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Question Number 167916 by bounhome last updated on 29/Mar/22

$$give:z=cos\left(\frac{2\pi }{2015}\right)+isin\left(\frac{2\pi }{2015}\right)$$$$findS=1+z+z^2+z^3+...+z^{2014}$$

Commented by benhamimed last updated on 29/Mar/22

$$z=e^{i\frac{2\pi }{2015}}$$$$S=\frac{e^{i\frac{2\pi }{2015}\times 2015}-1}{e^{i\frac{2\pi }{2015}}-1}=\frac{e^{i2\pi }-1}{e^{i\frac{2\pi }{2015}}-1}=0$$

Answered by dangduomg last updated on 29/Mar/22

$$S=1+z+z^2+z^3+...+z^{2014}$$$$=\frac{1-z^{2015}}{1-z}$$$$=\frac{1-{(\cos \left(\frac{2\pi }{2015}\right)+i\sin \left(\frac{2\pi }{2015}\right))}^{2015}}{1-(\cos \left(\frac{2\pi }{2015}\right)+i\sin \left(\frac{2\pi }{2015}\right))}$$$$=\frac{1-\cos \left(2\pi \right)-i\sin \left(2\pi \right)}{1-e^{i\frac{2\pi }{2015}}}$$$$=\frac{1-1}{1-e^{2\pi i/2015}}$$$$=0$$

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