# what does the positive and negative values concave mean?.

2 views (last 30 days)
Aisha Mohamed on 13 Sep 2022
Commented: Aisha Mohamed on 22 Sep 2022
I am studying the behavoure of this function
p=[(0.9000 - 0.0010i) (0.4243 + 0.0017i) (0.1000 + 0.3000i)];
and when I plotted the cos(phase(p(z))) I got the following graph
Although I described the graph according the color bar, I only described the color and I did not extract any information about the function, I am sure this graph tell us alot about the function cos(phase(p(z))), where(p(z)=f_k(z)).
For example ,
1-what does the positive and negative values concave mean?.
2- what does that mean when we find the two blue dots at the roots of the function (p(z))?
3- How can I describe this figure to get worth information about this function?
I appreciate any help.
Aisha Mohamed on 15 Sep 2022
Thanks Torsten. You are really genius, your answer gave me idea to start describing this figure. I hope to do this.
Aisha Mohamed on 19 Sep 2022
Hi Dearunder
In my previous question, when I asked about where the phase of $f_k(z)$ is discontinuous. I understood from the experts (Bruno Luong and Torsten, Thank you very much for your worthy efforts). that,
When the phase of $f_k(z)$ is plotted, all of the points of $x + iy$ lay on the projected curves. When the polyval of $z = (x + iy)$ is calculated at these points, it is found that the angle (polyval $(p, z))$ is equal to $\pi$ or $-\pi$. Here $f_k(z)=p(z)$ .
And when I asked about the dark blue area in the following figure, I have got that,
Dark blue colors indicate cos(angle(p(z)) = -1, thus angle(p(z)) = + or - pi
My question is,
Does that mean the blue area (in the above figure) is where the phase is discontinuous and to solve this
discontinuity we calculate the cos(phase(p(z)) in this area?
I appreciate any help

David Goodmanson on 20 Sep 2022
Edited: David Goodmanson on 20 Sep 2022
Hello Aisha,
You are looking at a polynomial function of z, with two roots as you stated. To see what is going on, it helps to look first at just the function f(z)=z and its single root at z=0. The upper left figure below shows the phase angle of z, which by convention is -pi<angle<pi. The angle is 0 along the pos real axis, runs ccw to +pi on the neg real axis and cw to -pi, also on the neg real axis. There is a 2pi discontinuity on neg real axis. That is the standard convention, but it's just a convention. You could have 0<angle<2pi if you wanted. Then the angle starts at 0 on the pos real axis and goes all the way around ccw to +2pi on pos real axis. Consequently there is a 2pi discontinuity on the pos real axis.
Once you take a cosine or sine of the phase angle, the discontinuity disappears because trig functions don't care about 2pi discontinuities. So you are not 'solving' the angle discontinuity, it's always there. You are merely inserting the angle into a function that doesn't care about it [as long as the discontinuity is an integral multiple of 2pi]. That's shown on the figure on the upper right, where yellow colors are pos cos and blue colors are neg cos as you mentioned. Note that as you circle the origin, you go from the top to the bottom of the colorbar and back again. I'll call that one circuit of the colorbar.
At z=0, the root of the function, all colors converge and the angle is undefined. That's like taking atan(y/x) where both x and y are zero.
The lower two figures are the function you have. There two roots, so two points where the angle is undefined. The phase angles associated with each point are additive. That means that on the lower right plot, if you take a large radius and go around both roots, you go through two circuits of the colorbar. Picking a smaller path that encircles one of the roots gives one circuit of the colorbar. The same kind of argument applies to lower left plot, where you go through either one or two angle discontinuities depending on the path.
##### 2 CommentsShow NoneHide None
Aisha Mohamed on 20 Sep 2022
Many Thanks David Goodmanson
It is really great explanation from great expert so please do not blame me if I am trying to understand every point in it, because you brought what I seek (I thank all experts in this group).
Just I can not understand the last part from your answer please
((The lower two figures are the function you have. There two roots, so two points where the angle is undefined. The phase angles associated with each point are additive. That means that on the lower right plot, if you take a large radius and go around both roots, you go through two circuits of the colorbar. Picking a smaller path that encircles one of the roots gives one circuit of the colorbar. The same kind of argument applies to lower left plot, where you go through either one or two angle discontinuities depending on the path.))
My questions are,
1-The zeros of this function are
r1= - 0.6114 + 0.4419i
r2= 0.1399 - 0.4443i
You said (There two roots, so two points where the angle is undefined) , did you mean because the phase(f_k(r1))=0) take two values 0 and 2pi so that means is undefined, and when the phase undefind how we plot cos(phase(f_k(r1)))
what did you mean by (The phase angles associated with each point are additive. )?
2- is there any relationship between the zeros of this function (above figure)and the two dark blue point in this figure?becauce I observed there are two of them and we have two zeros. If so what is this relationship?
I know they are negative values of the cos function, but why they look like points.
I will appreciate any help
Aisha Mohamed on 22 Sep 2022
Hi Dear
After studing your answer again. I think I understood some points in my last questions,
1- regarding to (There two roots, so two points where the angle is undefined.)
That because at f(z) = 0, the phase(f(z)) = phase (0) is undefined, that is why we find the two dark blue points at the location of the zeros of the function
2- is there any relationship between the zeros of this function (above figure) and the two dark blue point in this figure?becauce I observed there are two of them and we have two zeros. If so what is this relationship?
These two dark blue points lay at the location of the zeros of the function where the phase undefined.

### Categories

Find more on Blue in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!