where the jump of the phase function happen?

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Aisha Mohamed 2022-9-6

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评论： Aisha Mohamed 2023-4-10

I know that (if Iam correct), phase(f(z)) =arctan(f(z)) ,(where f(z) is complex number ) is multivalued function, that means for example if arctan(f(z))= x, there is infinite number of angles (x) has the same tan value = the value f(z). So if we want to make arctan continuous we have to ristrect the rang of this function in this interval (-pi/2 pi/2). where z=x+iy

My questions are:

1- How can I determined the points where the jump happen? are they when the real part of complex number=0?(but phase =arcta(y/x)= arctan (y/0) =pi/2)

2-why when I plotted the phase of the function f(z) in this interval (-pi/2 pi/2) , I still have the same jump which appear as discontinuity of the phase of the function f(z)?

I use this code

re_z = -pi/2:0.01:pi/2;

im_z = -pi/2:0.01:pi/2;

[re_z,im_z] = meshgrid(re_z,im_z);

z = re_z + 1i*im_z;

f_of_z_result = polyval(p,z);

figure();

subplot(2,2,1)

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

colorbar

title('phase(f_k(z))')

xlabel('Z_R')

ylabel('Z_I')

I appreciate any help

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Star Strider 2022-9-6

Reference — Why the figure of the phase discontinuous?

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Answer 1

Aisha Mohamed 2022-9-7

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Thanks Bruno

I know my a little information do not compare with expers like you and other experts in this group, but really I want to understad the worth information that you explane. Thank you so much. please allow me to ask more,

1- is this demi-line is the same for all second degree polynomail { y=0; x<=0, z=x+1i*y }?

2-please , what did you mean by your polynomial p(z) has two solutions of p(z)=x with x real and negative. ? I know second degree funtion has two root as a solution p(z)=0, but you mean p(z)=x with x real and negative. we have many values of z satisfay this condition not only two for example:

% -0.1320 - 0.0878i

% -0.1150 - 0.0756i

% -0.1067 - 0.0694i

2-if I have this polynomial (4th degree) p =[(0.6 - 0.7i) (-0.6000 + 0.0020i) (0.2449 + 0.0049i) (0.2000 + 0.0020i) (0.2 + 0.0010i) ] . is it has the same demi-line { y=0; x<=0, z=x+1i*y } and discontnue at four "places".

3 Becaufe of the discontinue at the negative values of x, can I avoied it by chosing this rang of z,

re_z = 0:0.01:pi/2;

im_z = 0:0.01:pi/2; and then I got this figure,

I appreciate any help.

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Bruno Luong 2022-9-7

编辑：Bruno Luong 2022-9-7

1- is this demi-line is the same for all second degree polynomail { y=0; x<=0, z=x+1i*y }?

This line is the place where there discontuinity of angle(z) it doesn't involves any polynomial yet.

2-please , what did you mean by your polynomial p(z) has two solutions of p(z)=x with x real and negative. ? I know second degree funtion has two root as a solution p(z)=0, but you mean p(z)=x with x real and negative. we have many values of z satisfay this condition not only two for example:

Or sorry, two z for each x. Becase it a solution of p(z)-x = 0

2-if I have this polynomial (4th degree) p =[(0.6 - 0.7i) (-0.6000 + 0.0020i) (0.2449 + 0.0049i) (0.2000 + 0.0020i) (0.2 + 0.0010i) ] . is it has the same demi-line { y=0; x<=0, z=x+1i*y } and discontnue at four "places".

Yes four (half) curves in complex plane, but again solution of P(z) = x with x real and x < 0. (Why you keep writing z = x + 1i*y, this is for discontuinity of angle(z) NOT for angle(P(z)).

3 Becaufe of the discontinue at the negative values of x, can I avoied it by chosing this rang of z,

You have to select a set in complex plane that is not crossed by any of the 4 above curves solution of { z : p(z) = x, x real and x <= 0 }. The geometry can be hard to described other than this algebric equation.

But yeah you can select a small rectangle/ball around a given point z0 so as it does cross any of the four curves, assuming the point z0 is not belong to the curve, meaning

|angle(polyval(p,z0))| is not equal to pi

Aisha Mohamed 2022-9-8

I have got excellent explination for this question: from the experts, Bruno Luong and Torsten,

Where the jump of the phase of complex polynomial happen? or How can I find the values that where the phase of complex polynomial is discontinuous?

and what I wrote here represent what I thought I understood, maybe not exactly what did they said (as I can not always find their worth information )

Is what I have undestoos and wrote here are correct, please?

1 -angle(z) = phase is always discontinue at the half line { y=0; x<=0, z=x+1i*y }. because the phase jumps from -pi for imaginary part y < 0 to +pi for y > 0.

2-there is a relationship between the degree of the polynomial and the the number of places that the angle(p(z) is discontnue at.For example the polynomial of second degree is discontinuse at two places both correposonds to p(z) real and negative.

3- According the following graph

3 Are the given points represent the places where the phase discontinuous? if so, why they do not lay on the line { y=0; x<=0, z=x+1i*y }? and what the line between two these points represent?

I really appreciate any help.

Bruno Luong 2022-9-11

编辑：Bruno Luong 2022-9-11

Correct statement:

angle(z) is discontinuous at { z : z real z <= 0}

For the second question: You again (for the 1000 times) confound the place of discontinuity of the angle(P(z)) and of angle(z).

Torsten 2022-9-11

Because the phase(p(z)) at z=(0+0i) equales 2.1055 not +pi or -pi.

If you look at the plot of the "discontinuity front", you can see that z=0 is not therein. So angle(p(z)) is continuous at z = 0. Again: angle(p(z)), not angle(z).

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Answer 2

Bruno Luong 2022-9-6

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编辑：Bruno Luong 2022-9-6

angle(z) is discontinue at the half line { y=0; x<=0, z=x+1i*y }.

The phase jumps from -pi for imaginary part y < 0 to +pi for y > 0. To make thing more complicated for y=0 in IEEE754 it can take the "IEEE-sign" of either +1 or -1, and the angle(z) returns +/-pi depending of the IEEE-sign of y (which is 0 mathematically).

In your case you have to determine when polyval(p,z)* is real and negative, which will implies angle discontinuity.

(*) you didn't tell us what is p.

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Bruno Luong 2022-9-6

在 MATLAB Online 中打开

1-What did you mean by the "IEEE-sign" please?

Illustration, the function getsgnbit returns IEEE-sign of x

getsgnbit=@(x)bitshift(typecast(x,'uint64'),-63);
zerop=1/inf,
zerop = 0
zerom=1/-inf,
zerom = 0
getsgnbit(zerop)
ans = uint64
    0
getsgnbit(zerom) % different than getsgnbit(zerop)
ans = uint64
    1
zp=complex(-1,zerop)
zp = -1.0000 + 0.0000i
zm=complex(-1,zerom)
zm = -1.0000 + -0.0000i
zp==zm
ans = logical
   1
angle(zp) % pi angle
ans = 3.1416
angle(zm) % -pi angle
ans = -3.1416

2-why the angle(z) is discontinue at the half line { y=0; x<=0, z=x+1i*y }.? can I say that:

On this half line phase =arctan(0/-x) =0

and phase =arctan(0/0) = undefine.

No: angle(z) is equivalent to atan2(imag(z),real(z))

3 -Also could you please how can I determine when polyval(p,z)* is real and negative?

p is is second degree polynomial so you should be able to solve

polyval(p,z) = x

for x that is real and x < 0, it must hacve 2 solutions z according to the fundamental algebra theorem. I believe that is why there are 2 lines with doscontinuity in your angle surface plot.

Aisha Mohamed 2022-9-7

When I took some values of polyval(p,z) such as (-0.0261 + 0.0053i) and (-0.1320 - 0.0878i) and studied the case when :

1-x<0 and y approch the zero from the positve side, I find the phase goes to pi,

and when x<0 and y approch the zero from the negative side, I find the phase goes to -pi

2-But when I took x>0 and y approch to zero from the positve and negative side, the phase goes to zero.

3- When (x<0 and y =0) the phase = pi and (x>0 and y =0), the phase = zero

Is this the explenation of what Bruno Luong said that ((angle(z) is discontinuies at the half line { y=0; x<=0, z=x+1i*y }.)). I am really still struggling with this point, how can I find and understand :

1- Why the jump happened in the { y=0; x<=0, z=x+1i*y }.?

2- from the following figure there are 2 lines with doscontinuity(exactly as what Bruno Luong said), is that mean there is another jump in x>0?

3- is the jump happen at the roots of this function (as seen in the figure I have two roots)? then, that means there is another discontinuity at x>0?

I appreciate any help

Bruno Luong 2022-9-12

编辑：Bruno Luong 2022-9-12

在 MATLAB Online 中打开

I also modify my code to avoid switching branches (the diagonal line):

p=[ ( 0.7756 - 0.4566i) ( 0.4347 + 0.0914i) (-0.1540 + 0.2600i) ];

x = linspace(-100,0,513);

a = p(1); b = p(2); c = p(3)-x;

% who recalls this formula?

delta = b^2-4*a*c;

z = (-b + [-1;1].*sqrt(delta))/(2*a);

% reorder the points so that each row contain the same "branch" of solution

cu = max(abs(z(:)));

extrapn = 2;

for j=2:size(z,2)

% prediction by fitting a polynomaial on old data

kp = max(j-extrapn,1):j;

zp = z(:,kp);

m = size(zp,2)-1;

order = m-1;

for i=1:size(zp,1)

Pi = polyfit(1:m, zp(i,1:m), order);

zp(i,end) = polyval(Pi,m+1);

end

M = matchpairs(abs(zp(:,end)-z(:,j).'),cu);

M = sortrows(M,2);

pj = M(:,1);

z(:,j) = z(pj,j);

end

z1 = z(1,:);

z2 = z(2,:);

% Group them as single complex curve

zjump = [z1, NaN, z2];

zjr = real(zjump);

zji = imag(zjump);

% Graphic

x=linspace(min(zjr),max(zjr));

y=linspace(min(zji),max(zji));

[X,Y]=meshgrid(x,y);

Z=X+1i*Y;

close all

surf(x,y,angle(polyval(p,Z)))

hold on

zlo = -5; % just below -pî, the lowest possible value of phase

plot3(zjr, zji, zlo*ones(size(zjr)), 'r', 'linewidth',2)

Bruno Luong 2022-9-12

编辑：Bruno Luong 2022-9-12

在 MATLAB Online 中打开

Here is the code for p(z) of polynomial of order > 2

clear
%p=[ (  0.7756 - 0.4566i)    ( 0.4347 + 0.0914i)   (-0.1540 + 0.2600i) ];
p = rand(1,7) + 1i*rand(1,7);
x = linspace(-100,0,513);
% Solve p(z) = x
npnts = length(x);
z = zeros(length(p)-1,npnts);
for k = 1:npnts
    pk = p;
    pk(end) = pk(end)-x(k);
    z(:,k) = roots(pk);
end
% reorder the points so that each row contain the same "branch" of solution
cu = max(abs(z(:)));
extrapn = 5; % how many points used to predict the next point in a branch
for j=2:npnts
    % prediction by fitting a polynomaial on old data
    kp = max(j-extrapn,1):j-1;
    m = size(kp,2);    
    % Build Vandermonde matrix: Vj = ((0:m-1)'/m).^(0:m-1);
    v = (0:m-1)'/m;
    Vj = zeros(m);
    vp = 1;
    Vj(:,1) = vp;
    for k=2:m
        vp = vp.*v;
        Vj(:,k) = vp;
    end
    zp = sum(Vj \ z(:,kp).',1);          
    M = matchpairs(abs(zp-z(:,j)),cu);
    M = sortrows(M,2);
    pj = M(:,1);
    z(:,j) = z(pj,j);
end
% Group them as single complex curve
z(:,end+1) = NaN;
zjump = reshape(z.', 1, []);
zjr = real(zjump);
zji = imag(zjump);
% Graphic
x=linspace(min(zjr),max(zjr));
y=linspace(min(zji),max(zji));
[X,Y]=meshgrid(x,y);
Z=X+1i*Y;
close all
surf(x,y,angle(polyval(p,Z)))
hold on
zlo = -10; %  just below -pî, the lowest possible value of phase
plot3(zjr, zji, zlo*ones(size(zjr)), 'r', 'linewidth',2);

Bruno Luong 2022-9-12

编辑：Bruno Luong 2022-9-13

在 MATLAB Online 中打开

One more simplification. The extrapolation can be computed with Pascal's triangle without using Vandermond matrix.

p = rand(1,7) + 1i*rand(1,7);
x = linspace(-100,0,513);
% Solve p(z) = x
npnts = length(x);
z = zeros(length(p)-1,npnts);
for k = 1:npnts
    pk = p;
    pk(end) = pk(end)-x(k);
    z(:,k) = roots(pk);
end
% reorder the points so that each row contain the same "branch" of solution
cu = max(abs(z(:)));
extrapn = 5; % how many points used to predict the next point in a branch
Pascal = pascal(extrapn+1,1);
for j=2:npnts
    % prediction by fitting a polynomaial on old data
    if j <= extrapn+1        
        a = -Pascal(j,j:-1:2).';
    end
    zp = z(:,max(j-extrapn,1):j-1)*a;
    % find permutation to match the prediction
    M = matchpairs(abs(zp.'-z(:,j)),cu);
    M = sortrows(M,2);
    z(:,j) = z(M(:,1),j);
end
% Group them as single complex curve
z(:,end+1) = NaN;
zjump = reshape(z.', 1, []);
zjr = real(zjump);
zji = imag(zjump);
% Graphic
x=linspace(min(zjr),max(zjr));
y=linspace(min(zji),max(zji));
[X,Y]=meshgrid(x,y);
Z=X+1i*Y;
close all
surf(x,y,angle(polyval(p,Z)))
hold on
zlo = -10; %  just below -pî, the lowest possible value of phase
plot3(zjr, zji, zlo*ones(size(zjr)), 'r', 'linewidth',2);

Torsten 2023-4-7

编辑：Torsten 2023-4-7

在 MATLAB Online 中打开

You took the extreme case x = 0 for the two points you selected. If you restrict yourself to the values of z for which p(z) is real and < 0 instead of <= 0, it works out fine.

syms z x
p = (-0.1540 + 0.2600*1i)+ ( 0.4347 + 0.0914*1i)*z+(  0.7756 - 0.4566*1i)*z.^2;
assume(x,'real')
ps = p - x;
sol = solve(ps==0,z);
rsol1 = matlabFunction(real(sol(1)));
isol1 = matlabFunction(imag(sol(1)));
rsol2 = matlabFunction(real(sol(2)));
isol2 = matlabFunction(imag(sol(2)));
format long
x = 0;
z1 = rsol1(x) + 1i*isol1(x);
z2 = rsol2(x) + 1i*isol2(x);
double(angle(subs(p,z,z1)))
ans = 
  -1.265169150966033
double(angle(subs(p,z,z2)))
ans = 
   1.651667331151838
x = -1e-8;
z1 = rsol1(x) + 1i*isol1(x);
z2 = rsol2(x) + 1i*isol2(x);
disp(pi)
   3.141592653589793
double(angle(subs(p,z,z1)))
ans = 
  -3.141592648089039
double(angle(subs(p,z,z2)))
ans = 
   3.141592641453825

Aisha Mohamed 2023-4-10

Thanks, Torsten you are really genius.

I used only the values of z where f(z)=x and x<0 , it works out fine.

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where the jump of the phase function happen?

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where the jump of the phase function happen?

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