How is it possibile for a complex analytic function not having a derivative in the region in which it is...
$begingroup$
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
edited Jan 8 at 9:47
Olof Rubin
878317
asked Jan 8 at 9:31
user632723user632723
1
$endgroup$
add a comment |
$begingroup$
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
edited Jan 8 at 9:47
Olof Rubin
878317
asked Jan 8 at 9:31
user632723user632723
1
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 8 at 9:47
Olof Rubin
878317
asked Jan 8 at 9:31
user632723user632723
1
$endgroup$
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
edited Jan 8 at 9:47
Olof Rubin
878317
asked Jan 8 at 9:31
user632723user632723
1
edited Jan 8 at 9:47
Olof Rubin
878317
asked Jan 8 at 9:31
user632723user632723
1
edited Jan 8 at 9:47
Olof Rubin
878317
edited Jan 8 at 9:47
Olof Rubin
878317
edited Jan 8 at 9:47
Olof Rubin
878317
878317
asked Jan 8 at 9:31
user632723user632723
1
asked Jan 8 at 9:31
user632723user632723
1
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 8 at 9:42
Olof RubinOlof Rubin
878317
$endgroup$
$begingroup$
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
|
show 1 more comment
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1 Answer
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$begingroup$
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.
answered Jan 8 at 9:42
Olof RubinOlof Rubin
878317
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Jan 8 at 9:42
Olof RubinOlof Rubin
878317
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Jan 8 at 9:42
Olof RubinOlof Rubin
878317
$endgroup$
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.
answered Jan 8 at 9:42
Olof RubinOlof Rubin
878317
edited Jan 8 at 9:45
answered Jan 8 at 9:42
Olof RubinOlof Rubin
878317
answered Jan 8 at 9:42
Olof RubinOlof Rubin
878317
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
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