Does this question about the sufficiency of the analyticity of f(z) make sense [duplicate]
$begingroup$
This question already has an answer here:
When is a function satisfying the Cauchy-Riemann equations holomorphic?
3 answers
Before I ask the question, I know for sure this following question makes sense:
Prove that a necessary condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
This question makes sense because if the Cauchy-Riemann equations are satisfied, $f(z)$ need not be analytic. Hence, why it's called a necessary condition.
But in order for $ f(z) $ to be analytic , there are 2 sufficient conditions, the Cauchy-Riemann equations must be satisfied and the partial derivatives must be continuous.
Now for the main question I want to ask:
Prove that a sufficient condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
I'm asking this because I had it on my exam yesterday, now how could this make sense, the Cauchy-Riemann equations being satisfied should not be enough, the question did not mention the continuity of the partial derivatives, thus I treated this question as if it was asking "Prove that a sufficient condition for the Cauchy-Riemann equations to be satisfied is that $f(z)$ must be analytic".
So am I right and should I argue with my professor about that?
My question is not about when a function is holomorphic, my question is if the following makes sense as a proof question or not:
Prove that a sufficient condition for f(z) to be analytic is that the Cauchy-Riemann equations are satisfied.
complex-analysis soft-question cauchy-riemann-equation
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
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marked as duplicate by SmileyCraft, KReiser, Lord Shark the Unknown, Leucippus, José Carlos Santos complex-analysis
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Jan 7 at 8:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
When is a function satisfying the Cauchy-Riemann equations holomorphic?
3 answers
Before I ask the question, I know for sure this following question makes sense:
Prove that a necessary condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
This question makes sense because if the Cauchy-Riemann equations are satisfied, $f(z)$ need not be analytic. Hence, why it's called a necessary condition.
But in order for $ f(z) $ to be analytic , there are 2 sufficient conditions, the Cauchy-Riemann equations must be satisfied and the partial derivatives must be continuous.
Now for the main question I want to ask:
Prove that a sufficient condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
I'm asking this because I had it on my exam yesterday, now how could this make sense, the Cauchy-Riemann equations being satisfied should not be enough, the question did not mention the continuity of the partial derivatives, thus I treated this question as if it was asking "Prove that a sufficient condition for the Cauchy-Riemann equations to be satisfied is that $f(z)$ must be analytic".
So am I right and should I argue with my professor about that?
My question is not about when a function is holomorphic, my question is if the following makes sense as a proof question or not:
Prove that a sufficient condition for f(z) to be analytic is that the Cauchy-Riemann equations are satisfied.
complex-analysis soft-question cauchy-riemann-equation
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
$endgroup$
marked as duplicate by SmileyCraft, KReiser, Lord Shark the Unknown, Leucippus, José Carlos Santos complex-analysis
Users with the complex-analysis badge can single-handedly close complex-analysis questions as duplicates and reopen them as needed.
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Jan 7 at 8:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
That's not my question.
$endgroup$
– khaled014z
Jan 6 at 23:18
$begingroup$
They discuss why the Cauchy Riemann equations are sufficient for a continuous function to be analytic. Hence, the question makes sense.
$endgroup$
– SmileyCraft
Jan 6 at 23:23
$begingroup$
Provided that the function and the partial derivatives are continuous, which are not mentioned in the question?
$endgroup$
– khaled014z
Jan 6 at 23:27
1
$begingroup$
I may have misunderstood your question then. Does this answer your question? "The function given by $f(z)=e^{-z^{-4}}$ for $zneq0$, $f(0)=0$ satisfies the Cauchy–Riemann equations everywhere but is not analytic (or even continuous) at $z=0$." en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem
$endgroup$
– SmileyCraft
Jan 6 at 23:31
$begingroup$
Yes that is precisely what I meant, the C.R condition is satisfied, the other condition for analyticity is not mentioned in the question, that's the problem.
$endgroup$
– khaled014z
Jan 7 at 0:04
add a comment |
$begingroup$
This question already has an answer here:
When is a function satisfying the Cauchy-Riemann equations holomorphic?
3 answers
Before I ask the question, I know for sure this following question makes sense:
Prove that a necessary condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
This question makes sense because if the Cauchy-Riemann equations are satisfied, $f(z)$ need not be analytic. Hence, why it's called a necessary condition.
But in order for $ f(z) $ to be analytic , there are 2 sufficient conditions, the Cauchy-Riemann equations must be satisfied and the partial derivatives must be continuous.
Now for the main question I want to ask:
Prove that a sufficient condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
I'm asking this because I had it on my exam yesterday, now how could this make sense, the Cauchy-Riemann equations being satisfied should not be enough, the question did not mention the continuity of the partial derivatives, thus I treated this question as if it was asking "Prove that a sufficient condition for the Cauchy-Riemann equations to be satisfied is that $f(z)$ must be analytic".
So am I right and should I argue with my professor about that?
My question is not about when a function is holomorphic, my question is if the following makes sense as a proof question or not:
Prove that a sufficient condition for f(z) to be analytic is that the Cauchy-Riemann equations are satisfied.
complex-analysis soft-question cauchy-riemann-equation
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
$endgroup$
This question already has an answer here:
When is a function satisfying the Cauchy-Riemann equations holomorphic?
3 answers
Before I ask the question, I know for sure this following question makes sense:
Prove that a necessary condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
This question makes sense because if the Cauchy-Riemann equations are satisfied, $f(z)$ need not be analytic. Hence, why it's called a necessary condition.
But in order for $ f(z) $ to be analytic , there are 2 sufficient conditions, the Cauchy-Riemann equations must be satisfied and the partial derivatives must be continuous.
Now for the main question I want to ask:
Prove that a sufficient condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
I'm asking this because I had it on my exam yesterday, now how could this make sense, the Cauchy-Riemann equations being satisfied should not be enough, the question did not mention the continuity of the partial derivatives, thus I treated this question as if it was asking "Prove that a sufficient condition for the Cauchy-Riemann equations to be satisfied is that $f(z)$ must be analytic".
So am I right and should I argue with my professor about that?
My question is not about when a function is holomorphic, my question is if the following makes sense as a proof question or not:
Prove that a sufficient condition for f(z) to be analytic is that the Cauchy-Riemann equations are satisfied.
This question already has an answer here:
When is a function satisfying the Cauchy-Riemann equations holomorphic?
3 answers
complex-analysis soft-question cauchy-riemann-equation
complex-analysis soft-question cauchy-riemann-equation
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
edited Jan 6 at 23:18
khaled014z
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
asked Jan 6 at 21:49
khaled014zkhaled014z
1679
1679
marked as duplicate by SmileyCraft, KReiser, Lord Shark the Unknown, Leucippus, José Carlos Santos complex-analysis
Users with the complex-analysis badge can single-handedly close complex-analysis questions as duplicates and reopen them as needed.
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Jan 7 at 8:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by SmileyCraft, KReiser, Lord Shark the Unknown, Leucippus, José Carlos Santos complex-analysis
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Jan 7 at 8:40
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
That's not my question.
$endgroup$
– khaled014z
Jan 6 at 23:18
$begingroup$
They discuss why the Cauchy Riemann equations are sufficient for a continuous function to be analytic. Hence, the question makes sense.
$endgroup$
– SmileyCraft
Jan 6 at 23:23
$begingroup$
Provided that the function and the partial derivatives are continuous, which are not mentioned in the question?
$endgroup$
– khaled014z
Jan 6 at 23:27
1
$begingroup$
I may have misunderstood your question then. Does this answer your question? "The function given by $f(z)=e^{-z^{-4}}$ for $zneq0$, $f(0)=0$ satisfies the Cauchy–Riemann equations everywhere but is not analytic (or even continuous) at $z=0$." en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem
$endgroup$
– SmileyCraft
Jan 6 at 23:31
$begingroup$
Yes that is precisely what I meant, the C.R condition is satisfied, the other condition for analyticity is not mentioned in the question, that's the problem.
$endgroup$
– khaled014z
Jan 7 at 0:04
add a comment |
$begingroup$
That's not my question.
$endgroup$
– khaled014z
Jan 6 at 23:18
$begingroup$
They discuss why the Cauchy Riemann equations are sufficient for a continuous function to be analytic. Hence, the question makes sense.
$endgroup$
– SmileyCraft
Jan 6 at 23:23
$begingroup$
Provided that the function and the partial derivatives are continuous, which are not mentioned in the question?
$endgroup$
– khaled014z
Jan 6 at 23:27
1
$begingroup$
I may have misunderstood your question then. Does this answer your question? "The function given by $f(z)=e^{-z^{-4}}$ for $zneq0$, $f(0)=0$ satisfies the Cauchy–Riemann equations everywhere but is not analytic (or even continuous) at $z=0$." en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem
$endgroup$
– SmileyCraft
Jan 6 at 23:31
$begingroup$
Yes that is precisely what I meant, the C.R condition is satisfied, the other condition for analyticity is not mentioned in the question, that's the problem.
$endgroup$
– khaled014z
Jan 7 at 0:04
$begingroup$
That's not my question.
$endgroup$
– khaled014z
Jan 6 at 23:18
$begingroup$
That's not my question.
$endgroup$
– khaled014z
Jan 6 at 23:18
$begingroup$
They discuss why the Cauchy Riemann equations are sufficient for a continuous function to be analytic. Hence, the question makes sense.
$endgroup$
– SmileyCraft
Jan 6 at 23:23
$begingroup$
They discuss why the Cauchy Riemann equations are sufficient for a continuous function to be analytic. Hence, the question makes sense.
$endgroup$
– SmileyCraft
Jan 6 at 23:23
$begingroup$
Provided that the function and the partial derivatives are continuous, which are not mentioned in the question?
$endgroup$
– khaled014z
Jan 6 at 23:27
$begingroup$
Provided that the function and the partial derivatives are continuous, which are not mentioned in the question?
$endgroup$
– khaled014z
Jan 6 at 23:27
1
1
$begingroup$
I may have misunderstood your question then. Does this answer your question? "The function given by $f(z)=e^{-z^{-4}}$ for $zneq0$, $f(0)=0$ satisfies the Cauchy–Riemann equations everywhere but is not analytic (or even continuous) at $z=0$." en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem
$endgroup$
– SmileyCraft
Jan 6 at 23:31
$begingroup$
I may have misunderstood your question then. Does this answer your question? "The function given by $f(z)=e^{-z^{-4}}$ for $zneq0$, $f(0)=0$ satisfies the Cauchy–Riemann equations everywhere but is not analytic (or even continuous) at $z=0$." en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem
$endgroup$
– SmileyCraft
Jan 6 at 23:31
$begingroup$
Yes that is precisely what I meant, the C.R condition is satisfied, the other condition for analyticity is not mentioned in the question, that's the problem.
$endgroup$
– khaled014z
Jan 7 at 0:04
$begingroup$
Yes that is precisely what I meant, the C.R condition is satisfied, the other condition for analyticity is not mentioned in the question, that's the problem.
$endgroup$
– khaled014z
Jan 7 at 0:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a diffrence between differentiability at a point and differentiability/analyticity in a neighborhood of a point. For example $f(x+iy)=sqrt {|xy|}$ satisfies C-R equations at $0$ but $f$ is not differentiable at $0$. However, in an open set analyticity is equivalent to validity of C-R equations. Continuity of partial derivatives need not be assumed. In fact C-R equations imply that $f$ has continuous partial derivatives of all orders.
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
$begingroup$
So that means that the question should still mention that the function is differentiable everywhere, correct? So that the cauchy riemann equations when satisfied, imply the analyticity of $ f(z) $ 100 % (even at zero)
$endgroup$
– khaled014z
Jan 6 at 23:49
$begingroup$
@khaled014z You are right. For differentiability in an open set the functions must have first partial derivatives at all points and these derivatives must satisfy C-R equations.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 23:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a diffrence between differentiability at a point and differentiability/analyticity in a neighborhood of a point. For example $f(x+iy)=sqrt {|xy|}$ satisfies C-R equations at $0$ but $f$ is not differentiable at $0$. However, in an open set analyticity is equivalent to validity of C-R equations. Continuity of partial derivatives need not be assumed. In fact C-R equations imply that $f$ has continuous partial derivatives of all orders.
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
$begingroup$
So that means that the question should still mention that the function is differentiable everywhere, correct? So that the cauchy riemann equations when satisfied, imply the analyticity of $ f(z) $ 100 % (even at zero)
$endgroup$
– khaled014z
Jan 6 at 23:49
$begingroup$
@khaled014z You are right. For differentiability in an open set the functions must have first partial derivatives at all points and these derivatives must satisfy C-R equations.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 23:56
add a comment |
$begingroup$
There is a diffrence between differentiability at a point and differentiability/analyticity in a neighborhood of a point. For example $f(x+iy)=sqrt {|xy|}$ satisfies C-R equations at $0$ but $f$ is not differentiable at $0$. However, in an open set analyticity is equivalent to validity of C-R equations. Continuity of partial derivatives need not be assumed. In fact C-R equations imply that $f$ has continuous partial derivatives of all orders.
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
$begingroup$
So that means that the question should still mention that the function is differentiable everywhere, correct? So that the cauchy riemann equations when satisfied, imply the analyticity of $ f(z) $ 100 % (even at zero)
$endgroup$
– khaled014z
Jan 6 at 23:49
$begingroup$
@khaled014z You are right. For differentiability in an open set the functions must have first partial derivatives at all points and these derivatives must satisfy C-R equations.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 23:56
add a comment |
$begingroup$
There is a diffrence between differentiability at a point and differentiability/analyticity in a neighborhood of a point. For example $f(x+iy)=sqrt {|xy|}$ satisfies C-R equations at $0$ but $f$ is not differentiable at $0$. However, in an open set analyticity is equivalent to validity of C-R equations. Continuity of partial derivatives need not be assumed. In fact C-R equations imply that $f$ has continuous partial derivatives of all orders.
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
$endgroup$
There is a diffrence between differentiability at a point and differentiability/analyticity in a neighborhood of a point. For example $f(x+iy)=sqrt {|xy|}$ satisfies C-R equations at $0$ but $f$ is not differentiable at $0$. However, in an open set analyticity is equivalent to validity of C-R equations. Continuity of partial derivatives need not be assumed. In fact C-R equations imply that $f$ has continuous partial derivatives of all orders.
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
answered Jan 6 at 23:29
Kavi Rama MurthyKavi Rama Murthy
59.8k42161
59.8k42161
$begingroup$
So that means that the question should still mention that the function is differentiable everywhere, correct? So that the cauchy riemann equations when satisfied, imply the analyticity of $ f(z) $ 100 % (even at zero)
$endgroup$
– khaled014z
Jan 6 at 23:49
$begingroup$
@khaled014z You are right. For differentiability in an open set the functions must have first partial derivatives at all points and these derivatives must satisfy C-R equations.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 23:56
add a comment |
$begingroup$
So that means that the question should still mention that the function is differentiable everywhere, correct? So that the cauchy riemann equations when satisfied, imply the analyticity of $ f(z) $ 100 % (even at zero)
$endgroup$
– khaled014z
Jan 6 at 23:49
$begingroup$
@khaled014z You are right. For differentiability in an open set the functions must have first partial derivatives at all points and these derivatives must satisfy C-R equations.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 23:56
$begingroup$
So that means that the question should still mention that the function is differentiable everywhere, correct? So that the cauchy riemann equations when satisfied, imply the analyticity of $ f(z) $ 100 % (even at zero)
$endgroup$
– khaled014z
Jan 6 at 23:49
$begingroup$
So that means that the question should still mention that the function is differentiable everywhere, correct? So that the cauchy riemann equations when satisfied, imply the analyticity of $ f(z) $ 100 % (even at zero)
$endgroup$
– khaled014z
Jan 6 at 23:49
$begingroup$
@khaled014z You are right. For differentiability in an open set the functions must have first partial derivatives at all points and these derivatives must satisfy C-R equations.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 23:56
$begingroup$
@khaled014z You are right. For differentiability in an open set the functions must have first partial derivatives at all points and these derivatives must satisfy C-R equations.
$endgroup$
– Kavi Rama Murthy
Jan 6 at 23:56
add a comment |
$begingroup$
That's not my question.
$endgroup$
– khaled014z
Jan 6 at 23:18
$begingroup$
They discuss why the Cauchy Riemann equations are sufficient for a continuous function to be analytic. Hence, the question makes sense.
$endgroup$
– SmileyCraft
Jan 6 at 23:23
$begingroup$
Provided that the function and the partial derivatives are continuous, which are not mentioned in the question?
$endgroup$
– khaled014z
Jan 6 at 23:27
1
$begingroup$
I may have misunderstood your question then. Does this answer your question? "The function given by $f(z)=e^{-z^{-4}}$ for $zneq0$, $f(0)=0$ satisfies the Cauchy–Riemann equations everywhere but is not analytic (or even continuous) at $z=0$." en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem
$endgroup$
– SmileyCraft
Jan 6 at 23:31
$begingroup$
Yes that is precisely what I meant, the C.R condition is satisfied, the other condition for analyticity is not mentioned in the question, that's the problem.
$endgroup$
– khaled014z
Jan 7 at 0:04