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Show that $Log((1+i)^2) = 2Log(1+i),$ while $Log((-1+i)^2) \not= 2Log(-1+i)$

Solution:

I am using $[0 , 2\pi]$ for the principle argument.

The first part worked out fine, I got:

$Log((1+i)^2) = log(2) + i \frac{\pi}{2}$

$2Log(1+i)= log(2) + i \frac{\pi}{2}$

But the second part is not working out because I found that the results are equal:

$Log((-1+i)^2) = log(2) + i \frac{3\pi}{2}$

$2Log((-1+i) = log(2) + i \frac{3\pi}{2}$

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    Many have noted that this problem might have a typo, and was meant to use $[0,2\pi)$ for the principle argument range. It is also possible that the second equation was meant to involve $-1-i$ or $1-i$ rather than $-1+i$.2012-04-21

1 Answers 1

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If you're using $[0,2\pi)$ as the range of the principal argument, then the results in the second part are indeed equal.

The exercise is probably meant to be solved with the principal argument having the range $(-\pi,\pi]$ -- which is somewhat more common, because it behaves nicer near the positive real axis. With that choice of principal argument you should get $\operatorname{Log}(-1+i)= \frac12\log2 + \frac34 \pi i$ but $\operatorname{Log}((-1+i)^2) = \log2 - \frac12\pi i$.