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 July 11th, 2016, 03:58 AM #1 Newbie   Joined: Jul 2016 From: Belgium Posts: 3 Thanks: 0 complex numbers Hello, I had a little trouble in finding the solution and the how-to-solve-it of ((1-i).(1-i)^(-1) )=1 Can someone help me, please? Kind regards, V.
 July 11th, 2016, 04:18 AM #2 Senior Member   Joined: Dec 2015 From: holland Posts: 163 Thanks: 37 Math Focus: tetration it is false 1/((1 - i )^2) is not 1
 July 11th, 2016, 04:29 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond It is true. $\displaystyle (1-i)(1-i)^{-1}=\dfrac{1-i}{1-i}=1$ Where's the difficulty? Thanks from topsquark
 July 11th, 2016, 05:11 AM #4 Newbie   Joined: Jul 2016 From: Belgium Posts: 3 Thanks: 0 I got that far too but I couldn't believe it would be that easy.. Just thought there was a catch somewhere..
July 11th, 2016, 08:38 AM   #5
Senior Member

Joined: Dec 2015
From: holland

Posts: 163
Thanks: 37

Math Focus: tetration
Quote:
 Originally Posted by greg1313 It is true. $\displaystyle (1-i)(1-i)^{-1}=\dfrac{1-i}{1-i}=1$ Where's the difficulty?
oh i see (( 1 - i )((1 - i )^{-1}))

Last edited by manus; July 11th, 2016 at 08:42 AM.

 July 11th, 2016, 10:07 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 If you want to do it the "hard way"- $\displaystyle (1- i)^{-1}= \frac{1}{1- i}\frac{1+ i}{1+ i}= \frac{1+ i}{(1- i)(1+ i)}= \frac{1+ i}{2}$. Then $\displaystyle (1- i)(1- i)^{-1}= \frac{(1- i)(1+ i)}{2}= \frac{2}{2}$.

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