Winding Numbers

Example

Calculate the winding number of the image of circles of radii 1/2, 1 and 2 under the map f(z)  =  ((z^3) + 2)/z.

Solution

The winding number equals the number of zeros - the number of poles inside the curve. The zeros of f(z) are the cube roots of -2. Since,

N[(2^(1/3))]

all three zeros lie inside the circle of radius 2, but none lie inside the circles of radii 1/2 or 1.  f(z) has a simple pole at the origin which lies in all three circles.
Therefore, the winding numbers for the images of the circles of radii 1/2 and 1 are both 0 - 1 = -1. The winding number for the image of the circle of radius 2 is 3  -  1  =  2.
Here are some graphical representations.

z  =  x  +  I y;
u = Re[((z^3)  +  2)/z] // ComplexExpand; 
v = Im[((z^3)  +  2)/z] // ComplexExpand;
x  =  r Cos[t];  y  =  r Sin[t];

The image of thecircle of radius 1/2 (animate the graphic by selecting it and pressingCTRL-Y).

r  =  .5;
Do[ParametricPlot[{u, v}, {t, 0, p}, 
        PlotRange -> {{-5, 5}, {-5, 5}}], {p, π/4, 
        2  π, π/4}]

The curve circles the origin counterclockwise once, so the winding number is -1.

The image of the circle of radius 1.

r  = 1;
Do[ParametricPlot[{u, v}, {t, 0, p}, 
        PlotRange -> {{-5, 5}, {-5, 5}}], {p, π/4, 
        2  π, π/4}]

The curve still circles the origin counterclockwise once, so the winding number is -1.

The image of the circle of radius 2.

r  = 2;
Do[ParametricPlot[{u, v}, {t, 0, p}, 
        PlotRange -> {{-5, 5}, {-5, 5}}], {p, π/4, 
        2  π, π/4}]

Now the curve circles the origin twice in a counter-clockwise direction, so the winding number is 2.
Next are images of the circles of r  =   1/2 to 2 in steps of .1. You can see parts of the curve cross over the origin when r passes through the cube root of 2, changing the winding number from -1 to 2.

Do[ParametricPlot[{u, v}, {t, 0, 2  π}, 
        PlotRange -> {{-5, 5}, {-5, 5}}], {r,  .5, 
        2,  .1}]

Example

Calculate the winding number of the image of circles of radii 1/4 and 1 under the map f(z)  =  (((2  z - 1))^7)/(z^3).

Solution

The winding number equals the number of zeros - the number of poles inside the curve. f(z) has a zero of order 7 at  1/2, and a pole of order three at the origin. Therefore, the winding number of the image of the circle of radius 1/4 is 0 - 3 = -3, and the winding number of the image of the circle of radius 1 is 7 - 3 = 4.

Here are some graphical representations.

z  =  x  +  I y;
u = Re[(((2  z - 1))^7)/(z^3)] // ComplexExpand; 
v = Im[(((2  z - 1))^7)/(z^3)] // ComplexExpand;
x  =  r Cos[t];  y  =  r Sin[t];

The image of the circle of radius 1/4.
The start of the curve shows that it is moving clockwise.

r  =  .25;
ParametricPlot[{u, v}, {t, 0, π/2}, 
      PlotRange -> {{-15, 10}, {-5, 5}}]

The curve wraps around twice near the origin:

r  =  .25;
ParametricPlot[{u, v}, {t, 0, 2  π}, 
      PlotRange -> {{-15, 10}, {-5, 5}}]

and once more much further away:

r  = 1;
ParametricPlot[{u, v}, {t, 0, 2  π}, 
      PlotRange -> {{-500, 2500}, {-1500, 1500}}]

Thus, the winding number is -3.

The image of the circle of radius 1.
The start of the curve shows that it is moving counter-clockwise.

r  = 1;
ParametricPlot[{u, v}, {t, 0, π/2}, 
      PlotRange -> {{-10, 15}, {-5, 5}}]

The curve wraps around twice near the origin:

r  = 1;
ParametricPlot[{u, v}, {t, 0, 2  π}, 
      PlotRange -> {{-10, 15}, {-5, 5}}]

once more further away:

r  = 1;
ParametricPlot[{u, v}, {t, 0, 2  π}, 
      PlotRange -> {{-300, 100}, {-100, 100}}]

and a final loop very far from the origin:

r  = 1;
ParametricPlot[{u, v}, {t, 0, 2  π}, 
      PlotRange -> {{-500, 2500}, {-1500, 1500}}]

Thus, the total winding number is 4.