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Question Number 72787 by Maclaurin Stickker last updated on 02/Nov/19
If z_1 =6(cos (π/4)+sin (π/4)) and z_2 =2(cos (π/5)+i×sin (π/5)) calculate (z_1 /z_2 ).
$${If}\:{z}_{\mathrm{1}} =\mathrm{6}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)\:{and} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{5}}+\mathrm{i}×\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\right)\:{calculate}\:\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }. \\ $$
Commented by MJS last updated on 02/Nov/19
z_1 ∈R???
$${z}_{\mathrm{1}} \:\in\mathbb{R}??? \\ $$
Answered by MJS last updated on 02/Nov/19
z_1 =6(cos (π/4) +sin (π/4))=6(√2) z_2 =2(cos (π/5)+i×sin (π/5))=2e^(i(π/5)) =((1+(√5))/2)+((√(10−2(√5)))/2)i ⇒ (z_1 /z_2 )=6(√2)×(1/2)e^(−i(π/5)) =3(√2)e^(−i(π/5)) =3(√2)(cos (π/5) −i×sin (π/5))= =((3(√2)(1+(√5)))/4)−((3(√(5−(√5))))/2)i or z_1 =6(cos (π/4) +i×sin (π/4))=6e^(i(π/4)) =3(√2)+3(√2)i z_2 =2(cos (π/5)+i×sin (π/5))=2e^(i(π/5)) =((1+(√5))/2)+((√(10−2(√5)))/2)i ⇒ (z_1 /z_2 )=6e^(i(π/4)) ×(1/2)e^(−i(π/5)) =3e^(i(π/(20))) =3(cos (π/(20)) +sin (π/(20)))= =((3((√2)(1+(√5))+2(√(5−(√5)))))/8)+((3((√2)(1+(√5))−2(√(5−(√5)))))/8)i
$${z}_{\mathrm{1}} =\mathrm{6}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)=\mathrm{6}\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{5}}+\mathrm{i}×\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\right)=\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{5}}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow\:\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }=\mathrm{6}\sqrt{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{5}}} =\mathrm{3}\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{5}}} =\mathrm{3}\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{5}}\:−\mathrm{i}×\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\right)= \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)}{\mathrm{4}}−\frac{\mathrm{3}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{or} \\ $$$${z}_{\mathrm{1}} =\mathrm{6}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:+\mathrm{i}×\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)=\mathrm{6e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{3}\sqrt{\mathrm{2}}\mathrm{i} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{5}}+\mathrm{i}×\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\right)=\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{5}}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow\:\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }=\mathrm{6e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} ×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{5}}} =\mathrm{3e}^{\mathrm{i}\frac{\pi}{\mathrm{20}}} =\mathrm{3}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{20}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{20}}\right)= \\ $$$$=\frac{\mathrm{3}\left(\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)+\mathrm{2}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}\right)}{\mathrm{8}}+\frac{\mathrm{3}\left(\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{2}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}\right)}{\mathrm{8}}\mathrm{i} \\ $$
Commented by Maclaurin Stickker last updated on 03/Nov/19
Thank you, sir MJS.
$${Thank}\:{you},\:{sir}\:{MJS}. \\ $$

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