1. ## imaginary part

Separate real and imaginary part of log (1+i)1-i.

2. Originally Posted by whackywheelz
Separate real and imaginary part of log (1+i)1-i.
Is this $log((1 + i)(1 - i))$?
(and I'm going to assume this is $log_e$)

$= log(1 + i) + log(1 - i)$

First:
$1 + i = \sqrt{2} \left [ cos \left ( \frac{pi}{4} \right ) + i~sin \left ( \frac{\pi}{4} \right ) \right ] = \sqrt{2}e^{i \pi/4}$

So
$log(1 + i) = log \left ( \sqrt{2}e^{i \pi/4} \right ) = log(\sqrt{2}) + log \left ( e^{i \pi/4} \right ) = \frac{log(2)}{2} + i\frac{\pi}{4}$

Similarly:
$log(1 - i) = \frac{log(2)}{2} - i\frac{\pi}{4}$

So
$log((1 + i)(1 - i)) = \frac{log(2)}{2} + \frac{log(2)}{2} = log(2)$

(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

3. ## correction

Separate real and imaginary part of log (1+i)^1-i.

i.e raised to (1-i). sorry bout that. the compose post view shows me correctly when i opy paste from word.

thanks Dan

4. Originally Posted by whackywheelz
Separate real and imaginary part of log (1+i)^1-i.

i.e raised to (1-i). sorry bout that. the compose post view shows me correctly when i opy paste from word.

thanks Dan
Ah, that makes a lot more sense now.

The concept is pretty much the same problem as I worked out before:
$log \left [(1 + i)^{1 - i} \right ]$

$= (1 - i) log(1 + i) = (1 - i) \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$

$= \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right ) - i \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$

So the imaginary part is
$- \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$

-Dan

5. ## thanks a ton!

just one more doubt about the same question.

which is the real part?

6. Originally Posted by whackywheelz
just one more doubt about the same question.

which is the real part?
The part that doesn't have an "i" in it, the $\frac{log(2)}{2} + \frac{\pi}{4}$. In this case the real and imaginary parts happen to be negatives of one another. This is not typical.

-Dan

7. Be careful the rule $\log (xy) = \log x + \log y \mbox{ and }\log x^y = y\log x$ does not work in $\mathbb{C}$ for $x,y\not = 0$ in general.