Analytic function $f(z)$ has zero of order $m$ at $z_0$, order of zero of $f'(z_0)$ and $f^2(z_0)$Open...
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Analytic function $f(z)$ has zero of order $m$ at $z_0$, order of zero of $f'(z_0)$ and $f^2(z_0)$
Open mapping principle complex?Proving function has simple pole and residueProve $f$ analytic on $D(z_0;R)setminus{z_0}$ implies $exists M, f(D(z_0;r)setminus{z_0})supset{zinmathbb{C}:|z|>M}$An analytic function on the open unit disk with a zero of order kProperty of a holomorphic function with a zero of order $n$ at $z_0$Prove that $f$ has a pole of order $m$ at $z_0 iff (z-z_0)^mf(z)$ is bounded but $(z-z_0)^{m-1}f(z)$ is notAnalytic function on unit disk has finitely many zerosPowers of a function being analyticDetermine an analytic function over the diskIf $f'$ has a zero of order $m$ at $z_0$ then there is $g$ s.t $f(z) - f(z_0) = g(z)^{k+1}$
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(This is HW.) Given $f(z)$ analytic on open disk $D$ containing $z_0$; $f(z)$ has a zero order $m$ at $z = z_0$. I need to find the order of zero at $z_0$ of $f'$ and $f^2$ (I assume $f^2$ is the second derivative of $f$, but I honestly don't know.)
This seems overly trivial to me. By plugging in the definition of the order of zeros, I easily obtain that $f'(z_0)$ has order $m-1$ and $f^2(z_0)$ with order $m - 2$ (just a change in index after the definition plugging). Am I missing something obvious?
complex-analysis
asked Mar 20 at 0:08
runeblazeruneblaze
404
$endgroup$
add a comment |
$begingroup$
(This is HW.) Given $f(z)$ analytic on open disk $D$ containing $z_0$; $f(z)$ has a zero order $m$ at $z = z_0$. I need to find the order of zero at $z_0$ of $f'$ and $f^2$ (I assume $f^2$ is the second derivative of $f$, but I honestly don't know.)
This seems overly trivial to me. By plugging in the definition of the order of zeros, I easily obtain that $f'(z_0)$ has order $m-1$ and $f^2(z_0)$ with order $m - 2$ (just a change in index after the definition plugging). Am I missing something obvious?
complex-analysis
asked Mar 20 at 0:08
runeblazeruneblaze
404
$endgroup$
$begingroup$
Or perhaps $f^2(z) = f(f(z))$...
$endgroup$
– Ertxiem
Mar 20 at 0:52
add a comment |
$begingroup$
(This is HW.) Given $f(z)$ analytic on open disk $D$ containing $z_0$; $f(z)$ has a zero order $m$ at $z = z_0$. I need to find the order of zero at $z_0$ of $f'$ and $f^2$ (I assume $f^2$ is the second derivative of $f$, but I honestly don't know.)
This seems overly trivial to me. By plugging in the definition of the order of zeros, I easily obtain that $f'(z_0)$ has order $m-1$ and $f^2(z_0)$ with order $m - 2$ (just a change in index after the definition plugging). Am I missing something obvious?
complex-analysis
asked Mar 20 at 0:08
runeblazeruneblaze
404
$endgroup$
(This is HW.) Given $f(z)$ analytic on open disk $D$ containing $z_0$; $f(z)$ has a zero order $m$ at $z = z_0$. I need to find the order of zero at $z_0$ of $f'$ and $f^2$ (I assume $f^2$ is the second derivative of $f$, but I honestly don't know.)
This seems overly trivial to me. By plugging in the definition of the order of zeros, I easily obtain that $f'(z_0)$ has order $m-1$ and $f^2(z_0)$ with order $m - 2$ (just a change in index after the definition plugging). Am I missing something obvious?
complex-analysis
complex-analysis
asked Mar 20 at 0:08
runeblazeruneblaze
404
asked Mar 20 at 0:08
runeblazeruneblaze
404
asked Mar 20 at 0:08
runeblazeruneblaze
404
asked Mar 20 at 0:08
runeblazeruneblaze
404
asked Mar 20 at 0:08
runeblazeruneblaze
404
404
$begingroup$
Or perhaps $f^2(z) = f(f(z))$...
$endgroup$
– Ertxiem
Mar 20 at 0:52
add a comment |
$begingroup$
Or perhaps $f^2(z) = f(f(z))$...
$endgroup$
– Ertxiem
Mar 20 at 0:52
$begingroup$
Or perhaps $f^2(z) = f(f(z))$...
$endgroup$
– Ertxiem
Mar 20 at 0:52
$begingroup$
Or perhaps $f^2(z) = f(f(z))$...
$endgroup$
– Ertxiem
Mar 20 at 0:52
add a comment |
1 Answer
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$begingroup$
Your approach seems fine, but I suspect that $f^2(z)=bigl(f(z)bigr)^2$. In that case, $f^2$ has a zero of order $2m$ at $z_0$.
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your approach seems fine, but I suspect that $f^2(z)=bigl(f(z)bigr)^2$. In that case, $f^2$ has a zero of order $2m$ at $z_0$.
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
Your approach seems fine, but I suspect that $f^2(z)=bigl(f(z)bigr)^2$. In that case, $f^2$ has a zero of order $2m$ at $z_0$.
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
Your approach seems fine, but I suspect that $f^2(z)=bigl(f(z)bigr)^2$. In that case, $f^2$ has a zero of order $2m$ at $z_0$.
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
Your approach seems fine, but I suspect that $f^2(z)=bigl(f(z)bigr)^2$. In that case, $f^2$ has a zero of order $2m$ at $z_0$.
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 20 at 0:12
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
add a comment |
add a comment |
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$begingroup$
Or perhaps $f^2(z) = f(f(z))$...
$endgroup$
– Ertxiem
Mar 20 at 0:52