Solving an equation

[enzor]

Hi, i was solving the next equation:

solve(abs((kc+g/u)/(1+kc))=1,kc)

and the solution I get is:

(2*i*_y*u-g-u)/(2*u)

Does anyone knows what _y means?

Thanks in advance!

[Albert Chan]

(01-27-2021 01:28 PM)enzor Wrote:  solve(abs((kc+g/u)/(1+kc))=1,kc)

and the solution I get is:

(2*i*_y*u-g-u)/(2*u)

Does anyone knows what _y means?

_y is just a notation for any constant. (note: above solution assumed g/u is real).

Let x=kc, g/u=m:

If x≠-1, |x+m| = |x+1|

Square both side, and simplify:

x² + 2mx + m² = x² + 2x + 1
2*(m-1)*x = 1-m² = -(m+1)*(m-1)
x = -(m+1)/2

Above solution, for x=-1 -> m=1 -> above solution work for x=-1 too.

If we assume m is real -> x is real.
We can add imaginary part to x, and still satisfy |x+m| = |x+1|

→ kc = x + _y*i = -(g/u+1)/2 + _y*i