\frac{\tan^{-1}_* \frac{im}{re} \cdot \log base - \log \left(\sqrt{re \cdot re + im \cdot im}\right) \cdot 0.0}{\log base \cdot \log base + 0.0 \cdot 0.0}\frac{1}{\frac{\log base}{\tan^{-1}_* \frac{im}{re}}}double f(double re, double im, double base) {
double r2527897 = im;
double r2527898 = re;
double r2527899 = atan2(r2527897, r2527898);
double r2527900 = base;
double r2527901 = log(r2527900);
double r2527902 = r2527899 * r2527901;
double r2527903 = r2527898 * r2527898;
double r2527904 = r2527897 * r2527897;
double r2527905 = r2527903 + r2527904;
double r2527906 = sqrt(r2527905);
double r2527907 = log(r2527906);
double r2527908 = 0.0;
double r2527909 = r2527907 * r2527908;
double r2527910 = r2527902 - r2527909;
double r2527911 = r2527901 * r2527901;
double r2527912 = r2527908 * r2527908;
double r2527913 = r2527911 + r2527912;
double r2527914 = r2527910 / r2527913;
return r2527914;
}
double f(double re, double im, double base) {
double r2527915 = 1.0;
double r2527916 = base;
double r2527917 = log(r2527916);
double r2527918 = im;
double r2527919 = re;
double r2527920 = atan2(r2527918, r2527919);
double r2527921 = r2527917 / r2527920;
double r2527922 = r2527915 / r2527921;
return r2527922;
}



Bits error versus re



Bits error versus im



Bits error versus base
Results
Initial program 32.1
Simplified0.4
Taylor expanded around 0 0.3
rmApplied clear-num0.5
Final simplification0.5
herbie shell --seed 2019171 +o rules:numerics
(FPCore (re im base)
:name "math.log/2 on complex, imaginary part"
(/ (- (* (atan2 im re) (log base)) (* (log (sqrt (+ (* re re) (* im im)))) 0.0)) (+ (* (log base) (log base)) (* 0.0 0.0))))