Hi Simran,
I understand you want to know the reason behind the transformation of w and plot the surface parametrization of the equation
using the transformation of w. Regarding the transformation, I believe that the unit disk
maps onto the left half of the plane . The proof is as follows: We know
. After substitution we get, ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1562894/image.png)
Now, Squaring both sides, we get
After simplification, we obtain ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1562904/image.png)
Whereas the range
mentioned in the caption of the right image specifies the area over which the function
is calculated but that doesn’t mention as the transformation to have a range of ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1562914/image.png)
To plot the function
, you can check the following steps used: 1.Taking z belong to the ball of radius 1 centered around origin,
We generate the points as follows.
theta=linspace(0,2*pi,np);
radius= linspace(0,1,np);
2.Create a mesh grid, which produces the polar coordinates of the disk.
[R,Angle] = meshgrid(radius,theta);
3.Now, Z corresponds to the location of point in a complex plane, and W is the transformation
Z=R.*(cos(Angle)+1i*sin(Angle));
4.Defining the function for each of the 3D coordinates using the function handle
xfunc=@(w) real(w-1./w)/4;
yfunc=@(w) imag (log (w))/2;
zfunc=@(w) imag (w+ 1./w)/4;
5.Now, we obtain the coordinate by substituting the values into the functions and get the plot using surf function
surf(xfunc(W), yfunc(W), zfunc(W));
6. To obtain the surface plot of a general surface D. You can generate the points corresponding to D following the steps from (1) to (3) to plot.
For understanding the function used, you check the following links:
I hope this resolves your query.
Thanks,
Rangesh.