Is $phi(z)$ is onto ? choose the correct statement ..

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My attempts : By mobius Transformation Formula $f(z) = frac az+bcz+ d$ now
here i take $a = c= 1$ as $|fracac|= 1$ , $f(z) = fracac frac(z+ fracba) (z+ fracdc)$ we can write $frac ac = e^itheta$ for some $theta$ $in mathbbR $ ,,,,$f(z) = e^itheta (frac z - beta z-alpha )$ as $ -1 le frac b d le 1$ now i know that $phi (z) = frac1+z1-z < 1$

now by open mapping theorem,,,option 3 is corrects..

is its correct...??

complex-analysis

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asked Jul 14 at 18:27

stupid

58419

add a comment | 

up vote
-1
down vote

favorite

My attempts : By mobius Transformation Formula $f(z) = frac az+bcz+ d$ now
here i take $a = c= 1$ as $|fracac|= 1$ , $f(z) = fracac frac(z+ fracba) (z+ fracdc)$ we can write $frac ac = e^itheta$ for some $theta$ $in mathbbR $ ,,,,$f(z) = e^itheta (frac z - beta z-alpha )$ as $ -1 le frac b d le 1$ now i know that $phi (z) = frac1+z1-z < 1$

now by open mapping theorem,,,option 3 is corrects..

is its correct...??

complex-analysis

share|cite|improve this question

asked Jul 14 at 18:27

stupid

58419

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

My attempts : By mobius Transformation Formula $f(z) = frac az+bcz+ d$ now
here i take $a = c= 1$ as $|fracac|= 1$ , $f(z) = fracac frac(z+ fracba) (z+ fracdc)$ we can write $frac ac = e^itheta$ for some $theta$ $in mathbbR $ ,,,,$f(z) = e^itheta (frac z - beta z-alpha )$ as $ -1 le frac b d le 1$ now i know that $phi (z) = frac1+z1-z < 1$

now by open mapping theorem,,,option 3 is corrects..

is its correct...??

complex-analysis

share|cite|improve this question

asked Jul 14 at 18:27

stupid

58419

My attempts : By mobius Transformation Formula $f(z) = frac az+bcz+ d$ now
here i take $a = c= 1$ as $|fracac|= 1$ , $f(z) = fracac frac(z+ fracba) (z+ fracdc)$ we can write $frac ac = e^itheta$ for some $theta$ $in mathbbR $ ,,,,$f(z) = e^itheta (frac z - beta z-alpha )$ as $ -1 le frac b d le 1$ now i know that $phi (z) = frac1+z1-z < 1$

now by open mapping theorem,,,option 3 is corrects..

is its correct...??

complex-analysis

share|cite|improve this question

asked Jul 14 at 18:27

stupid

58419

share|cite|improve this question

share|cite|improve this question

asked Jul 14 at 18:27

stupid

58419

asked Jul 14 at 18:27

stupid

58419

asked Jul 14 at 18:27

stupid

58419

58419

add a comment | 

add a comment | 

1 Answer
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up vote
1
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accepted

No, it is not correct. Note that$$varphi(z)=-1iff1+z=-1+ziff-1=1$$and therefore $-1$ does not belong to the range of $varphi$.

On the other hand, the fourth option is correct because, if $wneq-1$,$$varphi(z)=wiff z=fracw-1w+1.$$

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edited Jul 14 at 18:34

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u @ Jose carlos sir ..then which one is correct ??? can u elaborate the correct answer ?
  – stupid
  Jul 14 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes i got its @ Jose sir.....one more last doubts can u tell me why option 1 is not correct ??
  – stupid
  Jul 14 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because, for instance, $varphi(0)=1$.
  – José Carlos Santos
  Jul 14 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u i gots its @ jose sir
  – stupid
  Jul 14 at 18:44
  

  
  
  
  

 | 
show 1 more comment

Your Answer

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

No, it is not correct. Note that$$varphi(z)=-1iff1+z=-1+ziff-1=1$$and therefore $-1$ does not belong to the range of $varphi$.

On the other hand, the fourth option is correct because, if $wneq-1$,$$varphi(z)=wiff z=fracw-1w+1.$$

share|cite|improve this answer

edited Jul 14 at 18:34

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u @ Jose carlos sir ..then which one is correct ??? can u elaborate the correct answer ?
  – stupid
  Jul 14 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes i got its @ Jose sir.....one more last doubts can u tell me why option 1 is not correct ??
  – stupid
  Jul 14 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because, for instance, $varphi(0)=1$.
  – José Carlos Santos
  Jul 14 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u i gots its @ jose sir
  – stupid
  Jul 14 at 18:44
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

No, it is not correct. Note that$$varphi(z)=-1iff1+z=-1+ziff-1=1$$and therefore $-1$ does not belong to the range of $varphi$.

On the other hand, the fourth option is correct because, if $wneq-1$,$$varphi(z)=wiff z=fracw-1w+1.$$

share|cite|improve this answer

edited Jul 14 at 18:34

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u @ Jose carlos sir ..then which one is correct ??? can u elaborate the correct answer ?
  – stupid
  Jul 14 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes i got its @ Jose sir.....one more last doubts can u tell me why option 1 is not correct ??
  – stupid
  Jul 14 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because, for instance, $varphi(0)=1$.
  – José Carlos Santos
  Jul 14 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u i gots its @ jose sir
  – stupid
  Jul 14 at 18:44
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

up vote
1
down vote



accepted

No, it is not correct. Note that$$varphi(z)=-1iff1+z=-1+ziff-1=1$$and therefore $-1$ does not belong to the range of $varphi$.

On the other hand, the fourth option is correct because, if $wneq-1$,$$varphi(z)=wiff z=fracw-1w+1.$$

share|cite|improve this answer

edited Jul 14 at 18:34

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

No, it is not correct. Note that$$varphi(z)=-1iff1+z=-1+ziff-1=1$$and therefore $-1$ does not belong to the range of $varphi$.

On the other hand, the fourth option is correct because, if $wneq-1$,$$varphi(z)=wiff z=fracw-1w+1.$$

share|cite|improve this answer

edited Jul 14 at 18:34

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 18:34

edited Jul 14 at 18:34

edited Jul 14 at 18:34

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

answered Jul 14 at 18:30

José Carlos Santos

114k1698177

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u @ Jose carlos sir ..then which one is correct ??? can u elaborate the correct answer ?
  – stupid
  Jul 14 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes i got its @ Jose sir.....one more last doubts can u tell me why option 1 is not correct ??
  – stupid
  Jul 14 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because, for instance, $varphi(0)=1$.
  – José Carlos Santos
  Jul 14 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u i gots its @ jose sir
  – stupid
  Jul 14 at 18:44
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u @ Jose carlos sir ..then which one is correct ??? can u elaborate the correct answer ?
  – stupid
  Jul 14 at 18:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've edited my answer.
  – José Carlos Santos
  Jul 14 at 18:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes i got its @ Jose sir.....one more last doubts can u tell me why option 1 is not correct ??
  – stupid
  Jul 14 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because, for instance, $varphi(0)=1$.
  – José Carlos Santos
  Jul 14 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks u i gots its @ jose sir
  – stupid
  Jul 14 at 18:44
  

  
  
  
  

thanks u @ Jose carlos sir ..then which one is correct ??? can u elaborate the correct answer ?
– stupid
Jul 14 at 18:32

thanks u @ Jose carlos sir ..then which one is correct ??? can u elaborate the correct answer ?
– stupid
Jul 14 at 18:32

I've edited my answer.
– José Carlos Santos
Jul 14 at 18:35

I've edited my answer.
– José Carlos Santos
Jul 14 at 18:35

yes i got its @ Jose sir.....one more last doubts can u tell me why option 1 is not correct ??
– stupid
Jul 14 at 18:40

yes i got its @ Jose sir.....one more last doubts can u tell me why option 1 is not correct ??
– stupid
Jul 14 at 18:40

1

1

Because, for instance, $varphi(0)=1$.
– José Carlos Santos
Jul 14 at 18:41

Because, for instance, $varphi(0)=1$.
– José Carlos Santos
Jul 14 at 18:41

thanks u i gots its @ jose sir
– stupid
Jul 14 at 18:44

thanks u i gots its @ jose sir
– stupid
Jul 14 at 18:44

 | 
show 1 more comment

 

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