Determine the set of values of $exp(1/z)$ for $0<|z|<r$
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I can't do this exercise of Conway's Book: For $r>0$ let $A=<r$, determine the set $A$. Any hints?
complex-numbers complex-analysis
edited Aug 25 at 6:02
Did
243k23208443
add a comment |Â
up vote
5
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favorite
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I can't do this exercise of Conway's Book: For $r>0$ let $A=<r$, determine the set $A$. Any hints?
complex-numbers complex-analysis
edited Aug 25 at 6:02
Did
243k23208443
1
Start by looking at the set <r, then think Euler's formula...
– hardmath
Jan 9 '11 at 18:52
1
Another way to look at this is to notice that 0 is an essential singularity of the function, then use the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 19:25
2
It can't be all of the complex plane since 0 is never in the range of the exponential function. However, it is a dense subset of the complex plane by CW theorem.
– user3180
Jan 9 '11 at 19:41
Is it all the complex plane minus the origin? Because of the periodicity of the exponential. (Obs.: Sorry, I meant the complex plane minus the origin, I've excluded the previous message).
– user1971
Jan 9 '11 at 19:45
1
I think you're right. $A$ is also equivilent to >$$1/r$. The exponential function maps $C$ onto $C-0$. You can then use periodicity to argue that if $|z|<$$ 1/r$ then there exists $t$ such that $|t|>$$ 1/r$ and exp(z)=exp(t). Hence, the range or your function is the same as the range of the exponential function on all of $C$. Namely,$C-0$. The interesting thing is that this is true no mater how small $r$ is. That's the point of the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 20:15
add a comment |Â
up vote
5
down vote
favorite
2
up vote
5
down vote
favorite
2
2
I can't do this exercise of Conway's Book: For $r>0$ let $A=<r$, determine the set $A$. Any hints?
complex-numbers complex-analysis
edited Aug 25 at 6:02
Did
243k23208443
I can't do this exercise of Conway's Book: For $r>0$ let $A=<r$, determine the set $A$. Any hints?
complex-numbers complex-analysis
edited Aug 25 at 6:02
Did
243k23208443
edited Aug 25 at 6:02
Did
243k23208443
edited Aug 25 at 6:02
Did
243k23208443
edited Aug 25 at 6:02
Did
243k23208443
243k23208443
asked Jan 9 '11 at 18:35
user1971
1
Start by looking at the set <r, then think Euler's formula...
– hardmath
Jan 9 '11 at 18:52
1
Another way to look at this is to notice that 0 is an essential singularity of the function, then use the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 19:25
2
It can't be all of the complex plane since 0 is never in the range of the exponential function. However, it is a dense subset of the complex plane by CW theorem.
– user3180
Jan 9 '11 at 19:41
Is it all the complex plane minus the origin? Because of the periodicity of the exponential. (Obs.: Sorry, I meant the complex plane minus the origin, I've excluded the previous message).
– user1971
Jan 9 '11 at 19:45
1
I think you're right. $A$ is also equivilent to >$$1/r$. The exponential function maps $C$ onto $C-0$. You can then use periodicity to argue that if $|z|<$$ 1/r$ then there exists $t$ such that $|t|>$$ 1/r$ and exp(z)=exp(t). Hence, the range or your function is the same as the range of the exponential function on all of $C$. Namely,$C-0$. The interesting thing is that this is true no mater how small $r$ is. That's the point of the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 20:15
add a comment |Â
1
Start by looking at the set <r, then think Euler's formula...
– hardmath
Jan 9 '11 at 18:52
1
Another way to look at this is to notice that 0 is an essential singularity of the function, then use the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 19:25
2
It can't be all of the complex plane since 0 is never in the range of the exponential function. However, it is a dense subset of the complex plane by CW theorem.
– user3180
Jan 9 '11 at 19:41
Is it all the complex plane minus the origin? Because of the periodicity of the exponential. (Obs.: Sorry, I meant the complex plane minus the origin, I've excluded the previous message).
– user1971
Jan 9 '11 at 19:45
1
I think you're right. $A$ is also equivilent to >$$1/r$. The exponential function maps $C$ onto $C-0$. You can then use periodicity to argue that if $|z|<$$ 1/r$ then there exists $t$ such that $|t|>$$ 1/r$ and exp(z)=exp(t). Hence, the range or your function is the same as the range of the exponential function on all of $C$. Namely,$C-0$. The interesting thing is that this is true no mater how small $r$ is. That's the point of the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 20:15
1
1
Start by looking at the set <r, then think Euler's formula...
– hardmath
Jan 9 '11 at 18:52
Start by looking at the set <r, then think Euler's formula...
– hardmath
Jan 9 '11 at 18:52
1
1
Another way to look at this is to notice that 0 is an essential singularity of the function, then use the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 19:25
Another way to look at this is to notice that 0 is an essential singularity of the function, then use the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 19:25
2
2
It can't be all of the complex plane since 0 is never in the range of the exponential function. However, it is a dense subset of the complex plane by CW theorem.
– user3180
Jan 9 '11 at 19:41
It can't be all of the complex plane since 0 is never in the range of the exponential function. However, it is a dense subset of the complex plane by CW theorem.
– user3180
Jan 9 '11 at 19:41
Is it all the complex plane minus the origin? Because of the periodicity of the exponential. (Obs.: Sorry, I meant the complex plane minus the origin, I've excluded the previous message).
– user1971
Jan 9 '11 at 19:45
Is it all the complex plane minus the origin? Because of the periodicity of the exponential. (Obs.: Sorry, I meant the complex plane minus the origin, I've excluded the previous message).
– user1971
Jan 9 '11 at 19:45
1
1
I think you're right. $A$ is also equivilent to >$$1/r$. The exponential function maps $C$ onto $C-0$. You can then use periodicity to argue that if $|z|<$$ 1/r$ then there exists $t$ such that $|t|>$$ 1/r$ and exp(z)=exp(t). Hence, the range or your function is the same as the range of the exponential function on all of $C$. Namely,$C-0$. The interesting thing is that this is true no mater how small $r$ is. That's the point of the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 20:15
I think you're right. $A$ is also equivilent to >$$1/r$. The exponential function maps $C$ onto $C-0$. You can then use periodicity to argue that if $|z|<$$ 1/r$ then there exists $t$ such that $|t|>$$ 1/r$ and exp(z)=exp(t). Hence, the range or your function is the same as the range of the exponential function on all of $C$. Namely,$C-0$. The interesting thing is that this is true no mater how small $r$ is. That's the point of the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 20:15
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
$exp$ maps each horizontal strip of height $2pi$ onto $mathbbCsetminus 0$, and for all $rgt0$, $w: w=frac1z,0lt$ contains infinitely many such strips.
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
add a comment |Â
up vote
3
down vote
The Big Picard Theorem (http://en.wikipedia.org/wiki/Picard_theorem) is a great generalization of Casorati-Weierstrass which says that in the neighborhood of an essential singularity a holomorphic function assumes all complex values with possibly one exception. In this case, that is clearly $0$. So the set is $mathbbC-0$.
answered Jan 12 '11 at 9:42
Tony
3,3201235
1
The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer.
– Christian Blatter
Jan 12 '11 at 19:14
add a comment |Â
up vote
3
down vote
Let's prove that $A = mathbbC - 0$. Let's pick $r > 0$. First, we'll prove that $A subset mathbbC - 0$, then we'll show that $mathbbC - 0 subset A$.
- Let's show that $A subset mathbbC - 0$. This amounts to showing that $0$ does not belong to A. Given any $w, z$ such that $w = exp(1/z)$ and $0 < |z| < r$, we can write: $1/z = a + ib$ (with $a, b in mathbbR$). Then $|w| = e^a|e^ ib| > 0$. Thus, $0$ is not in A.
- Now, let's show that any non-zero complex number $x$ can be written as $x = exp(1/z)$ with $0 < |z| < r$. This will demonstrate that $A subset mathbbC - 0$. For that, we pick $x neq 0$. We write $x$ in exponential form ($rho, theta in mathbbR$):
begineqnarray
x & = & rho e^itheta\
& = & exp(ln(rho) + itheta)
endeqnarray
Let's pick $k in mathbbN$ such that $|ln(rho) + itheta + 2kpi| > 1/r$. Then $z = ln(rho) + itheta + 2kpi$ verifies $0 < |z| < r$ and we have $x = exp(1/z)$. Thus $x$ belongs to $A$.
We have shown that A both contains and is a subset of $mathbbC - 0$. Thus $A = mathbbC - 0$
answered Jan 12 '11 at 16:00
Régis B.
1313
in the first part of your argument, you considered $|r|$, why if on $A$ we have only $r$? I did not understand why you said that $|z|=exp(a) $ too
– Marcos Paulo
Aug 25 at 4:52
You're right: since $r > 0$ we have $|r| = r$, so I could have written r instead of |r|. Also, point 1 was incorrect. I fixed my answer.
– Régis B.
Aug 27 at 6:37
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$exp$ maps each horizontal strip of height $2pi$ onto $mathbbCsetminus 0$, and for all $rgt0$, $w: w=frac1z,0lt$ contains infinitely many such strips.
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
add a comment |Â
up vote
3
down vote
accepted
$exp$ maps each horizontal strip of height $2pi$ onto $mathbbCsetminus 0$, and for all $rgt0$, $w: w=frac1z,0lt$ contains infinitely many such strips.
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$exp$ maps each horizontal strip of height $2pi$ onto $mathbbCsetminus 0$, and for all $rgt0$, $w: w=frac1z,0lt$ contains infinitely many such strips.
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
$exp$ maps each horizontal strip of height $2pi$ onto $mathbbCsetminus 0$, and for all $rgt0$, $w: w=frac1z,0lt$ contains infinitely many such strips.
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
answered Jan 12 '11 at 14:28
Jonas Meyer
39.2k6141247
39.2k6141247
add a comment |Â
add a comment |Â
up vote
3
down vote
The Big Picard Theorem (http://en.wikipedia.org/wiki/Picard_theorem) is a great generalization of Casorati-Weierstrass which says that in the neighborhood of an essential singularity a holomorphic function assumes all complex values with possibly one exception. In this case, that is clearly $0$. So the set is $mathbbC-0$.
answered Jan 12 '11 at 9:42
Tony
3,3201235
1
The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer.
– Christian Blatter
Jan 12 '11 at 19:14
add a comment |Â
up vote
3
down vote
The Big Picard Theorem (http://en.wikipedia.org/wiki/Picard_theorem) is a great generalization of Casorati-Weierstrass which says that in the neighborhood of an essential singularity a holomorphic function assumes all complex values with possibly one exception. In this case, that is clearly $0$. So the set is $mathbbC-0$.
answered Jan 12 '11 at 9:42
Tony
3,3201235
1
The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer.
– Christian Blatter
Jan 12 '11 at 19:14
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The Big Picard Theorem (http://en.wikipedia.org/wiki/Picard_theorem) is a great generalization of Casorati-Weierstrass which says that in the neighborhood of an essential singularity a holomorphic function assumes all complex values with possibly one exception. In this case, that is clearly $0$. So the set is $mathbbC-0$.
answered Jan 12 '11 at 9:42
Tony
3,3201235
The Big Picard Theorem (http://en.wikipedia.org/wiki/Picard_theorem) is a great generalization of Casorati-Weierstrass which says that in the neighborhood of an essential singularity a holomorphic function assumes all complex values with possibly one exception. In this case, that is clearly $0$. So the set is $mathbbC-0$.
answered Jan 12 '11 at 9:42
Tony
3,3201235
answered Jan 12 '11 at 9:42
Tony
3,3201235
answered Jan 12 '11 at 9:42
Tony
3,3201235
answered Jan 12 '11 at 9:42
Tony
3,3201235
3,3201235
1
The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer.
– Christian Blatter
Jan 12 '11 at 19:14
add a comment |Â
1
The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer.
– Christian Blatter
Jan 12 '11 at 19:14
1
1
The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer.
– Christian Blatter
Jan 12 '11 at 19:14
The "Big Picard Theorem" is a very deep theorem about essential singularities of arbitrary holomorphic functions, and you will not see a proof of it in an undergraduate course. But here we have a very special function whose behavior we can describe in an elementary way. So there is no need to appeal to such heavy machinary. Just look at, e.g., the answer of Jonas Meyer.
– Christian Blatter
Jan 12 '11 at 19:14
add a comment |Â
up vote
3
down vote
Let's prove that $A = mathbbC - 0$. Let's pick $r > 0$. First, we'll prove that $A subset mathbbC - 0$, then we'll show that $mathbbC - 0 subset A$.
- Let's show that $A subset mathbbC - 0$. This amounts to showing that $0$ does not belong to A. Given any $w, z$ such that $w = exp(1/z)$ and $0 < |z| < r$, we can write: $1/z = a + ib$ (with $a, b in mathbbR$). Then $|w| = e^a|e^ ib| > 0$. Thus, $0$ is not in A.
- Now, let's show that any non-zero complex number $x$ can be written as $x = exp(1/z)$ with $0 < |z| < r$. This will demonstrate that $A subset mathbbC - 0$. For that, we pick $x neq 0$. We write $x$ in exponential form ($rho, theta in mathbbR$):
begineqnarray
x & = & rho e^itheta\
& = & exp(ln(rho) + itheta)
endeqnarray
Let's pick $k in mathbbN$ such that $|ln(rho) + itheta + 2kpi| > 1/r$. Then $z = ln(rho) + itheta + 2kpi$ verifies $0 < |z| < r$ and we have $x = exp(1/z)$. Thus $x$ belongs to $A$.
We have shown that A both contains and is a subset of $mathbbC - 0$. Thus $A = mathbbC - 0$
answered Jan 12 '11 at 16:00
Régis B.
1313
in the first part of your argument, you considered $|r|$, why if on $A$ we have only $r$? I did not understand why you said that $|z|=exp(a) $ too
– Marcos Paulo
Aug 25 at 4:52
You're right: since $r > 0$ we have $|r| = r$, so I could have written r instead of |r|. Also, point 1 was incorrect. I fixed my answer.
– Régis B.
Aug 27 at 6:37
add a comment |Â
up vote
3
down vote
Let's prove that $A = mathbbC - 0$. Let's pick $r > 0$. First, we'll prove that $A subset mathbbC - 0$, then we'll show that $mathbbC - 0 subset A$.
- Let's show that $A subset mathbbC - 0$. This amounts to showing that $0$ does not belong to A. Given any $w, z$ such that $w = exp(1/z)$ and $0 < |z| < r$, we can write: $1/z = a + ib$ (with $a, b in mathbbR$). Then $|w| = e^a|e^ ib| > 0$. Thus, $0$ is not in A.
- Now, let's show that any non-zero complex number $x$ can be written as $x = exp(1/z)$ with $0 < |z| < r$. This will demonstrate that $A subset mathbbC - 0$. For that, we pick $x neq 0$. We write $x$ in exponential form ($rho, theta in mathbbR$):
begineqnarray
x & = & rho e^itheta\
& = & exp(ln(rho) + itheta)
endeqnarray
Let's pick $k in mathbbN$ such that $|ln(rho) + itheta + 2kpi| > 1/r$. Then $z = ln(rho) + itheta + 2kpi$ verifies $0 < |z| < r$ and we have $x = exp(1/z)$. Thus $x$ belongs to $A$.
We have shown that A both contains and is a subset of $mathbbC - 0$. Thus $A = mathbbC - 0$
answered Jan 12 '11 at 16:00
Régis B.
1313
in the first part of your argument, you considered $|r|$, why if on $A$ we have only $r$? I did not understand why you said that $|z|=exp(a) $ too
– Marcos Paulo
Aug 25 at 4:52
You're right: since $r > 0$ we have $|r| = r$, so I could have written r instead of |r|. Also, point 1 was incorrect. I fixed my answer.
– Régis B.
Aug 27 at 6:37
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let's prove that $A = mathbbC - 0$. Let's pick $r > 0$. First, we'll prove that $A subset mathbbC - 0$, then we'll show that $mathbbC - 0 subset A$.
- Let's show that $A subset mathbbC - 0$. This amounts to showing that $0$ does not belong to A. Given any $w, z$ such that $w = exp(1/z)$ and $0 < |z| < r$, we can write: $1/z = a + ib$ (with $a, b in mathbbR$). Then $|w| = e^a|e^ ib| > 0$. Thus, $0$ is not in A.
- Now, let's show that any non-zero complex number $x$ can be written as $x = exp(1/z)$ with $0 < |z| < r$. This will demonstrate that $A subset mathbbC - 0$. For that, we pick $x neq 0$. We write $x$ in exponential form ($rho, theta in mathbbR$):
begineqnarray
x & = & rho e^itheta\
& = & exp(ln(rho) + itheta)
endeqnarray
Let's pick $k in mathbbN$ such that $|ln(rho) + itheta + 2kpi| > 1/r$. Then $z = ln(rho) + itheta + 2kpi$ verifies $0 < |z| < r$ and we have $x = exp(1/z)$. Thus $x$ belongs to $A$.
We have shown that A both contains and is a subset of $mathbbC - 0$. Thus $A = mathbbC - 0$
answered Jan 12 '11 at 16:00
Régis B.
1313
Let's prove that $A = mathbbC - 0$. Let's pick $r > 0$. First, we'll prove that $A subset mathbbC - 0$, then we'll show that $mathbbC - 0 subset A$.
- Let's show that $A subset mathbbC - 0$. This amounts to showing that $0$ does not belong to A. Given any $w, z$ such that $w = exp(1/z)$ and $0 < |z| < r$, we can write: $1/z = a + ib$ (with $a, b in mathbbR$). Then $|w| = e^a|e^ ib| > 0$. Thus, $0$ is not in A.
- Now, let's show that any non-zero complex number $x$ can be written as $x = exp(1/z)$ with $0 < |z| < r$. This will demonstrate that $A subset mathbbC - 0$. For that, we pick $x neq 0$. We write $x$ in exponential form ($rho, theta in mathbbR$):
begineqnarray
x & = & rho e^itheta\
& = & exp(ln(rho) + itheta)
endeqnarray
Let's pick $k in mathbbN$ such that $|ln(rho) + itheta + 2kpi| > 1/r$. Then $z = ln(rho) + itheta + 2kpi$ verifies $0 < |z| < r$ and we have $x = exp(1/z)$. Thus $x$ belongs to $A$.
We have shown that A both contains and is a subset of $mathbbC - 0$. Thus $A = mathbbC - 0$
answered Jan 12 '11 at 16:00
Régis B.
1313
edited Aug 27 at 6:36
answered Jan 12 '11 at 16:00
Régis B.
1313
answered Jan 12 '11 at 16:00
Régis B.
1313
answered Jan 12 '11 at 16:00
Régis B.
1313
1313
in the first part of your argument, you considered $|r|$, why if on $A$ we have only $r$? I did not understand why you said that $|z|=exp(a) $ too
– Marcos Paulo
Aug 25 at 4:52
You're right: since $r > 0$ we have $|r| = r$, so I could have written r instead of |r|. Also, point 1 was incorrect. I fixed my answer.
– Régis B.
Aug 27 at 6:37
add a comment |Â
in the first part of your argument, you considered $|r|$, why if on $A$ we have only $r$? I did not understand why you said that $|z|=exp(a) $ too
– Marcos Paulo
Aug 25 at 4:52
You're right: since $r > 0$ we have $|r| = r$, so I could have written r instead of |r|. Also, point 1 was incorrect. I fixed my answer.
– Régis B.
Aug 27 at 6:37
in the first part of your argument, you considered $|r|$, why if on $A$ we have only $r$? I did not understand why you said that $|z|=exp(a) $ too
– Marcos Paulo
Aug 25 at 4:52
in the first part of your argument, you considered $|r|$, why if on $A$ we have only $r$? I did not understand why you said that $|z|=exp(a) $ too
– Marcos Paulo
Aug 25 at 4:52
You're right: since $r > 0$ we have $|r| = r$, so I could have written r instead of |r|. Also, point 1 was incorrect. I fixed my answer.
– Régis B.
Aug 27 at 6:37
You're right: since $r > 0$ we have $|r| = r$, so I could have written r instead of |r|. Also, point 1 was incorrect. I fixed my answer.
– Régis B.
Aug 27 at 6:37
add a comment |Â
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1
Start by looking at the set <r, then think Euler's formula...
– hardmath
Jan 9 '11 at 18:52
1
Another way to look at this is to notice that 0 is an essential singularity of the function, then use the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 19:25
2
It can't be all of the complex plane since 0 is never in the range of the exponential function. However, it is a dense subset of the complex plane by CW theorem.
– user3180
Jan 9 '11 at 19:41
Is it all the complex plane minus the origin? Because of the periodicity of the exponential. (Obs.: Sorry, I meant the complex plane minus the origin, I've excluded the previous message).
– user1971
Jan 9 '11 at 19:45
1
I think you're right. $A$ is also equivilent to >$$1/r$. The exponential function maps $C$ onto $C-0$. You can then use periodicity to argue that if $|z|<$$ 1/r$ then there exists $t$ such that $|t|>$$ 1/r$ and exp(z)=exp(t). Hence, the range or your function is the same as the range of the exponential function on all of $C$. Namely,$C-0$. The interesting thing is that this is true no mater how small $r$ is. That's the point of the Casorati Weierstrass theorem.
– user3180
Jan 9 '11 at 20:15