How is it possibile for a complex analytic function not having a derivative in the region in which it is...












0












$begingroup$


There is a question asking if the function $f$, which is defined by $$f(z) = frac{bar{z}^2}{z},quad zneq 0$$ and $f(0) = 0$, is analytic at $z = 0$ and if there is $f'(0)$.



The answer is supposed to be it is analytic since CR equations hold but the limit depends on a variable so there is no derivative at $0$. We have also been given theorems implying that analytic property is equivalent to having a derivative which is also equivalent to CR equations which is not true I think. In an exam we are asked to find the set in which the function is not analytic and in the answer sheet, only the points that the function was not continuous have been shown. So, I am very confused about it. What are the conditions for a function to be analytic? Are there any tests to find out if a function is analytic?






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is the source of all this? It is full of errors. Your function is surely not analytic at $0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 9:39










  • $begingroup$
    It's based on my notes I take in complex functions theory 1 class 3rd year undergraduate. We were given a definition about series expansion but we apply the above theorems to the problems.
    $endgroup$
    – user632723
    Jan 8 at 9:40


















0












$begingroup$


There is a question asking if the function $f$, which is defined by $$f(z) = frac{bar{z}^2}{z},quad zneq 0$$ and $f(0) = 0$, is analytic at $z = 0$ and if there is $f'(0)$.



The answer is supposed to be it is analytic since CR equations hold but the limit depends on a variable so there is no derivative at $0$. We have also been given theorems implying that analytic property is equivalent to having a derivative which is also equivalent to CR equations which is not true I think. In an exam we are asked to find the set in which the function is not analytic and in the answer sheet, only the points that the function was not continuous have been shown. So, I am very confused about it. What are the conditions for a function to be analytic? Are there any tests to find out if a function is analytic?






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is the source of all this? It is full of errors. Your function is surely not analytic at $0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 9:39










  • $begingroup$
    It's based on my notes I take in complex functions theory 1 class 3rd year undergraduate. We were given a definition about series expansion but we apply the above theorems to the problems.
    $endgroup$
    – user632723
    Jan 8 at 9:40
















0












0








0





$begingroup$


There is a question asking if the function $f$, which is defined by $$f(z) = frac{bar{z}^2}{z},quad zneq 0$$ and $f(0) = 0$, is analytic at $z = 0$ and if there is $f'(0)$.



The answer is supposed to be it is analytic since CR equations hold but the limit depends on a variable so there is no derivative at $0$. We have also been given theorems implying that analytic property is equivalent to having a derivative which is also equivalent to CR equations which is not true I think. In an exam we are asked to find the set in which the function is not analytic and in the answer sheet, only the points that the function was not continuous have been shown. So, I am very confused about it. What are the conditions for a function to be analytic? Are there any tests to find out if a function is analytic?






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$




There is a question asking if the function $f$, which is defined by $$f(z) = frac{bar{z}^2}{z},quad zneq 0$$ and $f(0) = 0$, is analytic at $z = 0$ and if there is $f'(0)$.



The answer is supposed to be it is analytic since CR equations hold but the limit depends on a variable so there is no derivative at $0$. We have also been given theorems implying that analytic property is equivalent to having a derivative which is also equivalent to CR equations which is not true I think. In an exam we are asked to find the set in which the function is not analytic and in the answer sheet, only the points that the function was not continuous have been shown. So, I am very confused about it. What are the conditions for a function to be analytic? Are there any tests to find out if a function is analytic?






complex-analysis analytic-functions




complex-analysis analytic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 9:47









Olof Rubin

878317




878317










asked Jan 8 at 9:31









user632723user632723

1




1








  • 1




    $begingroup$
    What is the source of all this? It is full of errors. Your function is surely not analytic at $0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 9:39










  • $begingroup$
    It's based on my notes I take in complex functions theory 1 class 3rd year undergraduate. We were given a definition about series expansion but we apply the above theorems to the problems.
    $endgroup$
    – user632723
    Jan 8 at 9:40
















  • 1




    $begingroup$
    What is the source of all this? It is full of errors. Your function is surely not analytic at $0$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 9:39










  • $begingroup$
    It's based on my notes I take in complex functions theory 1 class 3rd year undergraduate. We were given a definition about series expansion but we apply the above theorems to the problems.
    $endgroup$
    – user632723
    Jan 8 at 9:40










1




1




$begingroup$
What is the source of all this? It is full of errors. Your function is surely not analytic at $0$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 9:39




$begingroup$
What is the source of all this? It is full of errors. Your function is surely not analytic at $0$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 9:39












$begingroup$
It's based on my notes I take in complex functions theory 1 class 3rd year undergraduate. We were given a definition about series expansion but we apply the above theorems to the problems.
$endgroup$
– user632723
Jan 8 at 9:40






$begingroup$
It's based on my notes I take in complex functions theory 1 class 3rd year undergraduate. We were given a definition about series expansion but we apply the above theorems to the problems.
$endgroup$
– user632723
Jan 8 at 9:40












1 Answer
1






active

oldest

votes


















0












$begingroup$

A function is analytic on an open set $D$ if its complex derivative exists in some neighbourhood of every point. That is if for every $zin D$ there exists some $r>0$ such that if $|zeta-z|<r$ then



$$lim_{hrightarrow 0}frac{f(zeta+h)-f(zeta)}{h}$$



exists. This is equivalent to satisfying the Cauchy-Riemann equations at $zeta = x+iy$: that is if $f(x,y) = u(x,y)+iv(x,y)$ then $u_x = v_y$ and $u_y = -v_x$. Your function is not differentiable in the complex sense on any open set. Since if it were then this would imply that $zf(z) = bar{z}^2$ was analytic which is simply not the case. For instance one can check Cauchy-Riemanns equation on the funciton



$$(x-iy)^2 = underbrace{x^2-y^2}_{u(x,y)}+i(underbrace{-2xy}_{v(x,y)})$$



and $u_x = 2x$ while $v_y = -2x$ and so they are not equal.



If $f$ was differentiable in the complex sense at $0$ then we would have that
$$lim_{hrightarrow 0}frac{f(h)-f(0)}{h} = lim_{hrightarrow 0}frac{bar{h}^2}{h^2}$$



exists.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, analytic property is equivalent to having a derivative which is also equivalent to CR is true?
    $endgroup$
    – user632723
    Jan 8 at 9:45










  • $begingroup$
    The analytic property necessitates that the function has complex derivative on an open connected set. This is equivalent to CR being satisfied on an open connected set. At one point is not enough
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:46










  • $begingroup$
    Yes there is no derivative there but CR holds true.
    $endgroup$
    – user632723
    Jan 8 at 9:46










  • $begingroup$
    CR holds at a point if and only if the function is complex differentiable at that point. Are you sure that $f$ satisfies $CR$ at $z = 0$? Could you post your computations
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:53










  • $begingroup$
    i65.tinypic.com/208wcox.jpg
    $endgroup$
    – user632723
    Jan 8 at 10:05














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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

A function is analytic on an open set $D$ if its complex derivative exists in some neighbourhood of every point. That is if for every $zin D$ there exists some $r>0$ such that if $|zeta-z|<r$ then



$$lim_{hrightarrow 0}frac{f(zeta+h)-f(zeta)}{h}$$



exists. This is equivalent to satisfying the Cauchy-Riemann equations at $zeta = x+iy$: that is if $f(x,y) = u(x,y)+iv(x,y)$ then $u_x = v_y$ and $u_y = -v_x$. Your function is not differentiable in the complex sense on any open set. Since if it were then this would imply that $zf(z) = bar{z}^2$ was analytic which is simply not the case. For instance one can check Cauchy-Riemanns equation on the funciton



$$(x-iy)^2 = underbrace{x^2-y^2}_{u(x,y)}+i(underbrace{-2xy}_{v(x,y)})$$



and $u_x = 2x$ while $v_y = -2x$ and so they are not equal.



If $f$ was differentiable in the complex sense at $0$ then we would have that
$$lim_{hrightarrow 0}frac{f(h)-f(0)}{h} = lim_{hrightarrow 0}frac{bar{h}^2}{h^2}$$



exists.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, analytic property is equivalent to having a derivative which is also equivalent to CR is true?
    $endgroup$
    – user632723
    Jan 8 at 9:45










  • $begingroup$
    The analytic property necessitates that the function has complex derivative on an open connected set. This is equivalent to CR being satisfied on an open connected set. At one point is not enough
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:46










  • $begingroup$
    Yes there is no derivative there but CR holds true.
    $endgroup$
    – user632723
    Jan 8 at 9:46










  • $begingroup$
    CR holds at a point if and only if the function is complex differentiable at that point. Are you sure that $f$ satisfies $CR$ at $z = 0$? Could you post your computations
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:53










  • $begingroup$
    i65.tinypic.com/208wcox.jpg
    $endgroup$
    – user632723
    Jan 8 at 10:05


















0












$begingroup$

A function is analytic on an open set $D$ if its complex derivative exists in some neighbourhood of every point. That is if for every $zin D$ there exists some $r>0$ such that if $|zeta-z|<r$ then



$$lim_{hrightarrow 0}frac{f(zeta+h)-f(zeta)}{h}$$



exists. This is equivalent to satisfying the Cauchy-Riemann equations at $zeta = x+iy$: that is if $f(x,y) = u(x,y)+iv(x,y)$ then $u_x = v_y$ and $u_y = -v_x$. Your function is not differentiable in the complex sense on any open set. Since if it were then this would imply that $zf(z) = bar{z}^2$ was analytic which is simply not the case. For instance one can check Cauchy-Riemanns equation on the funciton



$$(x-iy)^2 = underbrace{x^2-y^2}_{u(x,y)}+i(underbrace{-2xy}_{v(x,y)})$$



and $u_x = 2x$ while $v_y = -2x$ and so they are not equal.



If $f$ was differentiable in the complex sense at $0$ then we would have that
$$lim_{hrightarrow 0}frac{f(h)-f(0)}{h} = lim_{hrightarrow 0}frac{bar{h}^2}{h^2}$$



exists.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, analytic property is equivalent to having a derivative which is also equivalent to CR is true?
    $endgroup$
    – user632723
    Jan 8 at 9:45










  • $begingroup$
    The analytic property necessitates that the function has complex derivative on an open connected set. This is equivalent to CR being satisfied on an open connected set. At one point is not enough
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:46










  • $begingroup$
    Yes there is no derivative there but CR holds true.
    $endgroup$
    – user632723
    Jan 8 at 9:46










  • $begingroup$
    CR holds at a point if and only if the function is complex differentiable at that point. Are you sure that $f$ satisfies $CR$ at $z = 0$? Could you post your computations
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:53










  • $begingroup$
    i65.tinypic.com/208wcox.jpg
    $endgroup$
    – user632723
    Jan 8 at 10:05
















0












0








0





$begingroup$

A function is analytic on an open set $D$ if its complex derivative exists in some neighbourhood of every point. That is if for every $zin D$ there exists some $r>0$ such that if $|zeta-z|<r$ then



$$lim_{hrightarrow 0}frac{f(zeta+h)-f(zeta)}{h}$$



exists. This is equivalent to satisfying the Cauchy-Riemann equations at $zeta = x+iy$: that is if $f(x,y) = u(x,y)+iv(x,y)$ then $u_x = v_y$ and $u_y = -v_x$. Your function is not differentiable in the complex sense on any open set. Since if it were then this would imply that $zf(z) = bar{z}^2$ was analytic which is simply not the case. For instance one can check Cauchy-Riemanns equation on the funciton



$$(x-iy)^2 = underbrace{x^2-y^2}_{u(x,y)}+i(underbrace{-2xy}_{v(x,y)})$$



and $u_x = 2x$ while $v_y = -2x$ and so they are not equal.



If $f$ was differentiable in the complex sense at $0$ then we would have that
$$lim_{hrightarrow 0}frac{f(h)-f(0)}{h} = lim_{hrightarrow 0}frac{bar{h}^2}{h^2}$$



exists.






share|cite|improve this answer











$endgroup$



A function is analytic on an open set $D$ if its complex derivative exists in some neighbourhood of every point. That is if for every $zin D$ there exists some $r>0$ such that if $|zeta-z|<r$ then



$$lim_{hrightarrow 0}frac{f(zeta+h)-f(zeta)}{h}$$



exists. This is equivalent to satisfying the Cauchy-Riemann equations at $zeta = x+iy$: that is if $f(x,y) = u(x,y)+iv(x,y)$ then $u_x = v_y$ and $u_y = -v_x$. Your function is not differentiable in the complex sense on any open set. Since if it were then this would imply that $zf(z) = bar{z}^2$ was analytic which is simply not the case. For instance one can check Cauchy-Riemanns equation on the funciton



$$(x-iy)^2 = underbrace{x^2-y^2}_{u(x,y)}+i(underbrace{-2xy}_{v(x,y)})$$



and $u_x = 2x$ while $v_y = -2x$ and so they are not equal.



If $f$ was differentiable in the complex sense at $0$ then we would have that
$$lim_{hrightarrow 0}frac{f(h)-f(0)}{h} = lim_{hrightarrow 0}frac{bar{h}^2}{h^2}$$



exists.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 9:45

























answered Jan 8 at 9:42









Olof RubinOlof Rubin

878317




878317












  • $begingroup$
    So, analytic property is equivalent to having a derivative which is also equivalent to CR is true?
    $endgroup$
    – user632723
    Jan 8 at 9:45










  • $begingroup$
    The analytic property necessitates that the function has complex derivative on an open connected set. This is equivalent to CR being satisfied on an open connected set. At one point is not enough
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:46










  • $begingroup$
    Yes there is no derivative there but CR holds true.
    $endgroup$
    – user632723
    Jan 8 at 9:46










  • $begingroup$
    CR holds at a point if and only if the function is complex differentiable at that point. Are you sure that $f$ satisfies $CR$ at $z = 0$? Could you post your computations
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:53










  • $begingroup$
    i65.tinypic.com/208wcox.jpg
    $endgroup$
    – user632723
    Jan 8 at 10:05




















  • $begingroup$
    So, analytic property is equivalent to having a derivative which is also equivalent to CR is true?
    $endgroup$
    – user632723
    Jan 8 at 9:45










  • $begingroup$
    The analytic property necessitates that the function has complex derivative on an open connected set. This is equivalent to CR being satisfied on an open connected set. At one point is not enough
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:46










  • $begingroup$
    Yes there is no derivative there but CR holds true.
    $endgroup$
    – user632723
    Jan 8 at 9:46










  • $begingroup$
    CR holds at a point if and only if the function is complex differentiable at that point. Are you sure that $f$ satisfies $CR$ at $z = 0$? Could you post your computations
    $endgroup$
    – Olof Rubin
    Jan 8 at 9:53










  • $begingroup$
    i65.tinypic.com/208wcox.jpg
    $endgroup$
    – user632723
    Jan 8 at 10:05


















$begingroup$
So, analytic property is equivalent to having a derivative which is also equivalent to CR is true?
$endgroup$
– user632723
Jan 8 at 9:45




$begingroup$
So, analytic property is equivalent to having a derivative which is also equivalent to CR is true?
$endgroup$
– user632723
Jan 8 at 9:45












$begingroup$
The analytic property necessitates that the function has complex derivative on an open connected set. This is equivalent to CR being satisfied on an open connected set. At one point is not enough
$endgroup$
– Olof Rubin
Jan 8 at 9:46




$begingroup$
The analytic property necessitates that the function has complex derivative on an open connected set. This is equivalent to CR being satisfied on an open connected set. At one point is not enough
$endgroup$
– Olof Rubin
Jan 8 at 9:46












$begingroup$
Yes there is no derivative there but CR holds true.
$endgroup$
– user632723
Jan 8 at 9:46




$begingroup$
Yes there is no derivative there but CR holds true.
$endgroup$
– user632723
Jan 8 at 9:46












$begingroup$
CR holds at a point if and only if the function is complex differentiable at that point. Are you sure that $f$ satisfies $CR$ at $z = 0$? Could you post your computations
$endgroup$
– Olof Rubin
Jan 8 at 9:53




$begingroup$
CR holds at a point if and only if the function is complex differentiable at that point. Are you sure that $f$ satisfies $CR$ at $z = 0$? Could you post your computations
$endgroup$
– Olof Rubin
Jan 8 at 9:53












$begingroup$
i65.tinypic.com/208wcox.jpg
$endgroup$
– user632723
Jan 8 at 10:05






$begingroup$
i65.tinypic.com/208wcox.jpg
$endgroup$
– user632723
Jan 8 at 10:05




















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