Here is a slightly more interesting output
Note that uu here is the Heaviside step function but it's applied to the (shifted) modulus of a function rather than a variable. That is, we are interested in roots of g(z;k)≡f(z)+u(|f(z)|−k−1)kSin(z)g(z;k)≡f(z)+u(|f(z)|−k−1)kSin(z). This will yield the same roots as f(z)=x3−xf(z)=x3−x on the real line. This obtains from Rouché's theorem as long as |f|<k|f|<k, sin(z)sin(z) being of bounded modulus there. Elsewhere the value can be seen as depending on the hyperbolic sine of the imaginary part of zz, sinh(ℑz)sinh(Iz). Proof of which.