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 October 23rd, 2016, 07:29 PM #1 Member   Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0 sets of numbers (complex numbers) Show that (1+i/1-i)^4k=1 if k is a positive integer
 October 23rd, 2016, 07:52 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637 is this supposed to be $\left(\dfrac{1+i}{1-i}\right)^{4k}$ or $\left(1+\dfrac{i}{1-i}\right)^{4k}$ or possibly $\left(1 + \dfrac {i}{1} -i\right)^{4k}$
October 23rd, 2016, 08:55 PM   #3
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Quote:
 Originally Posted by romsek is this supposed to be $\left(\dfrac{1+i}{1-i}\right)^{4k}$
I guess it's supposed to be this

$\dfrac{1+i}{1-i} = \dfrac{1+i}{1-i}\dfrac{1+i}{1+i}=\dfrac{2i}{2}=i$

$i^4 = 1$

$i^{4k} = (i^4)^k = 1^k = 1$

$\left(\dfrac{1+i}{1-i}\right)^{4k} = 1$

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