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](https://i.stack.imgur.com/WFKIM.png)

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](http://en.wikipedia.org/wiki/Image_(mathematics)/ "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](http://users.dimi.uniud.it/~gianluca.gorni/Mma/Mma.html#CurvesGraphics "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]]

```