[SOLVED] Matlab symbolic "solve" gives unknown parameter "z"

Apr 2010
7
0
Hello

I am trying to symbolically solve for 'ws' in the following equations:

Code:
syms N ep es ws
t1=ep*es; t1_=sqrt(1-t1^2); t2=1/ws; t2_=sqrt(1-t2^2);
e1=0.5*(1-t1_^0.5)/(1+t1_^0.5); e2=0.5*(1-t2_^0.5)/(1+t2_^0.5);
q1=e1+2*e1^5; q2=e2+2*e2^5;
K1=pi/2*(1+2*q1+2*q1^4)^2; K2=pi/2*(1+2*q2+2*q2^4)^2;
K1_=K1/pi*log(1/q1); K2_=K2/pi*log(1/q2);
k=N*K1/K1_;
But when try to solve, I get this:

Code:
solve(k-K2/K2_,ws)
 
ans =
 
  1/((z^2 + 1)^(1/2)*(1 - z)^(1/2)*(z + 1)^(1/2))
 -1/((z^2 + 1)^(1/2)*(1 - z)^(1/2)*(z + 1)^(1/2))
What is that "z" parameter? "z <enter>" says "??? Undefined function or variable 'z'."
Also, I tried simplifying the equations, willingly accepting less precision in favor of a clear answer:

Code:
q1=e1; q2=e2;
K1=pi/2*(1+2*q1)^2; K2=pi/2*(1+2*q2)^2;
K1_=K1/pi*log(1/q1); K2_=K2/pi*log(1/q2);
k=N*K1/K1_;
...but I get the same answer!(?) Can anyone shed some light in this? I googled for some answers and the closest I could find was that "z" replaces some really complicated formulae, all looking the same, that would have only filled the display, so Matlab simplified them with "z", but it said that revealing it wasn't easy. That's it, not very helpful.


Regards,
Vlad.