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Question Number 72787 by Maclaurin Stickker last updated on 02/Nov/19

$$If{z}_1=6(\cos \frac{\pi }{4}+\sin \frac{\pi }{4})and$$$$z_2=2(\cos \frac{\pi }{5}+\mathrm{i}\times \sin \frac{\pi }{5})calculate\frac{z_1}{z_2}.$$

Commented by MJS last updated on 02/Nov/19

$$z_1\in \mathbb{R}???$$

Answered by MJS last updated on 02/Nov/19

$$z_1=6(\cos \frac{\pi }{4}+\sin \frac{\pi }{4})=6\sqrt{2}$$$$z_2=2(\cos \frac{\pi }{5}+\mathrm{i}\times \sin \frac{\pi }{5})={2\mathrm{e}}^{\mathrm{i}\frac{\pi }{5}}=\frac{1+\sqrt{5}}{2}+\frac{\sqrt{10-2\sqrt{5}}}{2}\mathrm{i}$$$$\Rightarrow \frac{z_1}{z_2}=6\sqrt{2}\times \frac{1}{2}{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{5}}=3\sqrt{2}{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{5}}=3\sqrt{2}(\cos \frac{\pi }{5}-\mathrm{i}\times \sin \frac{\pi }{5})=$$$$=\frac{3\sqrt{2}(1+\sqrt{5})}{4}-\frac{3\sqrt{5-\sqrt{5}}}{2}\mathrm{i}$$$$\mathrm{or}$$$$z_1=6(\cos \frac{\pi }{4}+\mathrm{i}\times \sin \frac{\pi }{4})={6\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}=3\sqrt{2}+3\sqrt{2}\mathrm{i}$$$$z_2=2(\cos \frac{\pi }{5}+\mathrm{i}\times \sin \frac{\pi }{5})={2\mathrm{e}}^{\mathrm{i}\frac{\pi }{5}}=\frac{1+\sqrt{5}}{2}+\frac{\sqrt{10-2\sqrt{5}}}{2}\mathrm{i}$$$$\Rightarrow \frac{z_1}{z_2}={6\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}\times \frac{1}{2}{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{5}}={3\mathrm{e}}^{\mathrm{i}\frac{\pi }{20}}=3(\cos \frac{\pi }{20}+\sin \frac{\pi }{20})=$$$$=\frac{3(\sqrt{2}(1+\sqrt{5})+2\sqrt{5-\sqrt{5}})}{8}+\frac{3(\sqrt{2}(1+\sqrt{5})-2\sqrt{5-\sqrt{5}})}{8}\mathrm{i}$$

Commented by Maclaurin Stickker last updated on 03/Nov/19

$$Thankyou,sirMJS.$$

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