Melissi Posted February 28, 2014 Report Share Posted February 28, 2014 Hi again! I'm (still) doing complex numbers revision and while that isn't going too badly altogether, I noticed something in the answers to the calculations in polar form: sometimes (seemingly for no reason) 2pi has been subtracted from the argument, whereas sometimes it hasn't been. This seems like a really small and insignificant question, but if complex numbers is one of the topics I can do reasonably well, then I don't want to lose marks on little things like this, so my question is: is there any particular reason why 2pi is subtracted only sometimes? There really isn't anything that I can see to tell the difference between questions! Thank you! Reply Link to post Share on other sites More sharing options...
Mikhael Posted February 28, 2014 Report Share Posted February 28, 2014 (edited) Would it be possible for you to post the question(s) and the answer(s) for the exercise here?Pictures would be highly recommended Edited February 28, 2014 by Mikhael Reply Link to post Share on other sites More sharing options...
Rahul Posted March 1, 2014 Report Share Posted March 1, 2014 Do keep in mind both sin(x) and cos(x) are cyclic mod 2pi, hence cis(x) and e^(i*x) are as well. So I would assume they subtract or add 2pi to the argument in order to keep the argument within a certain domain - I'm not sure if this is [0,2pi) or (-pi,pi]. That's my best guess! 3 Reply Link to post Share on other sites More sharing options...
Melissi Posted March 1, 2014 Author Report Share Posted March 1, 2014 Thanks! I did think of that, but the questions (1c and 2a in this picture, though there were others as well) don't give a range for the values to be in, which is why I was a bit confused. I suppose it's likely that they were keeping the answers within a range that just wasn't specified, though that's a bit annoying. Thank you anyway Reply Link to post Share on other sites More sharing options...
Dax Posted March 2, 2014 Report Share Posted March 2, 2014 Oh for this, because the Value of Cos and Sin are cyclic, they will repeat after every 2pi, so you add 2(k)pi to pi/5 where k = 0,1,2,3,4,5, and then you use the De-Moivers Theorem like normal. I think that should do it if I'm not wrongP.S - I was in the same position as you, I'm pretty sure about my answer but am kinda unsure as to how many values of k are necessary. 1 Reply Link to post Share on other sites More sharing options...
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