What is the value of $\ln(\ln(i))$?

Question

I came across this question while practicing some quant interview questions. It simply asks the value of the above expression, no additional information is given.

I tried googling for it, Google tells me the result is -

$$\ln(\ln(i)) = 0.451582705 + 1.57079633i$$

I have no clue how to get this. Any help would be appreciated.

Answer 1

Well, $e^{\pi i/2}=i$, so $\log i$ is $\frac{\pi}{2}i$ -- plus or minus some multiple of $2\pi i$, but let us work with $\frac\pi2i$ as the principal value to begin with. Then $$\log(\frac\pi2i) = \log(\frac\pi2) + \log i = \log(\frac\pi2) + \frac\pi2i $$

This matches Google's result, since $\log(\frac\pi2)\approx 0.452$ and $\frac\pi2\approx1.57$.

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In is full multivalued glory, the possible values are $$ \log\Bigl(\frac{4k_1+1+j}2\Bigr) + \log \pi + \frac{4k_2+1+j}2\pi i \qquad\qquad k_1 \in \mathbb N_0, k_2\in\mathbb Z, j\in\{0,2\}$$ (where the $j=2$ possibility is for the case that that we choose a $\log i$ on the negative imaginary axis for the first logarithm).

Answer 2

Hint: Given $z = r e^{i\theta}$, we have $$\log z = \log r +i\theta,$$ and, of course, $$\log r = \int_1^r \frac{1}{t} \mathrm{d}t.$$