On page 113 of Churchill in explaining the $\arcsin{(-i)}$ it comes across

$$ln(1-\sqrt{2})$$ which is fine but then it goes on to say that it is equal to


How do they justify this step. I thought it was a reflection across the real axis but there is no imaginary part to this complex number.


Something seems off about the sign. $\displaystyle \frac{1}{1 + \sqrt 2} = \frac{1 - \sqrt 2}{(1 + \sqrt 2)(1- \sqrt 2)} = \frac{1 - \sqrt 2}{1-2} = \sqrt 2 - 1$.

So $\displaystyle \ln(1 - \sqrt 2) = i\pi + \ln(\sqrt 2 - 1) = i\pi + \ln\frac{1}{1 + \sqrt 2}$.

Here, I'm only considering the principal value of the logarithm function for negative arguments.

  • $\begingroup$ Thats exactly it! I completely omitted the $i \pi$ $\endgroup$ – phandaman Jan 28 '15 at 6:07

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