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Question Number 159322 by Ar Brandon last updated on 15/Nov/21

P(z)=(1+i(√3))z^2 −(−4+4i)z+2icos((π/5))−2sin((π/5)) Let S denote the sum of roots of P(z) a) Express S in algebraic form then in exponential form. b. Deduce the exact values of cos(((5π)/(12))) and sin(((5π)/(12))).

$$\mathrm{P}\left(\mathrm{z}\right)=\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} −\left(−\mathrm{4}+\mathrm{4}{i}\right){z}+\mathrm{2}{i}\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{2sin}\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$$\mathrm{Let}\:{S}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{P}\left({z}\right) \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Express}\:{S}\:\mathrm{in}\:\mathrm{algebraic}\:\mathrm{form}\:\mathrm{then}\:\mathrm{in}\:\mathrm{exponential}\:\mathrm{form}. \\ $$$$\mathrm{b}.\:\mathrm{Deduce}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{values}\:\mathrm{of}\:\mathrm{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\:\mathrm{and}\:\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right). \\ $$

Answered by mindispower last updated on 16/Nov/21

S=((−4+4i)/(1+i(√3)))=(−1+i)(1−i(√3))=2(√2).e^(((3π)/4)i) .e^(−((iπ)/3)) =−1+(√3)+i(1+(√3))=2(√2).e^(((5π)/(12))i) ⇒e^(i((5π)/(12))) =(((√3)−1)/( 2(√2)))+((i(1+(√3)))/(2(√2))) cos(((5π)/(12)))=(((√3)−1)/(2(√2)))=(((√6)−(√2))/4) sin(((5π)/(12)))=(((√2)+(√6))/4)

$${S}=\frac{−\mathrm{4}+\mathrm{4}{i}}{\mathrm{1}+{i}\sqrt{\mathrm{3}}}=\left(−\mathrm{1}+{i}\right)\left(\mathrm{1}−{i}\sqrt{\mathrm{3}}\right)=\mathrm{2}\sqrt{\mathrm{2}}.{e}^{\frac{\mathrm{3}\pi}{\mathrm{4}}{i}} .{e}^{−\frac{{i}\pi}{\mathrm{3}}} \\ $$$$=−\mathrm{1}+\sqrt{\mathrm{3}}+{i}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)=\mathrm{2}\sqrt{\mathrm{2}}.{e}^{\frac{\mathrm{5}\pi}{\mathrm{12}}{i}} \\ $$$$\Rightarrow{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{12}}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}+\frac{{i}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${cos}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${sin}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{4}} \\ $$

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