Question Number 19455 by Tinkutara last updated on 11/Aug/17
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{z}\:=\:\mathrm{cos}\:\mathrm{6}°\:+\:{i}\:\mathrm{sin}\:\mathrm{6}°,\:\mathrm{then} \\ $$$$\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\frac{{iz}}{{z}^{\mathrm{4}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{3}} }{{z}^{\mathrm{8}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{7}} }{{z}^{\mathrm{16}} \:−\:\mathrm{1}}\:=\:\mathrm{0}. \\ $$
Answered by ajfour last updated on 16/Aug/17
$${z}^{\mathrm{15}} =\left(\mathrm{cos}\:\mathrm{6}°+{i}\mathrm{sin}\:\mathrm{6}°\right)^{\mathrm{15}} ={i}\:,\:{so} \\ $$$${l}.{h}.{s}.=\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{1}}−\frac{{iz}}{{z}^{\mathrm{4}} −\mathrm{1}}+\frac{{iz}^{\mathrm{3}} }{{z}^{\mathrm{8}} −\mathrm{1}}+\frac{{iz}^{\mathrm{7}} }{{i}\left({z}+{i}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} −{iz}+{i}^{\mathrm{2}} }{{z}^{\mathrm{4}} −\mathrm{1}}+\frac{{iz}^{\mathrm{4}} −{z}^{\mathrm{3}} +{i}−{z}^{\mathrm{7}} }{\left({z}^{\mathrm{8}} −\mathrm{1}\right)\left({z}+{i}\right)} \\ $$$$=\frac{{z}^{\mathrm{3}} +{i}^{\mathrm{3}} }{\left({z}^{\mathrm{4}} −\mathrm{1}\right)\left({z}+{i}\right)}+\frac{\left({z}^{\mathrm{4}} +\mathrm{1}\right)\left({i}−{z}^{\mathrm{3}} \right)}{\left({z}^{\mathrm{8}} −\mathrm{1}\right)\left({z}+{i}\right)} \\ $$$$=\frac{{z}^{\mathrm{3}} −{i}}{\left({z}^{\mathrm{4}} −\mathrm{1}\right)\left({z}+{i}\right)}+\frac{\left({i}−{z}^{\mathrm{3}} \right)}{\left({z}^{\mathrm{4}} −\mathrm{1}\right)\left({z}+{i}\right)}\:=\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 16/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$