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

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- Sep 1st 2007, 12:56 AMwhackywheelzimaginary part
Separate real and imaginary part of log (1+i)1-i.

- Sep 1st 2007, 01:13 AMtopsquark
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 - Sep 1st 2007, 01:53 AMwhackywheelzcorrection
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 - Sep 1st 2007, 04:37 AMtopsquark
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 - Sep 1st 2007, 04:56 AMwhackywheelzthanks a ton!
just one more doubt about the same question.

which is the real part?:confused: - Sep 1st 2007, 05:17 AMtopsquark
- Sep 1st 2007, 05:10 PMThePerfectHacker
Be careful the rule $\displaystyle \log (xy) = \log x + \log y \mbox{ and }\log x^y = y\log x$ does not work in $\displaystyle \mathbb{C}$ for $\displaystyle x,y\not = 0$ in general.