Mappings from the z plane into the w plane












0














I have solved the following problem , but I am not sure if my solution is correct : A Square $S$ has vertices $(0,0), (1,0), (1,1), (0,1)$.



Part 1: Determine the region in the $W$ plane which $S$ is mapped under the transformation $w=frac{1}{z+1}$.



Part 2: Same transformation but now change the vertices to $(1,1), (-1,1), (-1,-1), (1,-1)$.






complex-analysis







share|cite|improve this question
























  • So give your solution as part of your question. Also, give some thought to why the second question might be different.
    – Ted Shifrin
    Jan 18 '15 at 5:19










  • can you break up the transformation $w = frac{1}{z+1}$ into simpler transformations:(a) translation, (b) inversion, (c) reflection on the $x$-axis? it would be easier to find the image of the sides under the simple transformations.
    – abel
    Jan 18 '15 at 6:31










  • My solution which I want to check is this :For the first part I get Circles with Center 1/2 and -1/2 (on the v axis) , the radius for those circles is 1/2.I also get a circle of Center 1/4 on the u axis with radius 1/4. I shade the region inside the circles .For part 2 I get a circle with Center -1/2 on the v axis and one with center 1/2 on the u axis .The radius here is 1/2 .I shade these regions also
    – herashefat
    Jan 18 '15 at 7:21
















0














I have solved the following problem , but I am not sure if my solution is correct : A Square $S$ has vertices $(0,0), (1,0), (1,1), (0,1)$.



Part 1: Determine the region in the $W$ plane which $S$ is mapped under the transformation $w=frac{1}{z+1}$.



Part 2: Same transformation but now change the vertices to $(1,1), (-1,1), (-1,-1), (1,-1)$.






complex-analysis







share|cite|improve this question
























  • So give your solution as part of your question. Also, give some thought to why the second question might be different.
    – Ted Shifrin
    Jan 18 '15 at 5:19










  • can you break up the transformation $w = frac{1}{z+1}$ into simpler transformations:(a) translation, (b) inversion, (c) reflection on the $x$-axis? it would be easier to find the image of the sides under the simple transformations.
    – abel
    Jan 18 '15 at 6:31










  • My solution which I want to check is this :For the first part I get Circles with Center 1/2 and -1/2 (on the v axis) , the radius for those circles is 1/2.I also get a circle of Center 1/4 on the u axis with radius 1/4. I shade the region inside the circles .For part 2 I get a circle with Center -1/2 on the v axis and one with center 1/2 on the u axis .The radius here is 1/2 .I shade these regions also
    – herashefat
    Jan 18 '15 at 7:21














0












0








0


1





I have solved the following problem , but I am not sure if my solution is correct : A Square $S$ has vertices $(0,0), (1,0), (1,1), (0,1)$.



Part 1: Determine the region in the $W$ plane which $S$ is mapped under the transformation $w=frac{1}{z+1}$.



Part 2: Same transformation but now change the vertices to $(1,1), (-1,1), (-1,-1), (1,-1)$.






complex-analysis







share|cite|improve this question















I have solved the following problem , but I am not sure if my solution is correct : A Square $S$ has vertices $(0,0), (1,0), (1,1), (0,1)$.



Part 1: Determine the region in the $W$ plane which $S$ is mapped under the transformation $w=frac{1}{z+1}$.



Part 2: Same transformation but now change the vertices to $(1,1), (-1,1), (-1,-1), (1,-1)$.






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 '15 at 6:29









Daniel W. Farlow

17.5k114488




17.5k114488










asked Jan 18 '15 at 5:04









herashefat

3471216




3471216












  • So give your solution as part of your question. Also, give some thought to why the second question might be different.
    – Ted Shifrin
    Jan 18 '15 at 5:19










  • can you break up the transformation $w = frac{1}{z+1}$ into simpler transformations:(a) translation, (b) inversion, (c) reflection on the $x$-axis? it would be easier to find the image of the sides under the simple transformations.
    – abel
    Jan 18 '15 at 6:31










  • My solution which I want to check is this :For the first part I get Circles with Center 1/2 and -1/2 (on the v axis) , the radius for those circles is 1/2.I also get a circle of Center 1/4 on the u axis with radius 1/4. I shade the region inside the circles .For part 2 I get a circle with Center -1/2 on the v axis and one with center 1/2 on the u axis .The radius here is 1/2 .I shade these regions also
    – herashefat
    Jan 18 '15 at 7:21


















  • So give your solution as part of your question. Also, give some thought to why the second question might be different.
    – Ted Shifrin
    Jan 18 '15 at 5:19










  • can you break up the transformation $w = frac{1}{z+1}$ into simpler transformations:(a) translation, (b) inversion, (c) reflection on the $x$-axis? it would be easier to find the image of the sides under the simple transformations.
    – abel
    Jan 18 '15 at 6:31










  • My solution which I want to check is this :For the first part I get Circles with Center 1/2 and -1/2 (on the v axis) , the radius for those circles is 1/2.I also get a circle of Center 1/4 on the u axis with radius 1/4. I shade the region inside the circles .For part 2 I get a circle with Center -1/2 on the v axis and one with center 1/2 on the u axis .The radius here is 1/2 .I shade these regions also
    – herashefat
    Jan 18 '15 at 7:21
















So give your solution as part of your question. Also, give some thought to why the second question might be different.
– Ted Shifrin
Jan 18 '15 at 5:19




So give your solution as part of your question. Also, give some thought to why the second question might be different.
– Ted Shifrin
Jan 18 '15 at 5:19












can you break up the transformation $w = frac{1}{z+1}$ into simpler transformations:(a) translation, (b) inversion, (c) reflection on the $x$-axis? it would be easier to find the image of the sides under the simple transformations.
– abel
Jan 18 '15 at 6:31




can you break up the transformation $w = frac{1}{z+1}$ into simpler transformations:(a) translation, (b) inversion, (c) reflection on the $x$-axis? it would be easier to find the image of the sides under the simple transformations.
– abel
Jan 18 '15 at 6:31












My solution which I want to check is this :For the first part I get Circles with Center 1/2 and -1/2 (on the v axis) , the radius for those circles is 1/2.I also get a circle of Center 1/4 on the u axis with radius 1/4. I shade the region inside the circles .For part 2 I get a circle with Center -1/2 on the v axis and one with center 1/2 on the u axis .The radius here is 1/2 .I shade these regions also
– herashefat
Jan 18 '15 at 7:21




My solution which I want to check is this :For the first part I get Circles with Center 1/2 and -1/2 (on the v axis) , the radius for those circles is 1/2.I also get a circle of Center 1/4 on the u axis with radius 1/4. I shade the region inside the circles .For part 2 I get a circle with Center -1/2 on the v axis and one with center 1/2 on the u axis .The radius here is 1/2 .I shade these regions also
– herashefat
Jan 18 '15 at 7:21










1 Answer
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when we move from $(0,0) rightarrow (1,0)$ then $0leq xleq 1$ and $y=0$ under the given transformation $w=dfrac{1}{z+1}=dfrac{x+1-iy}{(x+1)^2+y^2}$



then $0leq xleq 1implies 1leq x+1leq 2implies 1geqdfrac{1}{x+1}geq dfrac{1}{2}$ Thus under the transformation if $w=u+iv$ then $dfrac{1}{2}leq uleq 1$ and $v=0$



again when $0leq yleq 1$



$dfrac{1}{1+y^2}-idfrac{y}{1+y^2}$ provided $x=0$



Hence the real part $0geq frac{-y}{1+y^2}geq dfrac{-1}{2}$



and the imaginary part $dfrac{1}{2}leq dfrac{1}{1+y^2}leq 1$.Now can you find the rest






share|cite|improve this answer





















  • please See my answer above ,I have found the regions which are circles, I am just not sure if my solution is correct
    – herashefat
    Jan 18 '15 at 7:39











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when we move from $(0,0) rightarrow (1,0)$ then $0leq xleq 1$ and $y=0$ under the given transformation $w=dfrac{1}{z+1}=dfrac{x+1-iy}{(x+1)^2+y^2}$



then $0leq xleq 1implies 1leq x+1leq 2implies 1geqdfrac{1}{x+1}geq dfrac{1}{2}$ Thus under the transformation if $w=u+iv$ then $dfrac{1}{2}leq uleq 1$ and $v=0$



again when $0leq yleq 1$



$dfrac{1}{1+y^2}-idfrac{y}{1+y^2}$ provided $x=0$



Hence the real part $0geq frac{-y}{1+y^2}geq dfrac{-1}{2}$



and the imaginary part $dfrac{1}{2}leq dfrac{1}{1+y^2}leq 1$.Now can you find the rest






share|cite|improve this answer





















  • please See my answer above ,I have found the regions which are circles, I am just not sure if my solution is correct
    – herashefat
    Jan 18 '15 at 7:39
















0














when we move from $(0,0) rightarrow (1,0)$ then $0leq xleq 1$ and $y=0$ under the given transformation $w=dfrac{1}{z+1}=dfrac{x+1-iy}{(x+1)^2+y^2}$



then $0leq xleq 1implies 1leq x+1leq 2implies 1geqdfrac{1}{x+1}geq dfrac{1}{2}$ Thus under the transformation if $w=u+iv$ then $dfrac{1}{2}leq uleq 1$ and $v=0$



again when $0leq yleq 1$



$dfrac{1}{1+y^2}-idfrac{y}{1+y^2}$ provided $x=0$



Hence the real part $0geq frac{-y}{1+y^2}geq dfrac{-1}{2}$



and the imaginary part $dfrac{1}{2}leq dfrac{1}{1+y^2}leq 1$.Now can you find the rest






share|cite|improve this answer





















  • please See my answer above ,I have found the regions which are circles, I am just not sure if my solution is correct
    – herashefat
    Jan 18 '15 at 7:39














0












0








0






when we move from $(0,0) rightarrow (1,0)$ then $0leq xleq 1$ and $y=0$ under the given transformation $w=dfrac{1}{z+1}=dfrac{x+1-iy}{(x+1)^2+y^2}$



then $0leq xleq 1implies 1leq x+1leq 2implies 1geqdfrac{1}{x+1}geq dfrac{1}{2}$ Thus under the transformation if $w=u+iv$ then $dfrac{1}{2}leq uleq 1$ and $v=0$



again when $0leq yleq 1$



$dfrac{1}{1+y^2}-idfrac{y}{1+y^2}$ provided $x=0$



Hence the real part $0geq frac{-y}{1+y^2}geq dfrac{-1}{2}$



and the imaginary part $dfrac{1}{2}leq dfrac{1}{1+y^2}leq 1$.Now can you find the rest






share|cite|improve this answer












when we move from $(0,0) rightarrow (1,0)$ then $0leq xleq 1$ and $y=0$ under the given transformation $w=dfrac{1}{z+1}=dfrac{x+1-iy}{(x+1)^2+y^2}$



then $0leq xleq 1implies 1leq x+1leq 2implies 1geqdfrac{1}{x+1}geq dfrac{1}{2}$ Thus under the transformation if $w=u+iv$ then $dfrac{1}{2}leq uleq 1$ and $v=0$



again when $0leq yleq 1$



$dfrac{1}{1+y^2}-idfrac{y}{1+y^2}$ provided $x=0$



Hence the real part $0geq frac{-y}{1+y^2}geq dfrac{-1}{2}$



and the imaginary part $dfrac{1}{2}leq dfrac{1}{1+y^2}leq 1$.Now can you find the rest







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 18 '15 at 6:11









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  • please See my answer above ,I have found the regions which are circles, I am just not sure if my solution is correct
    – herashefat
    Jan 18 '15 at 7:39


















  • please See my answer above ,I have found the regions which are circles, I am just not sure if my solution is correct
    – herashefat
    Jan 18 '15 at 7:39
















please See my answer above ,I have found the regions which are circles, I am just not sure if my solution is correct
– herashefat
Jan 18 '15 at 7:39




please See my answer above ,I have found the regions which are circles, I am just not sure if my solution is correct
– herashefat
Jan 18 '15 at 7:39


















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