why the figure of cos function does not symmetric around y axis?

I am tring to describe the figurs of these function .

fk(z) = (0.1000 + 0.3000i) + (0.4243 + 0.0017i)z + (0.9000- 0.0010i)z2

fb(z) = (0.1000 − 0.3000i)z−1 + (0.2121 − 0.0008i)z−2 + (0.9000 +0.0010i)z−3

but from the following figurs I faced many question:

1- Why cos(phase(f_k(z))) and cos(phase(f_b(1/z))) not symetric around the y axis?

2 - why there is overlab between the positiv and negative values of cos(phase(f_b(1/z))) in the origin?

3- what is the best points I have to describe them to cover these figures?

I will appriciate any help

[Torsten]

The plot for cos(angle(fb(1/z))) seems to be wrong.

Can you show the code how you created the plot ?

I get this:

x = -5:0.005:5;
y = -5:0.005:5;
[X,Y] = ndgrid(x,y);
fb = @(x,y)(0.1000 - 0.3000*1i)*(x+1i*y) + (0.2121 - 0.0008*1i)*(x+1i*y).^2 + (-0.9000 - 0.0010*1i)*(x+1i*y).^3;
Z = cos(angle(fb(X,Y)));
contourf(X,Y,Z)
caxis([-1 1])
colorbar 

[Paul]

Why are either expected to be symmetric around the y-axis, by which I presume means symmetric wrt imag(Z)?

fb is written as fb(z), but the question 1 asks about f_b(1/z). f_b(z) and f_b(1/z) are not the same. May be related to Torsten's observation.

[Aisha Mohamed]

Hi all and Thank you very much.

As Torsten asked, I used this code to get the plot both of cos(angle(f_k(z))) and cos(angle(f_b(1/z)))

p=[ ( 0.9000 - 0.0010i)  (0.4243 + 0.0017i) (0.1000 + 0.3000i) ]; 

p1=[(0.9000 + 0.0010i)  (0.2121 - 0.0008i) (0.1000 - 0.3000i)   (0)] ;

re_z = -6.005:.01:6.005;
im_z= -6.005:.01:6.005;

[re_z,im_z] = meshgrid(re_z,im_z);
z = re_z + 1i*im_z;

f_of_z_result = polyval(p,z); 

f_of_1_over_z_result = polyval(p1,1./z);

figure();
subplot(2,2,2)
surf(re_z,im_z,cos(angle(f_of_z_result)),'EdgeColor','none')
colorbar
title('cos(phase of f_k(z))')
xlabel('Z_R')  
ylabel('Z_I')
zlim([-5 5])   %adjust this value as needed
caxis([-1 1])    %adjust this value as needed 

subplot(2,2,4)
surf(re_z,im_z, cos(angle(f_of_1_over_z_result)),'EdgeColor','none')
colorbar
title(('phase of f(1/z)'))
title(('cos(phase of f_b(1/z))'))
xlabel('Z_R')  
ylabel('Z_I')
caxis([-1 1])    %adjust this value as needed 
zlim([-1 1])   %adjust this value as needed    
grid on

Is this wrong?

and as we know the cos figure is symmetric around y axis, why these figures do not?

I appreciate any help

[Aisha Mohamed]

When I plot these figures in plane I got the previous figures.

Is this wrong?

and as we know the cos figure is symmetric around y axis, why these figures do not?

I appreciate any help

Answer 1

[Chunru]

p=[ ( 0.9000 - 0.0010i)  (0.4243 + 0.0017i) (0.1000 + 0.3000i) ]; 
p1=[(0.9000 + 0.0010i)  (0.2121 - 0.0008i) (0.1000 - 0.3000i)   (0)] ;
re_z = -6.005:.01:6.005;
im_z= -6.005:.01:6.005;
[re_z,im_z] = meshgrid(re_z,im_z);

figure
z = re_z + 1i*im_z;
f_of_z_result = polyval(p,z); 
f_of_1_over_z_result = polyval(p1,1./z);

figure();
subplot(1,2,1)
surf(re_z,im_z,cos(angle(f_of_z_result)),'EdgeColor','none')
colorbar
title('cos(phase of f_k(z))')
xlabel('Z_R')  
ylabel('Z_I')
zlim([-5 5])   %adjust this value as needed
caxis([-1 1])    %adjust this value as needed 
view(2);
axis tight

subplot(1,2,2)
surf(re_z,im_z, cos(angle(f_of_1_over_z_result)),'EdgeColor','none')
colorbar
title(('phase of f(1/z)'))
title(('cos(phase of f_b(1/z))'))
xlabel('Z_R')  
ylabel('Z_I')
caxis([-1 1])    %adjust this value as needed 
zlim([-1 1])   %adjust this value as needed    
grid on
axis tight
view(2);

1- Why cos(phase(f_k(z))) and cos(phase(f_b(1/z))) not symetric around the y axis?

==> It is a general polynomial and cos(angle(f_x(x+iy))) is in general not equals to cos(angle(f_x(-x+iy))

2 - why there is overlab between the positiv and negative values of cos(phase(f_b(1/z))) in the origin?

==> What do you mean overlap? This is quite a general polynomial and it may not be straightforward to explain how the very nonlinear function cos(phase(f_b(1/z))) behavors.

3- what is the best points I have to describe them to cover these figures?

==> Without criteria given, it is not possible to tell what is the best.

[Torsten]

@Chunru

@Aisha Mohamed defined

fb(z) = (0.1000 − 0.3000i)z^−1 + (0.2121 − 0.0008i)z^−2 + (0.9000 +0.0010i)z^−3

So

fb(1/z) = (0.1000 − 0.3000i)z + (0.2121 − 0.0008i)z^2 + (0.9000 +0.0010i)z^3

Thus formally

f_of_1_over_z_result = polyval(p1,z);

in your code, not

f_of_1_over_z_result = polyval(p1,1./z);

Maybe @Aisha Mohamed wrote the above, but meant something else. I don't know.

[Aisha Mohamed]

Thank you Torsten. I find this code is logical and right to me. And I also find the following code from other MATLAB expert logical and right.

This is the other code,

% define the function f(z) in terms of its coefficients p:

% f(z) = z^2-2*z+1

p = [1 -2 1]; % polynomial coefficients p instead of anonymous function f

% evaluate f over all z in ([-1,1],[-1,1]):

f_of_z_result = polyval(p,z);

% evaluate f at 1/z:

f_of_1_over_z_result = polyval(p,1./z);

% plot the magnitude and phase of f(z) and f(1/z):

% (same as before)

figure();

subplot(2,2,1)

surf(re_z,im_z,abs(f_of_z_result),'EdgeColor','none')

title('|f(z)|')

subplot(2,2,2)

surf(re_z,im_z,angle(f_of_z_result),'EdgeColor','none')

title('phase of f(z)')

subplot(2,2,3)

surf(re_z,im_z,abs(f_of_1_over_z_result),'EdgeColor','none')

title('|f(1/z)|')

subplot(2,2,4)

surf(re_z,im_z,angle(f_of_1_over_z_result),'EdgeColor','none')

title('phase of f(1/z)')

I know you are experts and you know what you do, but I am still confuse which code I can use , specially I see some difference in the plot of $ cos(angle(f_b(1/z)) $ between these cods

May be there are some points are different in the two cods , and I could not recognize them

Could you please help me to get thesutible code from them to plot f(z) and f(1/z)?

I will appreciate any help.

[Torsten]

@Aisha Mohamed

Yes, as I suspected: you wrote something, but meant something else.

If you want to plot what is shown in your original graphics for cos(angle(fb(1/z))), you must define fb as

fb(z) = (0.1000 − 0.3000i)*z + (0.2121 − 0.0008i)*z^2+ (0.9000 +0.0010i)*z^3,

not as

fb(z) = (0.1000 − 0.3000i)z−1 + (0.2121 − 0.0008i)z−2 + (0.9000 +0.0010i)z−3

It's up to you to decide which of the definitions is correct. Depending on your decision, you get different plots.