mm} z neq 0 \
&0 &hspace{1mm} mbox{if} hspace{1mm} z = 0 \
end{align} right. $$



I have to show that this function satisfies the Cauchy-Riemann equation at $z=0$, but is not holomorphic.



I have been able to prove that it satisfies Cauchy-Riemann. But I am struggling with proving that it is not holomorphic at $z=0$.



My attempt:



Let $z=x+iy$ where $x,yinmathbb{R}$. Then
$$frac{1}{z}=frac{x-iy}{x^2+y^2}.$$
So
$$frac{1}{z^4}=frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}+ifrac{4xy^3-4x^3y}{(x^2+y^2)^4}.$$



Then
$$e^{-frac{1}{z^4}}=e^{-frac{x^4+y^4-6x^2y^2}{(x^2+y^2)^4}}left(cosleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)-i sinleft(frac{4xy^3-4x^3y}{(x^2+y^2)^4} right)right)$$



My idea is the following:




If we can show that $f$ is not continuous at $z=0$, then it will follow automatically that $f$ is not holomorphic.




To show that $f$ is not continuous, I am trying to show that $f$ has different limits as $(x,y)to (0,0)$ along different paths. I have tried the paths $y=x$, $y=-x$, $y=x^2$. But all these paths give the limit to be $0$, which is the value of $f$ at $z=0$.



(1) Is my idea wrong and the function is continuous at $z=0$?



(2) If $f$ is not continuous, which path can be used to prove it?






complex-analysis holomorphic-functions




complex-analysis holomorphic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 7 '17 at 8:06









José Carlos Santos

163k22131234




163k22131234










asked Sep 7 '17 at 7:43









learning_mathlearning_math

1,4531417




1,4531417








  • 1




    $begingroup$
    If it were holomorphic it would have a power-series expansion around zero. What can you say about the coefficients?
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:47






  • 1




    $begingroup$
    @uniquesolution I suppose that the OP meant to say that it is not differentiable at the origin.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:49






  • 1




    $begingroup$
    @Jose Carlos Santos But it is differentiable at the origin. In fact, it is differentiable infinitely many times at the origin, but it is not holmorphic there. (holomorphic=analytic). I suppose the OP meant to say exactly what (s)he said.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:53








  • 1




    $begingroup$
    @uniquesolution No, it is not differentiable at the origin. It is not even continuous there.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:55








  • 1




    $begingroup$
    Yes you are right -- I assumed - mistakenly - that $z$ is real.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 8:02














  • 1




    $begingroup$
    If it were holomorphic it would have a power-series expansion around zero. What can you say about the coefficients?
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:47






  • 1




    $begingroup$
    @uniquesolution I suppose that the OP meant to say that it is not differentiable at the origin.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:49






  • 1




    $begingroup$
    @Jose Carlos Santos But it is differentiable at the origin. In fact, it is differentiable infinitely many times at the origin, but it is not holmorphic there. (holomorphic=analytic). I suppose the OP meant to say exactly what (s)he said.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 7:53








  • 1




    $begingroup$
    @uniquesolution No, it is not differentiable at the origin. It is not even continuous there.
    $endgroup$
    – José Carlos Santos
    Sep 7 '17 at 7:55








  • 1




    $begingroup$
    Yes you are right -- I assumed - mistakenly - that $z$ is real.
    $endgroup$
    – uniquesolution
    Sep 7 '17 at 8:02








1




1




$begingroup$
If it were holomorphic it would have a power-series expansion around zero. What can you say about the coefficients?
$endgroup$
– uniquesolution
Sep 7 '17 at 7:47




$begingroup$
If it were holomorphic it would have a power-series expansion around zero. What can you say about the coefficients?
$endgroup$
– uniquesolution
Sep 7 '17 at 7:47




1




1




$begingroup$
@uniquesolution I suppose that the OP meant to say that it is not differentiable at the origin.
$endgroup$
– José Carlos Santos
Sep 7 '17 at 7:49




$begingroup$
@uniquesolution I suppose that the OP meant to say that it is not differentiable at the origin.
$endgroup$
– José Carlos Santos
Sep 7 '17 at 7:49




1




1




$begingroup$
@Jose Carlos Santos But it is differentiable at the origin. In fact, it is differentiable infinitely many times at the origin, but it is not holmorphic there. (holomorphic=analytic). I suppose the OP meant to say exactly what (s)he said.
$endgroup$
– uniquesolution
Sep 7 '17 at 7:53






$begingroup$
@Jose Carlos Santos But it is differentiable at the origin. In fact, it is differentiable infinitely many times at the origin, but it is not holmorphic there. (holomorphic=analytic). I suppose the OP meant to say exactly what (s)he said.
$endgroup$
– uniquesolution
Sep 7 '17 at 7:53






1




1




$begingroup$
@uniquesolution No, it is not differentiable at the origin. It is not even continuous there.
$endgroup$
– José Carlos Santos
Sep 7 '17 at 7:55






$begingroup$
@uniquesolution No, it is not differentiable at the origin. It is not even continuous there.
$endgroup$
– José Carlos Santos
Sep 7 '17 at 7:55






1




1




$begingroup$
Yes you are right -- I assumed - mistakenly - that $z$ is real.
$endgroup$
– uniquesolution
Sep 7 '17 at 8:02




$begingroup$
Yes you are right -- I assumed - mistakenly - that $z$ is real.
$endgroup$
– uniquesolution
Sep 7 '17 at 8:02










1 Answer
1






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$begingroup$

The function $f$ is not even continuous at $0$. If $z=rholeft(cosleft(fracpi8right)+sinleft(fracpi8right)iright)$ with $rho>0$, then $z^4=rho^4i$ and therefore$$bigl|f(z)bigr|=leftlvertexpleft(-frac1{z^4}right)rightrvert=leftlvertexpleft(frac i{rho^4}right)rightrvert=1.$$So, $f(z)$ doesn't even get near $0$ when $z$ belongs to the ray with origin at $0$ and passing through $cosleft(fracpi8right)+sinleft(fracpi8right)i$.






share|cite|improve this answer











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    $begingroup$

    The function $f$ is not even continuous at $0$. If $z=rholeft(cosleft(fracpi8right)+sinleft(fracpi8right)iright)$ with $rho>0$, then $z^4=rho^4i$ and therefore$$bigl|f(z)bigr|=leftlvertexpleft(-frac1{z^4}right)rightrvert=leftlvertexpleft(frac i{rho^4}right)rightrvert=1.$$So, $f(z)$ doesn't even get near $0$ when $z$ belongs to the ray with origin at $0$ and passing through $cosleft(fracpi8right)+sinleft(fracpi8right)i$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      The function $f$ is not even continuous at $0$. If $z=rholeft(cosleft(fracpi8right)+sinleft(fracpi8right)iright)$ with $rho>0$, then $z^4=rho^4i$ and therefore$$bigl|f(z)bigr|=leftlvertexpleft(-frac1{z^4}right)rightrvert=leftlvertexpleft(frac i{rho^4}right)rightrvert=1.$$So, $f(z)$ doesn't even get near $0$ when $z$ belongs to the ray with origin at $0$ and passing through $cosleft(fracpi8right)+sinleft(fracpi8right)i$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        The function $f$ is not even continuous at $0$. If $z=rholeft(cosleft(fracpi8right)+sinleft(fracpi8right)iright)$ with $rho>0$, then $z^4=rho^4i$ and therefore$$bigl|f(z)bigr|=leftlvertexpleft(-frac1{z^4}right)rightrvert=leftlvertexpleft(frac i{rho^4}right)rightrvert=1.$$So, $f(z)$ doesn't even get near $0$ when $z$ belongs to the ray with origin at $0$ and passing through $cosleft(fracpi8right)+sinleft(fracpi8right)i$.






        share|cite|improve this answer











        $endgroup$



        The function $f$ is not even continuous at $0$. If $z=rholeft(cosleft(fracpi8right)+sinleft(fracpi8right)iright)$ with $rho>0$, then $z^4=rho^4i$ and therefore$$bigl|f(z)bigr|=leftlvertexpleft(-frac1{z^4}right)rightrvert=leftlvertexpleft(frac i{rho^4}right)rightrvert=1.$$So, $f(z)$ doesn't even get near $0$ when $z$ belongs to the ray with origin at $0$ and passing through $cosleft(fracpi8right)+sinleft(fracpi8right)i$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 19:10

























        answered Sep 7 '17 at 7:55









        José Carlos SantosJosé Carlos Santos

        163k22131234




        163k22131234






























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