Where is the error in this argument

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SUMMARY

The discussion centers on the evaluation of logarithmic expressions involving complex numbers, specifically the logarithm of the imaginary unit \(i\) and its powers. The correct evaluation of \(\log[i^2]\) is established as \(i(\pi + 2\pi n)\) for \(n \in \mathbb{Z}\), while \(\log[\sqrt{i}]\) is shown to equal \(i(\frac{\pi}{4} + \pi n)\). The discrepancy arises from the incorrect interpretation of the logarithmic identities applied to complex numbers, particularly in the context of multi-valued functions in complex analysis.

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Bachelier
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It is possible to do this and it is correct:

<br /> \log \left[\sqrt{i}\right] = \log\left\{\exp\left[\frac{i}{2}\left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = \frac{i}{2}\left(\frac{\pi}{2} + 2\pi n\right) = i\left(\frac{\pi}{4} + \pi n\right)<br />

But:

<br /> \log \left[i^2 \right] = \log\left\{\exp\left[2i \left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = 2i \left(\frac{\pi}{2} + 2\pi n\right) = i\left(\pi + 4\pi n\right)<br />

yet

## \log \left[i^2 \right] = \log \left[-1 \right] = i\left(\pi + 2\pi n\right) \ for \ k \in \mathbb{Z}## which is the correct argument.
 
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Bachelier said:
It is possible to do this and it is correct:

<br /> \log \left[\sqrt{i}\right] = \log\left\{\exp\left[\frac{i}{2}\left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = \frac{i}{2}\left(\frac{\pi}{2} + 2\pi n\right) = i\left(\frac{\pi}{4} + \pi n\right)<br />

But:

<br /> \log \left[i^2 \right] = \log\left\{\exp\left[2i \left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = 2i \left(\frac{\pi}{2} + 2\pi n\right) = i\left(\pi + 4\pi n\right)<br />

yet

## \log \left[i^2 \right] = \log \left[-1 \right] = i\left(\pi + 2\pi n\right) \ for \ k \in \mathbb{Z}## which is the correct argument.

I'm not sure what your argument really is. Sure (sqrt(i))^4=(i)^2=(-1). (sqrt(-i))^4 is also (-1). So?
 

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