# imaginary part

• September 1st 2007, 12:56 AM
whackywheelz
imaginary part
Separate real and imaginary part of log (1+i)1-i.
• September 1st 2007, 01:13 AM
topsquark
Quote:

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
• September 1st 2007, 01:53 AM
whackywheelz
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
• September 1st 2007, 04:37 AM
topsquark
Quote:

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
• September 1st 2007, 04:56 AM
whackywheelz
thanks a ton!
just one more doubt about the same question.

which is the real part?:confused:
• September 1st 2007, 05:17 AM
topsquark
Quote:

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

which is the real part?:confused:

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
• September 1st 2007, 05:10 PM
ThePerfectHacker
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.