8 edited tags | link edited Aug 8 '15 at 18:53 Michael E2 159k1313 gold badges218218 silver badges518518 bronze badges Notice removed Draw attention by Community♦ occurred Sep 22 '14 at 8:38 Bounty Ended with Rahul's answer chosen by Community♦ occurred Sep 22 '14 at 8:38 7 corrected representation of eqn. 2 in the code. edited Sep 16 '14 at 6:13 Black Milk 31822 silver badges1414 bronze badges I'm working on a complex transformation sketch that I'd like to create via mathematica. I'm working with the function $$w=z^2\tag{0},$$ where its real and imaginary parts are $$u=x^2-y^2$$ and $$v=2xy$$, respectively. Now, I want to sketch the image under the given transformation of each branch of the hyperbola $$x^2-y^2=1\tag{1}$$ where the right and left branches (in blue) are oriented upward and downwards, respectively. For example, the blue branches of the hyperbola in (1) appear as such: Similarly, for the hyperbola $$2xy=2,\tag{2}$$the left and right branches (in orange) are oriented upwards and downwards, respectively. How do I sketch the images of these hyperbolas under the complex function $$w=z^2$$, while taking into account the orientation of their respective branches? By 'image' I mean to say the image of the function (0) applied to the hyperbolas (1) and (2). I expect the image of the hyperbola (under $$w$$) in (1), to be the line $$u=1$$, oriented updwards, and the image of the hyperbola (under $$w$$) in (2) to be the line $$v=2$$, oriented rightwards. Here's what I have so far; however, getting the orientation arrows requires that I load the CurvesGraphics6 package from here.  r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 21}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White, Oriented -> True]  [UPDATED CODE] Courtesy of Eldo and Mark McClure, I was able to apply orientation arrows to the Hyperbolas without using the aforementioned package. r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 21}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White] Show[r1, r1 /. Line[pts_] :> Arrow[pts, 2]]  I'm working on a complex transformation sketch that I'd like to create via mathematica. I'm working with the function $$w=z^2\tag{0},$$ where its real and imaginary parts are $$u=x^2-y^2$$ and $$v=2xy$$, respectively. Now, I want to sketch the image under the given transformation of each branch of the hyperbola $$x^2-y^2=1\tag{1}$$ where the right and left branches (in blue) are oriented upward and downwards, respectively. For example, the blue branches of the hyperbola in (1) appear as such: Similarly, for the hyperbola $$2xy=2,\tag{2}$$the left and right branches (in orange) are oriented upwards and downwards, respectively. How do I sketch the images of these hyperbolas under the complex function $$w=z^2$$, while taking into account the orientation of their respective branches? By 'image' I mean to say the image of the function (0) applied to the hyperbolas (1) and (2). I expect the image of the hyperbola (under $$w$$) in (1), to be the line $$u=1$$, oriented updwards, and the image of the hyperbola (under $$w$$) in (2) to be the line $$v=2$$, oriented rightwards. Here's what I have so far; however, getting the orientation arrows requires that I load the CurvesGraphics6 package from here.  r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White, Oriented -> True]  [UPDATED CODE] Courtesy of Eldo and Mark McClure, I was able to apply orientation arrows to the Hyperbolas without using the aforementioned package. r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White] Show[r1, r1 /. Line[pts_] :> Arrow[pts, 2]]  I'm working on a complex transformation sketch that I'd like to create via mathematica. I'm working with the function $$w=z^2\tag{0},$$ where its real and imaginary parts are $$u=x^2-y^2$$ and $$v=2xy$$, respectively. Now, I want to sketch the image under the given transformation of each branch of the hyperbola $$x^2-y^2=1\tag{1}$$ where the right and left branches (in blue) are oriented upward and downwards, respectively. For example, the blue branches of the hyperbola in (1) appear as such: Similarly, for the hyperbola $$2xy=2,\tag{2}$$the left and right branches (in orange) are oriented upwards and downwards, respectively. How do I sketch the images of these hyperbolas under the complex function $$w=z^2$$, while taking into account the orientation of their respective branches? By 'image' I mean to say the image of the function (0) applied to the hyperbolas (1) and (2). I expect the image of the hyperbola (under $$w$$) in (1), to be the line $$u=1$$, oriented updwards, and the image of the hyperbola (under $$w$$) in (2) to be the line $$v=2$$, oriented rightwards. Here's what I have so far; however, getting the orientation arrows requires that I load the CurvesGraphics6 package from here.  r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 1}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White, Oriented -> True]  [UPDATED CODE] Courtesy of Eldo and Mark McClure, I was able to apply orientation arrows to the Hyperbolas without using the aforementioned package. r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 1}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White] Show[r1, r1 /. Line[pts_] :> Arrow[pts, 2]]  6 added 24 characters in body edited Sep 16 '14 at 0:03 Black Milk 31822 silver badges1414 bronze badges I'm working on a complex transformation sketch that I'd like to create via mathematica. I'm working with the function $$w=z^2\tag{0},$$ where its real and imaginary parts are $$u=x^2-y^2$$ and $$v=2xy$$, respectively. Now, I want to sketch the image under the given transformation of each branch of the hyperbola $$x^2-y^2=1\tag{1}$$ where the right and left branches (in blue) are oriented upward and downwards, respectively. For example, the blue branches of the hyperbola in (1) appear as such: Similarly, for the hyperbola $$2xy=2,\tag{2}$$the left and right branches (in orange) are oriented upwards and downwards, respectively. How do I sketch the images of these hyperbolas under the complex function $$w=z^2$$, while taking into account the orientation of their respective branches? By 'image' I mean to say the image of the function (0) applied to the hyperbolas (1) and (2). I expect the image of the hyperbola (under $$w$$) in (1), to be the line $$u=1$$, oriented updwards, and the image of the hyperbola (under $$w$$) in (2) to be the line $$v=2$$, oriented rightwards. Here's what I have so far; however, getting the orientation arrows requires that I load the CurvesGraphics6 package from here.  r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White, Oriented -> True]  [UPDATED CODE] Courtesy of Eldo and Mark McClure, I was able to apply orientation arrows to the Hyperbolas without using the aforementioned package. r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White] Show[r1, r1 /. Line[pts_] :> Arrow[pts, 2]]  I'm working on a complex transformation sketch that I'd like to create via mathematica. I'm working with the function $$w=z^2\tag{0},$$ where its real and imaginary parts are $$u=x^2-y^2$$ and $$v=2xy$$, respectively. Now, I want to sketch the image under the given transformation of each branch of the hyperbola $$x^2-y^2=1\tag{1}$$ where the right and left branches (in blue) are oriented upward and downwards, respectively. For example, the blue branches of the hyperbola in (1) appear as such: Similarly, for the hyperbola $$2xy=2,\tag{2}$$the left and right branches (in orange) are oriented upwards and downwards, respectively. How do I sketch the images of these hyperbolas under the complex function $$w=z^2$$, while taking into account the orientation of their respective branches? By 'image' I mean to say the image of the function (0) applied to the hyperbolas (1) and (2). I expect the image of the hyperbola in (1), to be the line $$u=1$$, oriented updwards, and the image of the hyperbola in (2) to be the line $$v=2$$, oriented rightwards. Here's what I have so far; however, getting the orientation arrows requires that I load the CurvesGraphics6 package from here.  r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White, Oriented -> True]  [UPDATED CODE] Courtesy of Eldo and Mark McClure, I was able to apply orientation arrows to the Hyperbolas without using the aforementioned package. r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White] Show[r1, r1 /. Line[pts_] :> Arrow[pts, 2]]  I'm working on a complex transformation sketch that I'd like to create via mathematica. I'm working with the function $$w=z^2\tag{0},$$ where its real and imaginary parts are $$u=x^2-y^2$$ and $$v=2xy$$, respectively. Now, I want to sketch the image under the given transformation of each branch of the hyperbola $$x^2-y^2=1\tag{1}$$ where the right and left branches (in blue) are oriented upward and downwards, respectively. For example, the blue branches of the hyperbola in (1) appear as such: Similarly, for the hyperbola $$2xy=2,\tag{2}$$the left and right branches (in orange) are oriented upwards and downwards, respectively. How do I sketch the images of these hyperbolas under the complex function $$w=z^2$$, while taking into account the orientation of their respective branches? By 'image' I mean to say the image of the function (0) applied to the hyperbolas (1) and (2). I expect the image of the hyperbola (under $$w$$) in (1), to be the line $$u=1$$, oriented updwards, and the image of the hyperbola (under $$w$$) in (2) to be the line $$v=2$$, oriented rightwards. Here's what I have so far; however, getting the orientation arrows requires that I load the CurvesGraphics6 package from here.  r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White, Oriented -> True]  [UPDATED CODE] Courtesy of Eldo and Mark McClure, I was able to apply orientation arrows to the Hyperbolas without using the aforementioned package. r1 = ContourPlot[{x^2 - y^2 == 1, x*y == 2}, {x, -4, 4}, {y, -4, 4}, Axes -> True, Frame -> False, PlotRange -> Automatic, Background -> White] Show[r1, r1 /. Line[pts_] :> Arrow[pts, 2]]  5 added 180 characters in body, improved clarity of problem statement edited Sep 15 '14 at 22:53 Black Milk 31822 silver badges1414 bronze badges Tweeted twitter.com/#!/StackMma/status/511078075866439680 occurred Sep 14 '14 at 9:04 Notice added Draw attention by Black Milk occurred Sep 14 '14 at 6:43 Bounty Started worth 50 reputation by Black Milk occurred Sep 14 '14 at 6:43 4 deleted 131 characters in body edited Sep 14 '14 at 6:39 Black Milk 31822 silver badges1414 bronze badges 3 Provided more compatible code, and improved wording of question. edited Sep 10 '14 at 14:33 Black Milk 31822 silver badges1414 bronze badges 2 added 4 characters in body edited Sep 10 '14 at 14:13 Black Milk 31822 silver badges1414 bronze badges 1 asked Sep 10 '14 at 14:04 Black Milk 31822 silver badges1414 bronze badges