Complex Analysis: Prove that an entire function with $lim_{ztoinfty}f(z)=infty$ is surjective.
$begingroup$
If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.
This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.
complex-analysis
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
$endgroup$
add a comment |
$begingroup$
If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.
This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.
complex-analysis
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
$endgroup$
$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51
2
$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53
1
$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54
$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23
add a comment |
$begingroup$
If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.
This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.
complex-analysis
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
$endgroup$
If $f$ is an entire function with the property that $|f(z)|toinfty$
as $|z|toinfty$, verify that $f(mathbb{C})=mathbb{C}$.
This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.
complex-analysis
complex-analysis
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
asked Dec 8 '18 at 4:44
MephestophelesMephestopheles
83
83
$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51
2
$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53
1
$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54
$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23
add a comment |
$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51
2
$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53
1
$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54
$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23
$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51
$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51
2
2
$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53
$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53
1
1
$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54
$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54
$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23
$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
$$ f(z)=sum_{k=0}^infty a_k z^k.$$
At infinity, the Laurent series looks like
$$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.
Can you conclude using Rouché's Theorem?
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
$endgroup$
$begingroup$
Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
$endgroup$
– clark
Dec 8 '18 at 6:00
add a comment |
$begingroup$
Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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votes
$begingroup$
Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
$$ f(z)=sum_{k=0}^infty a_k z^k.$$
At infinity, the Laurent series looks like
$$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.
Can you conclude using Rouché's Theorem?
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
$endgroup$
$begingroup$
Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
$endgroup$
– clark
Dec 8 '18 at 6:00
add a comment |
$begingroup$
Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
$$ f(z)=sum_{k=0}^infty a_k z^k.$$
At infinity, the Laurent series looks like
$$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.
Can you conclude using Rouché's Theorem?
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
$endgroup$
$begingroup$
Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
$endgroup$
– clark
Dec 8 '18 at 6:00
add a comment |
$begingroup$
Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
$$ f(z)=sum_{k=0}^infty a_k z^k.$$
At infinity, the Laurent series looks like
$$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.
Can you conclude using Rouché's Theorem?
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
$endgroup$
Here's a way to look at it. As $ztoinfty$, $frac{1}{z}to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $frac{1}{w}$ in its Taylor series and examining the behavior as $wto 0$. If $fin mathcal{H}(mathbb{C})$ is entire and has $lim_{ztoinfty} lvert f(z)rvert=infty$, study the Taylor series at $0$
$$ f(z)=sum_{k=0}^infty a_k z^k.$$
At infinity, the Laurent series looks like
$$ f(w)=sum_{k=0}^infty a_k w^{-k}. $$
If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $infty$. But the $lvert f(z)rvert$ would not converge as $ztoinfty$. So, $a_k=0$ for $kge M$. This implies that $f(z)$ is a polynomial.
Can you conclude using Rouché's Theorem?
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
edited Dec 8 '18 at 5:51
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
answered Dec 8 '18 at 5:04
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,81241640
9,81241640
$begingroup$
Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
$endgroup$
– clark
Dec 8 '18 at 6:00
add a comment |
$begingroup$
Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
$endgroup$
– clark
Dec 8 '18 at 6:00
$begingroup$
Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
$endgroup$
– clark
Dec 8 '18 at 6:00
$begingroup$
Another way to see the last step. If $p(x)$ is a polynomial then $p(x)-a$ has a root by invoking the fundamental theorem of calculus.
$endgroup$
– clark
Dec 8 '18 at 6:00
add a comment |
$begingroup$
Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
$endgroup$
add a comment |
$begingroup$
Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
$endgroup$
add a comment |
$begingroup$
Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
$endgroup$
Since $f(z)toinfty$ as $ztoinfty$, we have $inf{lvert f(z)rvert:lvert zrvert=R}toinfty$ as $Rtoinfty$. By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $lvert arvert<inf{lvert f(z)rvert:lvert zrvert=R}$. Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0in f(mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
answered Dec 8 '18 at 5:56
user10354138user10354138
7,3772925
7,3772925
add a comment |
add a comment |
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$begingroup$
What about the exponential function?
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 4:51
2
$begingroup$
With that hypothesis, prove that $f$ has to be a polynomial.
$endgroup$
– Zircht
Dec 8 '18 at 4:53
1
$begingroup$
@JensSchwaiger The exponential map doesn't satisify the condition along the imaginary axis.
$endgroup$
– Zircht
Dec 8 '18 at 4:54
$begingroup$
I see! I'm still asleep.
$endgroup$
– Jens Schwaiger
Dec 8 '18 at 6:23