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 April 21st, 2017, 08:21 AM #1 Newbie   Joined: Apr 2017 From: London Posts: 4 Thanks: 0 Equivalence relation, complex numbers https://gyazo.com/c0663bbc6abe1224ba5ea32de4568cec Any ideas on how to progress with this? Do you cube and add 2pi to the arguments? For the second part is it acceptable to rewrite -2 in exponential form and then continue from there? Thanks
April 21st, 2017, 09:16 AM   #2
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 Originally Posted by faker97 https://gyazo.com/c0663bbc6abe1224ba5ea32de4568cec Any ideas on how to progress with this? Do you cube and add 2pi to the arguments? For the second part is it acceptable to rewrite -2 in exponential form and then continue from there? Thanks
Hint: $1 \sim e^{\frac{2\pi}{3}} \sim e^{\frac{4\pi}{3}}$.

 April 21st, 2017, 09:49 AM #3 Newbie   Joined: Apr 2017 From: London Posts: 4 Thanks: 0 Thanks for the reply. So for e^(ipi/4) you would get e^(11pi/12), e^(19pi/12) and so on?
 April 21st, 2017, 11:35 AM #4 Senior Member   Joined: Aug 2012 Posts: 2,342 Thanks: 731 Hint on the other one. Why doesn't the idea of rotations work with ${-2}$?

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