How these figures changed?

I plotted these functions in the extended complex plane and some experts in this community helped me to analyze these figures and we got these explanations (Thanks to all the experts who helped me a lot):
1-In the yellow area the cos(phase(f(z))) =1 that means the phase(f(z)) =0.
2-In the blue area the cos(phase(f(z))) =-1 that means the
What I learned before that this happens at the points when on the positive real axis and this happen at the points when on the negative real axis.
My question
why as time passes these colors (yellow for ( the phase(f(z)) =0) ) and blue when change their position as seen in the following figure?
Does that mean the real and imaginary axes rotated with time and changed their positions?
I appreciate any help

5 Comments

Hi all
Just I want if any expert from this community try to help me to understand this issue?
I will appriciate any help
Torsten
Torsten on 25 Sep 2023
Edited: Torsten on 25 Sep 2023
Most probably because your polynomials do not simply depend on z, but also on t. Thus if t changes, the polynomial changes as well.
Of course it would help if we knew what f_k(z) and f_b(z) are ... :-)
And what | f > and < f | means.
The symbol |f> is Bra-ket notation.
Aisha Mohamed
Aisha Mohamed on 25 Sep 2023
Moved: Dyuman Joshi on 25 Sep 2023
Thank you so much Torsten, you are absolutly right.
I must to add some important information.
I use analytic function in extended complex plane to represent the states |f> and <f|in Hilbert space
In this representation, functions $f_k(z)$ and $f_b(z) $ are calculated by calculating the vector $|f(t)>$ by using some rotation operator and then the function $f_k(z)$ and $f_b(z) $ and their phase at each time $t$ are find and plotted by using MATLAB functions.
As we know the phase of the function =0 on for the point lie on the positve real axis and ( $phase(f_k(z)) =\pm \pi$ ) on the negative real axis
My question is why as time passes the places (yellow for ( the phase(f_k(z)) =0) ) and blue when ( $phase(f_k(z)) =\pm \pi$ )
change?
I appriciate all efforts.
yes, The symbol |f> is Bra-ket notation

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