Find a conformal map from onto an annulus

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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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    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














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

















  • 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












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













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












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 17:06
























asked Jul 24 at 0:33









Rachel.L

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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












  • 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















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