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July 11th, 2016, 04:58 AM   #1
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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.
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July 11th, 2016, 05:18 AM   #2
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Math Focus: tetration
it is false 1/((1 - i )^2) is not 1
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July 11th, 2016, 05:29 AM   #3
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It is true.

$\displaystyle (1-i)(1-i)^{-1}=\dfrac{1-i}{1-i}=1$

Where's the difficulty?
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July 11th, 2016, 06:11 AM   #4
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I got that far too but I couldn't believe it would be that easy.. Just thought there was a catch somewhere..
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July 11th, 2016, 09:38 AM   #5
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Originally Posted by greg1313 View Post
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 09:42 AM.
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July 11th, 2016, 11:07 AM   #6
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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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