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% Let us show some complex mappings. The main tool is "cplot".
% First we define a complex grid around the unit circle.
>r=exp(linspace(0.000001,1,20)); phi=linspace(0.0000001,2*pi,120)';
>z=r*exp(phi*I);
% The we plot it, and add a x-y-ticks.
>setplot(-4,4,-4,4); cplot(z); xplot(); wait(20);
% Press the TAB key to see the graphics, if the text window
% hides it.
%
% If the plot does not look circular, you should resize
% the graphics window.
%
% The following is the Joukowski mapping, which maps the
% outside of the unit circle onto the outside of the unit
% interval [-1,1] one-to-one.
>function J(z)
$return (z+1/z)/2
$endfunction
% We map the grid and plot the result.
>setplot(-1.7,1.7,-1.7,1.7); cplot(J(z)); xplot(); wait(20);
% This is the inverse of the Joukowski mapping. We have
% to define cases here.
>function invJ(z)
$w=sqrt(z^2-1);
$return z+(re(z)>0)*w-(re(z)<=0)*w;
$endfunction
% Using these functions, we can map the outside of the
% unit circle to the outside of a cross.
>setplot(-2,2,-2,2); cplot(J(I*invJ(J(z)*sqrt(2)))); xplot(); wait(20);
% The following maps the inside of the unit circle to itself,
% taking 0 to a.
>function K(z,a)
$return (z+a)/(1+conj(a)*z)
$endfunction
% Let us view the image of our grid, when the outside of
% the circle is mapped to the inside and then 0 to 0.5.
>setplot(-1,1,-1,1); cplot(K(1/z,0.5)); xplot(); wait(20);
% The following function maps the outside of the unit circle
% to the upper half plane.
>function H(z)
$return (1i*z+1)/(1i+z)
$endfunction
% We can view the image of our grid. Note that infinity
% is mapped to i.
>setplot(-3,3,0,6); cplot(H(z)); xplot(); wait(20);
% This is the inverse function, mapping the upper half
% plane to the outside of the unit circle.
>function invH(z)
$return (1i*z-1)/(1i-z)
$endfunction
% This function maps the outside of the unit circle onto
% itself, and takes a to infinity.
>function L(z,a)
$return (a/conj(a))*(conj(a)*z-1)/(a-z)
$endfunction
% In the following picture, 5 goes to infinity.
>setplot(-6,6,-6,6); cplot(L(z,5)); xplot(); wait(20);
% Now we map the outside of the unit circle to the outside
% of the arc with angle Pi.
>setplot(-2,2,-1.5,2.5); cplot(invH(J(L(z,-invJ(1i))))); xplot(); wait(20);
>w=invH(J(L(z,-invJ(1i))));
>zoom(4); viewheight(35); viewangle(40); solid(re(w)-0.7,im(w),abs(z)-3);
>setplot(-2,2,-2,2);
% The following picture shows the level lines of the Green's
% function for two intervals and its harmonic conjugate.
>w=sqrt(J(-z)-J(-exp(1))); cplot(w); hold on; wait(20);
>w1=sqrt(J(-exp(1)*z)-J(-exp(1))); cplot(w1); cplot(-w); cplot(-w1); hold off; wait(20);
>xplot(); wait(20);
>
>
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