Problem with the solutions of two nonlinear equations

I have a problem with the solutions of the blow two nonlinear equations.I used 'ezplot' function to see the intersection of these two equations and I saw there are numerous intersections between them. So, there are numerous solutions to the system of these two equations,but the solutions are finite. I used 'solve' function to solve this system,but it gives it only one solution! How can I see all the solutions? (the two variables are l1 and l2)

equation1=(11.390625*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+5.0625*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2)^(1/2)-.1

equation2=1/2*(22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2*(-.1111111111e-7*(1+tan(10/3*pi*l1)^2)*l1+.2407407408e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.1203703704e-7*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)*(.1083333333e-7*(1+tan(10/3*pi*l1)^2)*l1*tan(10/3*pi*l2)+.1083333333e-7*tan(10/3*pi*l1)*(1+tan(10/3*pi*l2)^2)*l2+.1000000000e-7*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2-22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2*(.4900520833e-6*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2450260417e-6*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2-.3046875000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1-.1015625000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l2)^2)*l2+.2700000000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2+.3637500000e-6*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1+.1350000000e-6*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^3+10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2*(-.5555555557e-8*(1+tan(10/3*pi*l1)^2)*l1+.1203703704e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.6018518520e-8*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))*(.4041666667e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.2020833333e-7*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2-.3250000000e-7*(1+tan(10/3*pi*l1)^2)*l1-.1500000000e-7*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2-10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2*(.4900520833e-6*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2450260417e-6*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2-.3046875000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1-.1015625000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l2)^2)*l2+.2700000000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2+.3637500000e-6*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1+.1350000000e-6*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^3)/(11.390625*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+5.0625*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2)^(1/2)

Answer 1

[Walter Roberson]

The number of values where they intersect is definitely not finite if you include complex numbers, and it looks to me as if it might not be finite even if you restrict to real values.

You have two expressions in two unknowns. You ask when they equal each other. That is equivalent to asking when the first one minus the second one is zero. But that subtraction is a single expression in two variables, so there are an infinite number of complex solutions.

To establish that there are a finite number of real-valued solutions, it would be necessary to prove that there is some combination of values beyond which one of the expressions is strictly greater than the other. The second expression stays close to zero but divergently so going increasingly positive and negative. Still there might be a way to prove that the first is always greater than the second outside of a finite region where the zeros could be counted. At the moment I would not rule it out but it looks unlikely from the graphs I do.

[Mark]

You are right;the real solutions are finite. Here, there is the image of plot of two equations with ezplot function. As it can be seen, the real intersections are finite. So what should I do?

[Walter Roberson]

ezplot plots over a finite subset of the variable domains. You asked for all of the intersections so the behaviour has to be considered as the values tend towards infinity.

[Mark]

Can I use the functions like 'numeric::solver' with 'allrealroots' condition on it? in matlab help, I saw that this function operates only for one equation. Does it operate for two equations or not?

[Walter Roberson]

There are an infinite number of zeros. The best you could hope for is to parameterize the solutions. Unfortunately, the formal analysis of equation2 is difficult, so I will have to illustrate it experimentally:

equation1=@(l1,l2) (11.390625*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+5.0625*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2)^(1/2)-.1;

equation2=@(l1,l2) 1/2*(22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2*(-.1111111111e-7*(1+tan(10/3*pi*l1)^2)*l1+.2407407408e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.1203703704e-7*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)*(.1083333333e-7*(1+tan(10/3*pi*l1)^2)*l1*tan(10/3*pi*l2)+.1083333333e-7*tan(10/3*pi*l1)*(1+tan(10/3*pi*l2)^2)*l2+.1000000000e-7*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2-22.781250*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2*(.4900520833e-6*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2450260417e-6*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2-.3046875000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1-.1015625000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l2)^2)*l2+.2700000000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2+.3637500000e-6*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1+.1350000000e-6*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^3+10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2*(-.5555555557e-8*(1+tan(10/3*pi*l1)^2)*l1+.1203703704e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.6018518520e-8*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))*(.4041666667e-7*tan(10/3*pi*l1)*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1+.2020833333e-7*tan(10/3*pi*l1)^2*(1+tan(10/3*pi*l2)^2)*l2-.3250000000e-7*(1+tan(10/3*pi*l1)^2)*l1-.1500000000e-7*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2-10.1250*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2*(.4900520833e-6*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2450260417e-6*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2-.3046875000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l1)^2)*l1-.1015625000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l2)^2)*l2+.2700000000e-6*tan(10/3*pi*l1)^3*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)*tan(10/3*pi*l2)^2*(1+tan(10/3*pi*l1)^2)*l1+.2700000000e-6*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2+.3637500000e-6*tan(10/3*pi*l1)*(1+tan(10/3*pi*l1)^2)*l1+.1350000000e-6*tan(10/3*pi*l2)*(1+tan(10/3*pi*l2)^2)*l2)/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^3)/(11.390625*(-3.333333334*tan(10/3*pi*l1)+3.611111112*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(3.25*tan(10/3*pi*l1)*tan(10/3*pi*l2)+1.5*tan(10/3*pi*l1)^2-1.5)^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2+5.0625*(-1.666666667*tan(10/3*pi*l1)+1.805555556*tan(10/3*pi*l1)^2*tan(10/3*pi*l2))^2*(6.0625*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)-9.750*tan(10/3*pi*l1)-4.50*tan(10/3*pi*l2))^2/(36.75390625*tan(10/3*pi*l1)^4*tan(10/3*pi*l2)^2-30.4687500*tan(10/3*pi*l1)^3*tan(10/3*pi*l2)+20.2500*tan(10/3*pi*l1)^4+40.5000*tan(10/3*pi*l1)^2*tan(10/3*pi*l2)^2+54.562500*tan(10/3*pi*l1)^2+20.2500*tan(10/3*pi*l2)^2+20.2500)^2)^(1/2);

V1 = str2func(vectorize(func2str(equation1)));
V2 = str2func(vectorize(func2str(equation2)));
E3 = @(l1,l2) V1(l1,l2) - V2(l1,l2);

l1 = linspace(0,3/10,500);
l2 = linspace(-1/10,4/10,500);
[L1,L2] = ndgrid(l1,l2);
R = E3(L1,L2);
S = sign(R);
imagesc(l1,l2,S.');
colormap(jet(3));
xlabel('L1');
ylabel('L2');

In the resulting diagram, the areas of negative value will be blue, the places with zeros will be cyan, and the areas with positive values will be yellow.

If you were correct that there are a finite number of zeros then if we just happened to sample at the zeros, the those positions would show up as isolated cyan points and everything else would be solid blue (negative) or solid yellow (positive). Instead we see:

Every point along the edge of the yellow ovoid is a zero. There is clearly a region of positive values, and a region of negative values, and by the Mean Value Theorem, there must be a 0 at every transition between the positive and negative region. Regions have an infinite number of points on their boundaries.

I was not able to do a formal analysis of equation2: doing so consistently hung my analysis program.

I was able to do a formal analysis of equation1. If you use a symbolic solver and let l1, l2, and equation1 all be symbolic, then you can solve() the expression in terms of l1 and l2. The result will have 4 solution subsets. 2 of the solution subsets have values of l2 for every value of l1.

The 3rd and 4th solution subsets are each a pair of specific points, parameterized by equation1. That is, for each value of equation1, there is a specific pair of points that is the solution. I was able to show that the pairs had real-valued coordinates for every equation1 value between -1/10 and +553/970. Another way of expressing that is that the value calculated on the right of that equation can definitely assume any value between -1/10 and +553/970, with exact coordinates being available for each value (it's a relatively straight-looking line.)

The second expression has values that are much smaller, roughly -1.3E-9 to +1.3E-9 over the region examined. The second expression has the effect of blurring the boundary a bit on solving the first expression for 0 -- but because the first expression has negative and positive regions, but the Mean Value Theorem you know that it must pass through whatever the value of the second expression is.

[Mark]

Thanks a lot for your explanation. But, the fact is that these two equations are related to a paper with results that are shown. The two equations are: (1) magnitude(C(z,theta1,theta2))=0.1 and (2) d/dw(magnitude(C(z,theta1,theta2)))=0 , with C(z,theta1,theta2)=A/B and both of equations are at the operating frequency equals to 500 MHz , where

A=j*(2*(z-1/z)*tan(theta1)+(1/z^2-z^2)*tan(theta1)^2*tan(theta2));

B=2-2*(z+1/z)*tan(theta1)*tan(theta2)-2*tan(theta1)^2+j*(2*(z+1/z)*tan(theta1)+2*tan(theta2)-(1/z^2+z^2)*tan(theta1)^2*tan(theta2));

where theta1=beta*l1 and theta2=beta*l2 and beta=w/c where c=3*10^8 and w is angular frequency.

The two equations that I showed you, are these two above equations with z=1.5 and w=500*10^6 . I wanted to solve that equations for one value of 'z' and then expand the solution for all values of z to derive the paper's result that is shown below.

According to your explanations, solving the two equations for all values of 'z' such as 1.2 to 3 as shown in the result of the paper, is difficult! So,how can I solve this problem?!

[Walter Roberson]

The z value range 1.2 to 3 is over continuous real values, is it not? Not discrete valued? So the number of points in the range 1.2 to 3 is uncountably infinite. If there happened to be one solution for each particular z value, that would still lead to uncountably infinite solutions. But you are expecting a finite countable number of solutions. The best you could hope for would be a finite countable number of parameterized solutions.

If what I saw in my investigations holds for a wider area, there may be a countable infinity number of parameterized solutions (the first equation that you gave repeats infinitely at intervals of 3/10 in L1 and so can be expressed in terms of a value plus an arbitrary integer times 3/10, but there are a countable infinity of integers, each leading to a region that has an uncountable infinity of boundary points.)

[Mark]

The z value range is discrete and according to the paper, there is one solution for each particular z for real values of l1 and l2. With these investigations, how the authors of the paper have been able to derive these results?!

[Walter Roberson]

Assuming that the "j" there is sqrt(-1), then the magnitude of C is abs(A/B) . It is possible to solve for that being equal to 1/10, solving for L1. The result is a quartic in L2, so there are 4 exact solutions.

If you use the given z value, then: if you substitute the solutions for L1 in to diff(A/B,w), then there are no solutions for that being 0. There are places where diff(A/B,w) has its real component be 0, and there are places where its imaginary component is 0, but there are no places where both are 0. To determine this, take abs() of the diff() and plot that abs() over -10 to +10 and it can be seen that the magnitudes wobble but tend away from 0. The minima is on one of the roots, at about +/- 0.5858 , but that minima is on the order of 1.2535E-10 for the magnitude.

The second equation might have zeros, but it does not have them in the places where the magnitude is 1/10.

[Walter Roberson]

It looks like the magnitude of the second expression (the differential) is 0 only when L1 = 0, which zeros out all the other terms. However, when L1 = 0, the magnitude of C is 0 rather than 1/10.