Separate real and imaginary part of log (1+i)1-i.
Is this $\displaystyle log((1 + i)(1 - i))$?
(and I'm going to assume this is $\displaystyle log_e$)
$\displaystyle = log(1 + i) + log(1 - i)$
First:
$\displaystyle 1 + i = \sqrt{2} \left [ cos \left ( \frac{pi}{4} \right ) + i~sin \left ( \frac{\pi}{4} \right ) \right ] = \sqrt{2}e^{i \pi/4} $
So
$\displaystyle log(1 + i) = log \left ( \sqrt{2}e^{i \pi/4} \right ) = log(\sqrt{2}) + log \left ( e^{i \pi/4} \right ) = \frac{log(2)}{2} + i\frac{\pi}{4}$
Similarly:
$\displaystyle log(1 - i) = \frac{log(2)}{2} - i\frac{\pi}{4}$
So
$\displaystyle log((1 + i)(1 - i)) = \frac{log(2)}{2} + \frac{log(2)}{2} = log(2)$
(Of course the short way is to note that (1 + i)(1 - i) = 2, but that would spoil all the fun, wouldn't it? )
-Dan
Ah, that makes a lot more sense now.
The concept is pretty much the same problem as I worked out before:
$\displaystyle log \left [(1 + i)^{1 - i} \right ] $
$\displaystyle = (1 - i) log(1 + i) = (1 - i) \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$
$\displaystyle = \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right ) - i \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$
So the imaginary part is
$\displaystyle - \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$
-Dan