Mixing
Show the set is connected and open (Complex Analysis) [closed]
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I am new to Complex Algebra, and I am stuck how to show if a set is connected and open. Let $phi = {z: 1 < |z| < 2 $ and $Re(z) > -0.5}$. I would appreciate any help, but I would love if someone could show detailed explanation.
complex-analysis complex-numbers
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
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closed as off-topic by Martin R, max_zorn, Cesareo, Shailesh, user91500 Jan 27 at 7:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, max_zorn, Cesareo, Shailesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am new to Complex Algebra, and I am stuck how to show if a set is connected and open. Let $phi = {z: 1 < |z| < 2 $ and $Re(z) > -0.5}$. I would appreciate any help, but I would love if someone could show detailed explanation.
complex-analysis complex-numbers
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
$endgroup$
closed as off-topic by Martin R, max_zorn, Cesareo, Shailesh, user91500 Jan 27 at 7:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, max_zorn, Cesareo, Shailesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am new to Complex Algebra, and I am stuck how to show if a set is connected and open. Let $phi = {z: 1 < |z| < 2 $ and $Re(z) > -0.5}$. I would appreciate any help, but I would love if someone could show detailed explanation.
complex-analysis complex-numbers
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
$endgroup$
I am new to Complex Algebra, and I am stuck how to show if a set is connected and open. Let $phi = {z: 1 < |z| < 2 $ and $Re(z) > -0.5}$. I would appreciate any help, but I would love if someone could show detailed explanation.
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
edited Jan 27 at 3:38
J. W. Tanner
3,6551320
3,6551320
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
asked Jan 26 at 20:22
Elvin JafaliElvin Jafali
214
214
closed as off-topic by Martin R, max_zorn, Cesareo, Shailesh, user91500 Jan 27 at 7:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, max_zorn, Cesareo, Shailesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Martin R, max_zorn, Cesareo, Shailesh, user91500 Jan 27 at 7:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, max_zorn, Cesareo, Shailesh, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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The fact that it is open is a consequence of the fact that it is the intersection of two open sets
$$
U_1=f_1^{-1}(]1,2[) mbox{ with } f_1(z)=|z| mbox{ continuous }
$$
and
$$
U_2=f_2^{-1}(]1/2,+infty[) mbox{ with } f_2(z)=Re e(z) mbox{ continuous }
$$
If you draw this $phi$, you will see how to show that it is arcwise connected (take two points and "pass" on the right).
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The fact that it is open is a consequence of the fact that it is the intersection of two open sets
$$
U_1=f_1^{-1}(]1,2[) mbox{ with } f_1(z)=|z| mbox{ continuous }
$$
and
$$
U_2=f_2^{-1}(]1/2,+infty[) mbox{ with } f_2(z)=Re e(z) mbox{ continuous }
$$
If you draw this $phi$, you will see how to show that it is arcwise connected (take two points and "pass" on the right).
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
add a comment |
$begingroup$
The fact that it is open is a consequence of the fact that it is the intersection of two open sets
$$
U_1=f_1^{-1}(]1,2[) mbox{ with } f_1(z)=|z| mbox{ continuous }
$$
and
$$
U_2=f_2^{-1}(]1/2,+infty[) mbox{ with } f_2(z)=Re e(z) mbox{ continuous }
$$
If you draw this $phi$, you will see how to show that it is arcwise connected (take two points and "pass" on the right).
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
add a comment |
$begingroup$
The fact that it is open is a consequence of the fact that it is the intersection of two open sets
$$
U_1=f_1^{-1}(]1,2[) mbox{ with } f_1(z)=|z| mbox{ continuous }
$$
and
$$
U_2=f_2^{-1}(]1/2,+infty[) mbox{ with } f_2(z)=Re e(z) mbox{ continuous }
$$
If you draw this $phi$, you will see how to show that it is arcwise connected (take two points and "pass" on the right).
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
The fact that it is open is a consequence of the fact that it is the intersection of two open sets
$$
U_1=f_1^{-1}(]1,2[) mbox{ with } f_1(z)=|z| mbox{ continuous }
$$
and
$$
U_2=f_2^{-1}(]1/2,+infty[) mbox{ with } f_2(z)=Re e(z) mbox{ continuous }
$$
If you draw this $phi$, you will see how to show that it is arcwise connected (take two points and "pass" on the right).
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 27 at 3:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
2,644919
add a comment |
add a comment |
