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


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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Is that a typo for the second case? Note that the left hand side is supposed to be $2\operatorname{Log}(-1+i)$, not $2\operatorname{Log}(1+i)$. Are you sure you're supposed to use $[0,2\pi]$ for the principal argument? (It's usually $[-\pi, \pi]$.) – mrf Apr 21 '12 at 13:09
There is a typo, but if you use $[0,2\pi]$, then the results are going to be the same, since you don't cross the branch cut, when you square. You are probably supposed to use the branch $\theta\in [-\pi,\pi]$. – Alex B. Apr 21 '12 at 13:15
Actually, neither $[0,2\pi]$ nor $[-\pi,\pi]$ make sense as defining a branch for the logarithm. However $[0,2\pi)$ and $(-\pi,\pi]$ do (and the latter even makes the exercise solvable). – Henning Makholm Apr 21 '12 at 13:18
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 $-1+i$. – alex.jordan Apr 21 '12 at 16:35
up vote 2 down vote accepted

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 $\operatorname{Log}((-1+i)^2) = \log2 - \frac12\pi i$.

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