8 edited tags
| link
    Notice removed Draw attention by Community
    Bounty Ended with Rahul's answer chosen by Community
7 corrected representation of eqn. 2 in the code.
source | link

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:

enter image description here

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:

enter image description here

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:

enter image description here

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
source | link

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:

enter image description here

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:

enter image description here

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:

enter image description here

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
source | link
    Tweeted twitter.com/#!/StackMma/status/511078075866439680
    Notice added Draw attention by Black Milk
    Bounty Started worth 50 reputation by Black Milk
4 deleted 131 characters in body
source | link
3 Provided more compatible code, and improved wording of question.
source | link
2 added 4 characters in body
source | link
1
source | link