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October 23rd, 2016, 07:29 PM   #1
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sets of numbers (complex numbers)

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(1+i/1-i)^4k=1 if k is a positive integer
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October 23rd, 2016, 07:52 PM   #2
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is this supposed to be

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

or

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

or possibly

$\left(1 + \dfrac {i}{1} -i\right)^{4k}$
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October 23rd, 2016, 08:55 PM   #3
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Quote:
Originally Posted by romsek View Post
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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