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Find the number of zeroes of $6z^3 + e^z + 1$ in the unit disc $|z|<1$
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I have studied Rouche's theorem and applied it to polynomial expressions but I don't seem to understand the problem in expressions with an exponential term. My approach to the above question is as follows-
Let $f(z) = 6z^3$ and $g(z) = e^z + 1$. Now $|e^z| < 1$ hence $|g(z)| < 2$ and $|f(z)| <6.$
Thus $|f(z)| > |g(z)|$ and so $f + g$ which is the required expression should have same number of zeros by Rouche's theorem which is $3$ as $6z^3$ has a zero at origin with multiplicity $3.$
I am afraid I might be fundamentally wrong somewhere. If yes, please explain where and how. Thanks.
complex-analysis roots rouches-theorem
edited Jul 16 at 3:40
saulspatz
10.7k21323
asked Jul 16 at 3:24
deadcode
228
 |Â
show 1 more comment
up vote
1
down vote
favorite
I have studied Rouche's theorem and applied it to polynomial expressions but I don't seem to understand the problem in expressions with an exponential term. My approach to the above question is as follows-
Let $f(z) = 6z^3$ and $g(z) = e^z + 1$. Now $|e^z| < 1$ hence $|g(z)| < 2$ and $|f(z)| <6.$
Thus $|f(z)| > |g(z)|$ and so $f + g$ which is the required expression should have same number of zeros by Rouche's theorem which is $3$ as $6z^3$ has a zero at origin with multiplicity $3.$
I am afraid I might be fundamentally wrong somewhere. If yes, please explain where and how. Thanks.
complex-analysis roots rouches-theorem
edited Jul 16 at 3:40
saulspatz
10.7k21323
asked Jul 16 at 3:24
deadcode
228
3
Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$.
– copper.hat
Jul 16 at 3:33
Please use MathJax
– saulspatz
Jul 16 at 3:36
Oh sorry. So |e^z| is less than or equal to 1? (I don't know how to write that in mathJax, any edit is welcome) but does that change the rest of my approach?
– deadcode
Jul 16 at 3:36
All I did to edit your code was to put '$' signs around what you had written. That's a big part of it.
– saulspatz
Jul 16 at 3:39
Well, your reasoning is not entirely clear. I will elaborate slightly in my answer.
– copper.hat
Jul 16 at 3:40
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have studied Rouche's theorem and applied it to polynomial expressions but I don't seem to understand the problem in expressions with an exponential term. My approach to the above question is as follows-
Let $f(z) = 6z^3$ and $g(z) = e^z + 1$. Now $|e^z| < 1$ hence $|g(z)| < 2$ and $|f(z)| <6.$
Thus $|f(z)| > |g(z)|$ and so $f + g$ which is the required expression should have same number of zeros by Rouche's theorem which is $3$ as $6z^3$ has a zero at origin with multiplicity $3.$
I am afraid I might be fundamentally wrong somewhere. If yes, please explain where and how. Thanks.
complex-analysis roots rouches-theorem
edited Jul 16 at 3:40
saulspatz
10.7k21323
asked Jul 16 at 3:24
deadcode
228
I have studied Rouche's theorem and applied it to polynomial expressions but I don't seem to understand the problem in expressions with an exponential term. My approach to the above question is as follows-
Let $f(z) = 6z^3$ and $g(z) = e^z + 1$. Now $|e^z| < 1$ hence $|g(z)| < 2$ and $|f(z)| <6.$
Thus $|f(z)| > |g(z)|$ and so $f + g$ which is the required expression should have same number of zeros by Rouche's theorem which is $3$ as $6z^3$ has a zero at origin with multiplicity $3.$
I am afraid I might be fundamentally wrong somewhere. If yes, please explain where and how. Thanks.
complex-analysis roots rouches-theorem
edited Jul 16 at 3:40
saulspatz
10.7k21323
asked Jul 16 at 3:24
deadcode
228
edited Jul 16 at 3:40
saulspatz
10.7k21323
edited Jul 16 at 3:40
saulspatz
10.7k21323
edited Jul 16 at 3:40
saulspatz
10.7k21323
10.7k21323
asked Jul 16 at 3:24
deadcode
228
asked Jul 16 at 3:24
deadcode
228
asked Jul 16 at 3:24
deadcode
228
228
3
Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$.
– copper.hat
Jul 16 at 3:33
Please use MathJax
– saulspatz
Jul 16 at 3:36
Oh sorry. So |e^z| is less than or equal to 1? (I don't know how to write that in mathJax, any edit is welcome) but does that change the rest of my approach?
– deadcode
Jul 16 at 3:36
All I did to edit your code was to put '$' signs around what you had written. That's a big part of it.
– saulspatz
Jul 16 at 3:39
Well, your reasoning is not entirely clear. I will elaborate slightly in my answer.
– copper.hat
Jul 16 at 3:40
 |Â
show 1 more comment
3
Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$.
– copper.hat
Jul 16 at 3:33
Please use MathJax
– saulspatz
Jul 16 at 3:36
Oh sorry. So |e^z| is less than or equal to 1? (I don't know how to write that in mathJax, any edit is welcome) but does that change the rest of my approach?
– deadcode
Jul 16 at 3:36
All I did to edit your code was to put '$' signs around what you had written. That's a big part of it.
– saulspatz
Jul 16 at 3:39
Well, your reasoning is not entirely clear. I will elaborate slightly in my answer.
– copper.hat
Jul 16 at 3:40
3
3
Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$.
– copper.hat
Jul 16 at 3:33
Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$.
– copper.hat
Jul 16 at 3:33
Please use MathJax
– saulspatz
Jul 16 at 3:36
Please use MathJax
– saulspatz
Jul 16 at 3:36
Oh sorry. So |e^z| is less than or equal to 1? (I don't know how to write that in mathJax, any edit is welcome) but does that change the rest of my approach?
– deadcode
Jul 16 at 3:36
Oh sorry. So |e^z| is less than or equal to 1? (I don't know how to write that in mathJax, any edit is welcome) but does that change the rest of my approach?
– deadcode
Jul 16 at 3:36
All I did to edit your code was to put '$' signs around what you had written. That's a big part of it.
– saulspatz
Jul 16 at 3:39
All I did to edit your code was to put '$' signs around what you had written. That's a big part of it.
– saulspatz
Jul 16 at 3:39
Well, your reasoning is not entirely clear. I will elaborate slightly in my answer.
– copper.hat
Jul 16 at 3:40
Well, your reasoning is not entirely clear. I will elaborate slightly in my answer.
– copper.hat
Jul 16 at 3:40
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Note that on the boundary of the disc you have $|f(z)|=|6 z^3| = 6$ and $|g(z)|=|e^z+1| le e+1 < 4$.
Hence $|f(z)+g(z)-f(z)| = |g(z)| < 4 < 6 = |f(z)|$ on the unit disc (strict inequality is important) and so
Rouché tells us that $f+g,f$ have the same number of zeros inside the disc.
Hence $f+g$ has 3 zeros inside the disc.
answered Jul 16 at 3:36
copper.hat
122k557156
So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help
– deadcode
Jul 16 at 3:48
And I had the idea $|e^z + 1|$ $le$ $|e^z| + 1$ $le$ $2$ . And 2<3 or 2 < 5 etc
– deadcode
Jul 16 at 3:50
@deadcode: The latter statement is incorrect as I pointed out in my first comment above.
– copper.hat
Jul 16 at 3:52
Wolfram just shows the real zero, there are two complex (conjugates) zeros.
– copper.hat
Jul 16 at 3:58
1
If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726pm i0.65950$.
– copper.hat
Jul 16 at 4:14
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Note that on the boundary of the disc you have $|f(z)|=|6 z^3| = 6$ and $|g(z)|=|e^z+1| le e+1 < 4$.
Hence $|f(z)+g(z)-f(z)| = |g(z)| < 4 < 6 = |f(z)|$ on the unit disc (strict inequality is important) and so
Rouché tells us that $f+g,f$ have the same number of zeros inside the disc.
Hence $f+g$ has 3 zeros inside the disc.
answered Jul 16 at 3:36
copper.hat
122k557156
So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help
– deadcode
Jul 16 at 3:48
And I had the idea $|e^z + 1|$ $le$ $|e^z| + 1$ $le$ $2$ . And 2<3 or 2 < 5 etc
– deadcode
Jul 16 at 3:50
@deadcode: The latter statement is incorrect as I pointed out in my first comment above.
– copper.hat
Jul 16 at 3:52
Wolfram just shows the real zero, there are two complex (conjugates) zeros.
– copper.hat
Jul 16 at 3:58
1
If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726pm i0.65950$.
– copper.hat
Jul 16 at 4:14
add a comment |Â
up vote
3
down vote
accepted
Note that on the boundary of the disc you have $|f(z)|=|6 z^3| = 6$ and $|g(z)|=|e^z+1| le e+1 < 4$.
Hence $|f(z)+g(z)-f(z)| = |g(z)| < 4 < 6 = |f(z)|$ on the unit disc (strict inequality is important) and so
Rouché tells us that $f+g,f$ have the same number of zeros inside the disc.
Hence $f+g$ has 3 zeros inside the disc.
answered Jul 16 at 3:36
copper.hat
122k557156
So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help
– deadcode
Jul 16 at 3:48
And I had the idea $|e^z + 1|$ $le$ $|e^z| + 1$ $le$ $2$ . And 2<3 or 2 < 5 etc
– deadcode
Jul 16 at 3:50
@deadcode: The latter statement is incorrect as I pointed out in my first comment above.
– copper.hat
Jul 16 at 3:52
Wolfram just shows the real zero, there are two complex (conjugates) zeros.
– copper.hat
Jul 16 at 3:58
1
If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726pm i0.65950$.
– copper.hat
Jul 16 at 4:14
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Note that on the boundary of the disc you have $|f(z)|=|6 z^3| = 6$ and $|g(z)|=|e^z+1| le e+1 < 4$.
Hence $|f(z)+g(z)-f(z)| = |g(z)| < 4 < 6 = |f(z)|$ on the unit disc (strict inequality is important) and so
Rouché tells us that $f+g,f$ have the same number of zeros inside the disc.
Hence $f+g$ has 3 zeros inside the disc.
answered Jul 16 at 3:36
copper.hat
122k557156
Note that on the boundary of the disc you have $|f(z)|=|6 z^3| = 6$ and $|g(z)|=|e^z+1| le e+1 < 4$.
Hence $|f(z)+g(z)-f(z)| = |g(z)| < 4 < 6 = |f(z)|$ on the unit disc (strict inequality is important) and so
Rouché tells us that $f+g,f$ have the same number of zeros inside the disc.
Hence $f+g$ has 3 zeros inside the disc.
answered Jul 16 at 3:36
copper.hat
122k557156
edited Jul 16 at 3:46
answered Jul 16 at 3:36
copper.hat
122k557156
answered Jul 16 at 3:36
copper.hat
122k557156
answered Jul 16 at 3:36
copper.hat
122k557156
122k557156
So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help
– deadcode
Jul 16 at 3:48
And I had the idea $|e^z + 1|$ $le$ $|e^z| + 1$ $le$ $2$ . And 2<3 or 2 < 5 etc
– deadcode
Jul 16 at 3:50
@deadcode: The latter statement is incorrect as I pointed out in my first comment above.
– copper.hat
Jul 16 at 3:52
Wolfram just shows the real zero, there are two complex (conjugates) zeros.
– copper.hat
Jul 16 at 3:58
1
If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726pm i0.65950$.
– copper.hat
Jul 16 at 4:14
add a comment |Â
So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help
– deadcode
Jul 16 at 3:48
And I had the idea $|e^z + 1|$ $le$ $|e^z| + 1$ $le$ $2$ . And 2<3 or 2 < 5 etc
– deadcode
Jul 16 at 3:50
@deadcode: The latter statement is incorrect as I pointed out in my first comment above.
– copper.hat
Jul 16 at 3:52
Wolfram just shows the real zero, there are two complex (conjugates) zeros.
– copper.hat
Jul 16 at 3:58
1
If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726pm i0.65950$.
– copper.hat
Jul 16 at 4:14
So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help
– deadcode
Jul 16 at 3:48
So I was on the right track. Wolfram showed 1 root and hence I was not sure. Thanks for the help
– deadcode
Jul 16 at 3:48
And I had the idea $|e^z + 1|$ $le$ $|e^z| + 1$ $le$ $2$ . And 2<3 or 2 < 5 etc
– deadcode
Jul 16 at 3:50
And I had the idea $|e^z + 1|$ $le$ $|e^z| + 1$ $le$ $2$ . And 2<3 or 2 < 5 etc
– deadcode
Jul 16 at 3:50
@deadcode: The latter statement is incorrect as I pointed out in my first comment above.
– copper.hat
Jul 16 at 3:52
@deadcode: The latter statement is incorrect as I pointed out in my first comment above.
– copper.hat
Jul 16 at 3:52
Wolfram just shows the real zero, there are two complex (conjugates) zeros.
– copper.hat
Jul 16 at 3:58
Wolfram just shows the real zero, there are two complex (conjugates) zeros.
– copper.hat
Jul 16 at 3:58
1
1
If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726pm i0.65950$.
– copper.hat
Jul 16 at 4:14
If you use Newtons method, you can approximate the zeros at $-0.63418$, $0.27726pm i0.65950$.
– copper.hat
Jul 16 at 4:14
add a comment |Â
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3
Umm, $e^0 = 1$, how did you get $|e^z| < 1$? On the unit disc you have $|e^z| < e$.
– copper.hat
Jul 16 at 3:33
Please use MathJax
– saulspatz
Jul 16 at 3:36
Oh sorry. So |e^z| is less than or equal to 1? (I don't know how to write that in mathJax, any edit is welcome) but does that change the rest of my approach?
– deadcode
Jul 16 at 3:36
All I did to edit your code was to put '$' signs around what you had written. That's a big part of it.
– saulspatz
Jul 16 at 3:39
Well, your reasoning is not entirely clear. I will elaborate slightly in my answer.
– copper.hat
Jul 16 at 3:40