Need Help with a Simple Symbolic Equation Solve in Matlab 2015a
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I am trying the following code in Matlab 2015a. It cannot solve it. Correct solution should be simply k. Could you tell me what is wrong?
syms Z k
assume(Z,'real')
assume(k,'real')
[Z]=solve(abs(k/(k+1i*Z))==1/sqrt(2));
Z=????
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Roger Stafford
2015-7-24
Without your assumptions that k and Z are real, the solution for any given k would be Z = w*k where w = x+1i*y is any point on the circle in the complex plane:
x^2 + (y-1)^2 = 2
In other words there would be an infinite continuum of possible solutions and 'solve' could not give you an answer.
Given that k and Z must be real, the solution is thereby restricted to the two points, Z = k and Z = -k, where the circle crosses the real line, so that 'solve' could and should have given you that answer. I suspect that the combination of the given assumptions together with the 'abs' operator confused Mathworks' 'solve' to the point where it was unable to give this simple answer.
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Sean de Wolski
2015-7-24
One equation with two unknowns. What do you expect?
Torsten's suggestion gives you the result for Z:
ZZ = solve(abs(k/(k+1i*Z))==1/sqrt(2),Z);
ZZ = k*1i - 2^(1/2)*exp(z*2i)*abs(k)*1i
But it will of course include k because you have two unknowns.
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