Need Help with a Simple Symbolic Equation Solve in Matlab 2015a

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=????

[Roger Stafford]

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.

[S H]

Even if I try the following code, still Matlab answer does not make any sense:

clear all

syms Z k

[sol]=solve(abs(k/(k+1i*Z))==1/sqrt(2),Z);

.

.

.

and the answer will be k*1i + 2^(1/2)*abs(k)*1i which cannot be interpreted in anyhow as the correct answer of k.

Answer 1

[Torsten]

Maybe

[Z]=solve(abs(k/(k+1i*Z))==1/sqrt(2),Z);

?

Best wishes

Torsten.

Answer 2

[S H]

Still does not work or better said, Matlab cannot solve this simple equation.

Answer 3

[Sean de Wolski]

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.

Answer 4

[S H]

It is not one equation with two unknowns as laid out clearly by Matlab symbolic documentation. The solve function should be able to find one symbolic variable in terms of other symbolic variables. In this case, Matlab provides a ridiculous answer of Z = k*1i - 2^(1/2)*exp(z*2i)*abs(k)*1i. But the correct answer must be simply Z=k. What needs to be answered is how Matlab cannot solve such a simple equation while Symbolic documentation shows the solve function is capable of finding all complex trigonometric functions in terms of symbolic variables. If there is no way that Matlab 2015a and older versions can find Z=k as the answer, then I consider this as a bug or premature Matlab symbolic tool.

[Walter Roberson]

No the answer is Z = +/- k

[S H]

This is overdue, but it seems the later versions of Matlab can solve this equation properly.