Zack Mora

2022-03-31

$\sum _{k=1}^{10}\left(\mathrm{sin}\frac{2\pi k}{11}-i\mathrm{cos}\frac{2\pi k}{11}\right)=$?

pastuh7vka

$\sum _{k=1}^{10}\left(\mathrm{sin}\frac{2\pi k}{11}-i\mathrm{cos}\frac{2\pi k}{11}\right)=-i\sum _{k=1}^{10}{e}^{\frac{i2\pi k}{11}}=-i×{e}^{\frac{i2\pi }{11}}×\frac{{\left\{{e}^{\frac{i2\pi }{11}}\right\}}^{10}-1}{{e}^{\frac{i2\pi }{11}}-1}$
$=-i{e}^{\frac{i2\pi }{11}}\frac{{e}^{\frac{i20\pi }{11}}-1}{{e}^{\frac{i2\pi }{11}}-1}$
$-i{e}^{\frac{i2\pi }{11}}\frac{{e}^{\frac{i20\pi }{11}}-1}{{e}^{\frac{i2\pi }{11}}-1}=-i\frac{{e}^{\frac{i2\pi }{11}}{e}^{\frac{i20\pi }{11}}-{e}^{\frac{i2\pi }{11}}}{{e}^{\frac{i2\pi }{11}}-1}=-i\frac{1-{e}^{\frac{i2\pi }{11}}}{{e}^{\frac{i2\pi }{11}}-1}=i$

Tristatex9tw

$-i{e}^{\frac{i2\pi }{11}}\frac{{e}^{\frac{i20\pi }{11}}-1}{{e}^{\frac{i2\pi }{11}}-1}=-i\frac{\mathrm{exp}\left(\frac{i22\pi }{11}\right)-\mathrm{exp}\left(\frac{i2\pi }{11}\right)}{\mathrm{exp}\left(\frac{i2\pi }{11}\right)-1}$
$=-i\frac{\mathrm{exp}\left(i2\pi \right)-\mathrm{exp}\left(\frac{i2\pi }{11}\right)}{\mathrm{exp}\left(\frac{i2\pi }{11}\right)-1}$
$=-i\frac{1-\mathrm{exp}\left(\frac{i2\pi }{11}\right)}{\mathrm{exp}\left(\frac{i2\pi }{11}\right)-1}$
$=-i\left(-1\right)$
=i

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