Average Error: 0.8 → 0.5
Time: 2.8s
Precision: 64
\[\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\]
↓
\[\begin{array}{l} \mathbf{if}\;\frac{\tan^{-1}_* \frac{im}{re}}{\log 10} \le -0.6821881769209206:\\ \;\;\;\;\sqrt[3]{{\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)\right)\\ \end{array}\]
\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}
↓
\begin{array}{l}
\mathbf{if}\;\frac{\tan^{-1}_* \frac{im}{re}}{\log 10} \le -0.6821881769209206:\\
\;\;\;\;\sqrt[3]{{\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)}^{3}}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)\right)\\

\end{array}
double code(double re, double im) {
	return ((double) (((double) atan2(im, re)) / ((double) log(10.0))));
}
↓
double code(double re, double im) {
	double VAR;
	if ((((double) (((double) atan2(im, re)) / ((double) log(10.0)))) <= -0.6821881769209206)) {
		VAR = ((double) cbrt(((double) pow(((double) (((double) atan2(im, re)) / ((double) log(10.0)))), 3.0))));
	} else {
		VAR = ((double) expm1(((double) log1p(((double) (((double) atan2(im, re)) / ((double) log(10.0))))))));
	}
	return VAR;
}

Error

Bits error versus re

Bits error versus im

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes

  2. if (/ (atan2 im re) (log 10.0)) < -0.6821881769209206

    1. Initial program 1.0

      \[\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\]
    2. Using strategy rm

    3. Applied add-cbrt-cube1.6

      \[\leadsto \frac{\tan^{-1}_* \frac{im}{re}}{\color{blue}{\sqrt[3]{\left(\log 10 \cdot \log 10\right) \cdot \log 10}}}\]

    4. Applied add-cbrt-cube1.0

      \[\leadsto \frac{\color{blue}{\sqrt[3]{\left(\tan^{-1}_* \frac{im}{re} \cdot \tan^{-1}_* \frac{im}{re}\right) \cdot \tan^{-1}_* \frac{im}{re}}}}{\sqrt[3]{\left(\log 10 \cdot \log 10\right) \cdot \log 10}}\]

    5. Applied cbrt-undiv0.7

      \[\leadsto \color{blue}{\sqrt[3]{\frac{\left(\tan^{-1}_* \frac{im}{re} \cdot \tan^{-1}_* \frac{im}{re}\right) \cdot \tan^{-1}_* \frac{im}{re}}{\left(\log 10 \cdot \log 10\right) \cdot \log 10}}}\]

    6. Simplified0.0

      \[\leadsto \sqrt[3]{\color{blue}{{\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)}^{3}}}\]

    if -0.6821881769209206 < (/ (atan2 im re) (log 10.0))

    1. Initial program 0.8

      \[\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\]
    2. Using strategy rm

    3. Applied expm1-log1p-u0.8

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\tan^{-1}_* \frac{im}{re}}{\log 10} \le -0.6821881769209206:\\ \;\;\;\;\sqrt[3]{{\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\tan^{-1}_* \frac{im}{re}}{\log 10}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020120 +o rules:numerics
(FPCore (re im)
  :name "math.log10 on complex, imaginary part"
  :precision binary64
  (/ (atan2 im re) (log 10)))