The Mobius Transformation is a Conformal Mapping.

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So the Mobius Transformation $S(z) = dfracaz+bcz+d$ is analytic on its domain $left~ z neq -dfracdcright$



And $$S^'(z) = dfracad-bc(cz+d)^2 neq 0$$ because of the condition $ad-bc neq 0$, such condition will imply $S^'(z)$ is never zero.



Hence $S(z)$ is a conformal map on the $mathbbC setminus -leftdfracdcright$



However, I think the Professor mentioned something that Mobius Transformation is conformal on the whole of $mathbbC$. I do not quite understand where I went wrong with the Theorem. I follow closely on this theorem:




If $D$ is an open subset of the complex plane $mathbbC$, then a function
$f: D rightarrow mathbbC$ is a conformal mapping if and only if it is
holomorphic (analytic) on its domain and its derivative $f^'(z)$ is
everywhere non-zero on $D$.





complex-analysis




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  • 3




    I think the best way to find out is to ask your professor what he/she meant. What you think was mentioned is not necessarily what was said... But if I may guess, it's that they are conformal on the whole extended complex plane (i.e., as maps from the Riemann sphere to itself).
    – Hans Lundmark
    20 hours ago














up vote
1
down vote

favorite












So the Mobius Transformation $S(z) = dfracaz+bcz+d$ is analytic on its domain $left~ z neq -dfracdcright$



And $$S^'(z) = dfracad-bc(cz+d)^2 neq 0$$ because of the condition $ad-bc neq 0$, such condition will imply $S^'(z)$ is never zero.



Hence $S(z)$ is a conformal map on the $mathbbC setminus -leftdfracdcright$



However, I think the Professor mentioned something that Mobius Transformation is conformal on the whole of $mathbbC$. I do not quite understand where I went wrong with the Theorem. I follow closely on this theorem:




If $D$ is an open subset of the complex plane $mathbbC$, then a function
$f: D rightarrow mathbbC$ is a conformal mapping if and only if it is
holomorphic (analytic) on its domain and its derivative $f^'(z)$ is
everywhere non-zero on $D$.





complex-analysis




share|cite|improve this question

















  • 3




    I think the best way to find out is to ask your professor what he/she meant. What you think was mentioned is not necessarily what was said... But if I may guess, it's that they are conformal on the whole extended complex plane (i.e., as maps from the Riemann sphere to itself).
    – Hans Lundmark
    20 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











So the Mobius Transformation $S(z) = dfracaz+bcz+d$ is analytic on its domain $left~ z neq -dfracdcright$



And $$S^'(z) = dfracad-bc(cz+d)^2 neq 0$$ because of the condition $ad-bc neq 0$, such condition will imply $S^'(z)$ is never zero.



Hence $S(z)$ is a conformal map on the $mathbbC setminus -leftdfracdcright$



However, I think the Professor mentioned something that Mobius Transformation is conformal on the whole of $mathbbC$. I do not quite understand where I went wrong with the Theorem. I follow closely on this theorem:




If $D$ is an open subset of the complex plane $mathbbC$, then a function
$f: D rightarrow mathbbC$ is a conformal mapping if and only if it is
holomorphic (analytic) on its domain and its derivative $f^'(z)$ is
everywhere non-zero on $D$.





complex-analysis




share|cite|improve this question













So the Mobius Transformation $S(z) = dfracaz+bcz+d$ is analytic on its domain $left~ z neq -dfracdcright$



And $$S^'(z) = dfracad-bc(cz+d)^2 neq 0$$ because of the condition $ad-bc neq 0$, such condition will imply $S^'(z)$ is never zero.



Hence $S(z)$ is a conformal map on the $mathbbC setminus -leftdfracdcright$



However, I think the Professor mentioned something that Mobius Transformation is conformal on the whole of $mathbbC$. I do not quite understand where I went wrong with the Theorem. I follow closely on this theorem:




If $D$ is an open subset of the complex plane $mathbbC$, then a function
$f: D rightarrow mathbbC$ is a conformal mapping if and only if it is
holomorphic (analytic) on its domain and its derivative $f^'(z)$ is
everywhere non-zero on $D$.






complex-analysis





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edited 20 hours ago









pointguard0

498215




498215









asked 20 hours ago









ilovewt

821313




821313







  • 3




    I think the best way to find out is to ask your professor what he/she meant. What you think was mentioned is not necessarily what was said... But if I may guess, it's that they are conformal on the whole extended complex plane (i.e., as maps from the Riemann sphere to itself).
    – Hans Lundmark
    20 hours ago












  • 3




    I think the best way to find out is to ask your professor what he/she meant. What you think was mentioned is not necessarily what was said... But if I may guess, it's that they are conformal on the whole extended complex plane (i.e., as maps from the Riemann sphere to itself).
    – Hans Lundmark
    20 hours ago







3




3




I think the best way to find out is to ask your professor what he/she meant. What you think was mentioned is not necessarily what was said... But if I may guess, it's that they are conformal on the whole extended complex plane (i.e., as maps from the Riemann sphere to itself).
– Hans Lundmark
20 hours ago




I think the best way to find out is to ask your professor what he/she meant. What you think was mentioned is not necessarily what was said... But if I may guess, it's that they are conformal on the whole extended complex plane (i.e., as maps from the Riemann sphere to itself).
– Hans Lundmark
20 hours ago










1 Answer
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The statement is false. Consider the exponential function on a horizontal strip like $D=<pi+epsilon$; it is holomorphic, the derivative never vanishes, but it is not invertible on $D$.
The correct statement is that if a function is holomorphic and the derivative never vanishes, then the function is locally conformal.






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  • So is my solution correct?
    – ilovewt
    18 hours ago










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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










The statement is false. Consider the exponential function on a horizontal strip like $D=<pi+epsilon$; it is holomorphic, the derivative never vanishes, but it is not invertible on $D$.
The correct statement is that if a function is holomorphic and the derivative never vanishes, then the function is locally conformal.






share|cite|improve this answer





















  • So is my solution correct?
    – ilovewt
    18 hours ago














up vote
2
down vote



accepted










The statement is false. Consider the exponential function on a horizontal strip like $D=<pi+epsilon$; it is holomorphic, the derivative never vanishes, but it is not invertible on $D$.
The correct statement is that if a function is holomorphic and the derivative never vanishes, then the function is locally conformal.






share|cite|improve this answer





















  • So is my solution correct?
    – ilovewt
    18 hours ago












up vote
2
down vote



accepted







up vote
2
down vote



accepted






The statement is false. Consider the exponential function on a horizontal strip like $D=<pi+epsilon$; it is holomorphic, the derivative never vanishes, but it is not invertible on $D$.
The correct statement is that if a function is holomorphic and the derivative never vanishes, then the function is locally conformal.






share|cite|improve this answer













The statement is false. Consider the exponential function on a horizontal strip like $D=<pi+epsilon$; it is holomorphic, the derivative never vanishes, but it is not invertible on $D$.
The correct statement is that if a function is holomorphic and the derivative never vanishes, then the function is locally conformal.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered 19 hours ago









Joe

6,77621028




6,77621028











  • So is my solution correct?
    – ilovewt
    18 hours ago
















  • So is my solution correct?
    – ilovewt
    18 hours ago















So is my solution correct?
– ilovewt
18 hours ago




So is my solution correct?
– ilovewt
18 hours ago












 

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