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This might get fixed in the future, but at the time of this writing, Wolfram Alpha gets apparently sometimes confused by logarithms of complex numbers:

Wolfram Alpha -- $\log(1+ \frac{1}{2}i) - \log(1 - \frac{1}{2} i)$Wolfram Alpha -- $\log(1+ \frac{1}{2}i) - \log(1 - \frac{1}{2} i)$

For reference, should the problem get fixed: it claims that $2i = 2i\cot^{-1}(2) \approx 0.9272$.

Curiously, the numerical approximation is correct, but the symbolic form seems to be wrong.

This might get fixed in the future, but at the time of this writing, Wolfram Alpha gets apparently sometimes confused by logarithms of complex numbers:

Wolfram Alpha -- $\log(1+ \frac{1}{2}i) - \log(1 - \frac{1}{2} i)$

For reference, should the problem get fixed: it claims that $2i = 2i\cot^{-1}(2) \approx 0.9272$.

Curiously, the numerical approximation is correct, but the symbolic form seems to be wrong.

This might get fixed in the future, but at the time of this writing, Wolfram Alpha gets apparently sometimes confused by logarithms of complex numbers:

Wolfram Alpha -- $\log(1+ \frac{1}{2}i) - \log(1 - \frac{1}{2} i)$

For reference, should the problem get fixed: it claims that $2i = 2i\cot^{-1}(2) \approx 0.9272$.

Curiously, the numerical approximation is correct, but the symbolic form seems to be wrong.

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This might get fixed in the future, but at the time of this writing, Wolfram Alpha gets apparently sometimes confused by logarithms of complex numbers:

Wolfram Alpha -- $\log(1+ \frac{1}{2}i) - \log(1 - \frac{1}{2} i)$

For reference, should the problem get fixed: it claims that $2i = 2i\cot^{-1}(2) \approx 0.9272$.

Curiously, the numerical approximation is correct, but the symbolic form seems to be wrong.