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 October 13th, 2016, 12:09 PM #1 Newbie   Joined: Oct 2016 From: Sweden Posts: 2 Thanks: 0 Quick question about complex numbers So, I just started learning about complex numbers, and there has been one thing that has boggled my mind. I know how to convert simple complex numbers from rectangular form to polar form. This might sound stupid but, how do I write $\displaystyle 16 - 0i$ in polar form (since there are no angles to figure out)? Last edited by skipjack; October 24th, 2016 at 07:15 AM.
 October 13th, 2016, 02:56 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,458 Thanks: 1340 $16 + i0 = \sqrt{16^2 + 0^2}~ e^{i \arctan(0/16)} = 16 e^{i0} = 16$ Thanks from topsquark and Yoshiio
 October 13th, 2016, 04:10 PM #3 Global Moderator   Joined: May 2007 Posts: 6,766 Thanks: 697 Positive real number - angle = 0, negative real number - angle = $\displaystyle \pi$. Thanks from topsquark and Yoshiio
 October 14th, 2016, 02:27 AM #4 Newbie   Joined: Oct 2016 From: Sweden Posts: 2 Thanks: 0 Thank you so much guys!

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