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"Welcome to $K_1( \mathbb{C}(t))$!"

The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by taking the real or, in this case, imaginary part of the formula, i.e., symmetrizing under Gal(C/R) which replaces logarithm with arctangent).

Specializing the fact to $f(z)=g(iz)/g(-iz)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio.

(Remember also that $\arctan t = \arg (q+itq)$ for real $t$ and $q$, so that $\log f(z)$ can be evaluated without factorization, by computing real and imaginary parts of $g(iz)$.)

"Welcome to $K_1( \mathbb{C}(t))$!"

The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by taking the real or, in this case, imaginary part of the formula, i.e., symmetrizing under Gal(C/R) which replaces logarithm with arctangent).

Specializing the fact to $f(x)=g(ix)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio. The imaginary part of $\log(f)$ is $(1/2i)\log(g(ix)/g(-ix))$, which can be expanded as a sum over the roots and differentiated.

(Remember also that $\arctan t = \arg (q+itq)$ for real $t$ and $q$, so that $\log f(x)$ can be evaluated without factorization, by computing real and imaginary parts of $g(ix)$. Equating the two expressions for the imaginary part of ($d\log(f)$) gives the formula in the question.)

"Welcome to $K_1( \mathbb{C}(t))$!"

The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by using arctangent in place of log, i.e., taking invariants under Gal(C/R)).

Specializing the fact to $f(z)=g(iz)/g(-iz)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio.

(Remember also that $\arctan t = \arg (q+itq)$ for real $t$ and $q$, so that $\log f(z)$ can be evaluated without factorization, by computing real and imaginary parts of $g(iz)$.)

"Welcome to $K_1( \mathbb{C}(t))$!"

The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by taking the real or, in this case, imaginary part of the formula, i.e., symmetrizing under Gal(C/R) which replaces logarithm with arctangent).

Specializing the fact to $f(z)=g(iz)/g(-iz)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio.

(Remember also that $\arctan t = \arg (q+itq)$ for real $t$ and $q$, so that $\log f(z)$ can be evaluated without factorization, by computing real and imaginary parts of $g(iz)$.)

"Welcome to $K_1( \mathbb{C}(t))$!"

The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by using arctangent in place of log, i.e., taking invariants under Gal(C/R)).

Specializing the fact to $f(z)=g(iz)/g(-iz)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio.

(Remember also that $\arctan t = \arg (q+itq)$ for real $q$, so that $\log f(z)$ can be evaluated without factorization, by computing real and imaginary parts of $g(iz)$.)

"Welcome to $K_1( \mathbb{C}(t))$!"

The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by using arctangent in place of log, i.e., taking invariants under Gal(C/R)).

Specializing the fact to $f(z)=g(iz)/g(-iz)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio.

(Remember also that $\arctan t = \arg (q+itq)$ for real $t$ and $q$, so that $\log f(z)$ can be evaluated without factorization, by computing real and imaginary parts of $g(iz)$.)