Find a conformal map from onto an annulus

Clash Royale CLAN TAG#URR8PPP

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Let $D = zin mathbbC :$ Re $ z geq 0 setminus z-2$.

1. Find a conformal map from D onto an annulus $r < |z| <1$.


2. Find a continuous bounded function on $barD$ which is harmonic in $D$, vanishes on the imaginary axis, and takes value $1$ on $|z-2|=1$.

For 1, I tried to find a linear fractional transformation but it doesn't work for all $r$.

For 2, I considered a holomorphic function $f$ such that $u = $Re $f$ is harmonic, $u(z) = 0$ on the imaginary axis and $u(z) = 1$ on $|z-2|=1$, but couldn't find such f. How can I find such f?

complex-analysis harmonic-functions holomorphic-functions

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edited Jul 24 at 17:06

asked Jul 24 at 0:33

Rachel.L

305

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  1
  

  
  
  
  
  
  

  
  And you can't take a holomorphic function on $overlineD$ for 2 either, since vanishing on the imaginary axis would imply vanishing on $overlineD$ by analytic continuation.
  – Dzoooks
  Jul 24 at 0:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks An annulus is not simply connected.
  – Lord Shark the Unknown
  Jul 24 at 1:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean $D=cdotssetminusz-2$?
  – Lord Shark the Unknown
  Jul 24 at 1:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks the region with radius between r and 1 (? or .1) is not simply connected. And a harmonic function is not holomorphic.
  – Empy2
  Jul 24 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For 2, find a harmonic function that is 0 on $|z|=r$, 1 on $|z|=1$, and combine it with the map to make it a function on the original region.
  – Empy2
  Jul 24 at 3:29
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

1

Let $D = zin mathbbC :$ Re $ z geq 0 setminus z-2$.

1. Find a conformal map from D onto an annulus $r < |z| <1$.


2. Find a continuous bounded function on $barD$ which is harmonic in $D$, vanishes on the imaginary axis, and takes value $1$ on $|z-2|=1$.

For 1, I tried to find a linear fractional transformation but it doesn't work for all $r$.

For 2, I considered a holomorphic function $f$ such that $u = $Re $f$ is harmonic, $u(z) = 0$ on the imaginary axis and $u(z) = 1$ on $|z-2|=1$, but couldn't find such f. How can I find such f?

complex-analysis harmonic-functions holomorphic-functions

share|cite|improve this question

edited Jul 24 at 17:06

asked Jul 24 at 0:33

Rachel.L

305

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And you can't take a holomorphic function on $overlineD$ for 2 either, since vanishing on the imaginary axis would imply vanishing on $overlineD$ by analytic continuation.
  – Dzoooks
  Jul 24 at 0:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks An annulus is not simply connected.
  – Lord Shark the Unknown
  Jul 24 at 1:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean $D=cdotssetminusz-2$?
  – Lord Shark the Unknown
  Jul 24 at 1:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks the region with radius between r and 1 (? or .1) is not simply connected. And a harmonic function is not holomorphic.
  – Empy2
  Jul 24 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For 2, find a harmonic function that is 0 on $|z|=r$, 1 on $|z|=1$, and combine it with the map to make it a function on the original region.
  – Empy2
  Jul 24 at 3:29
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Let $D = zin mathbbC :$ Re $ z geq 0 setminus z-2$.

1. Find a conformal map from D onto an annulus $r < |z| <1$.


2. Find a continuous bounded function on $barD$ which is harmonic in $D$, vanishes on the imaginary axis, and takes value $1$ on $|z-2|=1$.

For 1, I tried to find a linear fractional transformation but it doesn't work for all $r$.

For 2, I considered a holomorphic function $f$ such that $u = $Re $f$ is harmonic, $u(z) = 0$ on the imaginary axis and $u(z) = 1$ on $|z-2|=1$, but couldn't find such f. How can I find such f?

complex-analysis harmonic-functions holomorphic-functions

share|cite|improve this question

edited Jul 24 at 17:06

asked Jul 24 at 0:33

Rachel.L

305

Let $D = zin mathbbC :$ Re $ z geq 0 setminus z-2$.

1. Find a conformal map from D onto an annulus $r < |z| <1$.


2. Find a continuous bounded function on $barD$ which is harmonic in $D$, vanishes on the imaginary axis, and takes value $1$ on $|z-2|=1$.

For 1, I tried to find a linear fractional transformation but it doesn't work for all $r$.

For 2, I considered a holomorphic function $f$ such that $u = $Re $f$ is harmonic, $u(z) = 0$ on the imaginary axis and $u(z) = 1$ on $|z-2|=1$, but couldn't find such f. How can I find such f?

complex-analysis harmonic-functions holomorphic-functions

share|cite|improve this question

edited Jul 24 at 17:06

asked Jul 24 at 0:33

Rachel.L

305

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 17:06

edited Jul 24 at 17:06

edited Jul 24 at 17:06

asked Jul 24 at 0:33

Rachel.L

305

asked Jul 24 at 0:33

Rachel.L

305

asked Jul 24 at 0:33

Rachel.L

305

305

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And you can't take a holomorphic function on $overlineD$ for 2 either, since vanishing on the imaginary axis would imply vanishing on $overlineD$ by analytic continuation.
  – Dzoooks
  Jul 24 at 0:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks An annulus is not simply connected.
  – Lord Shark the Unknown
  Jul 24 at 1:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean $D=cdotssetminusz-2$?
  – Lord Shark the Unknown
  Jul 24 at 1:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks the region with radius between r and 1 (? or .1) is not simply connected. And a harmonic function is not holomorphic.
  – Empy2
  Jul 24 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For 2, find a harmonic function that is 0 on $|z|=r$, 1 on $|z|=1$, and combine it with the map to make it a function on the original region.
  – Empy2
  Jul 24 at 3:29
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And you can't take a holomorphic function on $overlineD$ for 2 either, since vanishing on the imaginary axis would imply vanishing on $overlineD$ by analytic continuation.
  – Dzoooks
  Jul 24 at 0:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks An annulus is not simply connected.
  – Lord Shark the Unknown
  Jul 24 at 1:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you mean $D=cdotssetminusz-2$?
  – Lord Shark the Unknown
  Jul 24 at 1:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzoooks the region with radius between r and 1 (? or .1) is not simply connected. And a harmonic function is not holomorphic.
  – Empy2
  Jul 24 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For 2, find a harmonic function that is 0 on $|z|=r$, 1 on $|z|=1$, and combine it with the map to make it a function on the original region.
  – Empy2
  Jul 24 at 3:29
  

  
  
  
  

1

1

And you can't take a holomorphic function on $overlineD$ for 2 either, since vanishing on the imaginary axis would imply vanishing on $overlineD$ by analytic continuation.
– Dzoooks
Jul 24 at 0:47

And you can't take a holomorphic function on $overlineD$ for 2 either, since vanishing on the imaginary axis would imply vanishing on $overlineD$ by analytic continuation.
– Dzoooks
Jul 24 at 0:47

@Dzoooks An annulus is not simply connected.
– Lord Shark the Unknown
Jul 24 at 1:44

@Dzoooks An annulus is not simply connected.
– Lord Shark the Unknown
Jul 24 at 1:44

Do you mean $D=cdotssetminusz-2$?
– Lord Shark the Unknown
Jul 24 at 1:45

Do you mean $D=cdotssetminusz-2$?
– Lord Shark the Unknown
Jul 24 at 1:45

@Dzoooks the region with radius between r and 1 (? or .1) is not simply connected. And a harmonic function is not holomorphic.
– Empy2
Jul 24 at 1:59

@Dzoooks the region with radius between r and 1 (? or .1) is not simply connected. And a harmonic function is not holomorphic.
– Empy2
Jul 24 at 1:59

1

1

For 2, find a harmonic function that is 0 on $|z|=r$, 1 on $|z|=1$, and combine it with the map to make it a function on the original region.
– Empy2
Jul 24 at 3:29

For 2, find a harmonic function that is 0 on $|z|=r$, 1 on $|z|=1$, and combine it with the map to make it a function on the original region.
– Empy2
Jul 24 at 3:29

 | 
show 1 more comment

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