How to show that all the zeros of complex function are on unit circle.
$begingroup$
Prove that if $be^{a+1}<1$ where $a$ and $b$ are positive and real, then the function $$z^ne^{-a}-be^z$$ has $n$ zeroes in the unit circle.
I have no idea how to start this problem?
complex-analysis exponential-function roots
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
$endgroup$
add a comment |
$begingroup$
Prove that if $be^{a+1}<1$ where $a$ and $b$ are positive and real, then the function $$z^ne^{-a}-be^z$$ has $n$ zeroes in the unit circle.
I have no idea how to start this problem?
complex-analysis exponential-function roots
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
$endgroup$
add a comment |
$begingroup$
Prove that if $be^{a+1}<1$ where $a$ and $b$ are positive and real, then the function $$z^ne^{-a}-be^z$$ has $n$ zeroes in the unit circle.
I have no idea how to start this problem?
complex-analysis exponential-function roots
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
$endgroup$
Prove that if $be^{a+1}<1$ where $a$ and $b$ are positive and real, then the function $$z^ne^{-a}-be^z$$ has $n$ zeroes in the unit circle.
I have no idea how to start this problem?
complex-analysis exponential-function roots
complex-analysis exponential-function roots
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
asked Dec 1 '18 at 17:10
Mittal GMittal G
1,203516
1,203516
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $h(z)=z^ne^{-a}-b,e^z$ and $f(z)=z^ne^{-a}$. Then, if $|z|=1$ we have
$$
|f(z)-h(z)|=b|e^z|le e,ble e^{-a}=|f(z)|.
$$
By Rouche's Theorem, $f$ and $h$ have the same number of zeros in ${|z|<1}$.
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
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add a comment |
$begingroup$
Use Rouches theorem
Consider $f(z)=z^ne^{-a}$ and $g(z)=-be^z{}$
On unit circle
$|f(z)|<|g(z)|$
By given condition
$be<e^{-a}$.
So Same number of zero
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $h(z)=z^ne^{-a}-b,e^z$ and $f(z)=z^ne^{-a}$. Then, if $|z|=1$ we have
$$
|f(z)-h(z)|=b|e^z|le e,ble e^{-a}=|f(z)|.
$$
By Rouche's Theorem, $f$ and $h$ have the same number of zeros in ${|z|<1}$.
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
Let $h(z)=z^ne^{-a}-b,e^z$ and $f(z)=z^ne^{-a}$. Then, if $|z|=1$ we have
$$
|f(z)-h(z)|=b|e^z|le e,ble e^{-a}=|f(z)|.
$$
By Rouche's Theorem, $f$ and $h$ have the same number of zeros in ${|z|<1}$.
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
Let $h(z)=z^ne^{-a}-b,e^z$ and $f(z)=z^ne^{-a}$. Then, if $|z|=1$ we have
$$
|f(z)-h(z)|=b|e^z|le e,ble e^{-a}=|f(z)|.
$$
By Rouche's Theorem, $f$ and $h$ have the same number of zeros in ${|z|<1}$.
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
Let $h(z)=z^ne^{-a}-b,e^z$ and $f(z)=z^ne^{-a}$. Then, if $|z|=1$ we have
$$
|f(z)-h(z)|=b|e^z|le e,ble e^{-a}=|f(z)|.
$$
By Rouche's Theorem, $f$ and $h$ have the same number of zeros in ${|z|<1}$.
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 1 '18 at 18:05
Julián AguirreJulián Aguirre
68.1k24094
68.1k24094
add a comment |
add a comment |
$begingroup$
Use Rouches theorem
Consider $f(z)=z^ne^{-a}$ and $g(z)=-be^z{}$
On unit circle
$|f(z)|<|g(z)|$
By given condition
$be<e^{-a}$.
So Same number of zero
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
$endgroup$
add a comment |
$begingroup$
Use Rouches theorem
Consider $f(z)=z^ne^{-a}$ and $g(z)=-be^z{}$
On unit circle
$|f(z)|<|g(z)|$
By given condition
$be<e^{-a}$.
So Same number of zero
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
$endgroup$
add a comment |
$begingroup$
Use Rouches theorem
Consider $f(z)=z^ne^{-a}$ and $g(z)=-be^z{}$
On unit circle
$|f(z)|<|g(z)|$
By given condition
$be<e^{-a}$.
So Same number of zero
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
$endgroup$
Use Rouches theorem
Consider $f(z)=z^ne^{-a}$ and $g(z)=-be^z{}$
On unit circle
$|f(z)|<|g(z)|$
By given condition
$be<e^{-a}$.
So Same number of zero
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
answered Dec 1 '18 at 18:08
SRJSRJ
1,6121520
1,6121520
add a comment |
add a comment |
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