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Question Number 199642 by mokys last updated on 06/Nov/23

Commented by mokys last updated on 07/Nov/23

$$wherssteps$$

Commented by Frix last updated on 07/Nov/23

$${\mathrm{e}}^{-\frac{5\pi }{12}\mathrm{i}}\sqrt{2}$$

Answered by Frix last updated on 07/Nov/23

$$z=\frac{\sqrt{3}-\mathrm{i}}{1+\mathrm{i}}=\frac{(\sqrt{3}-\mathrm{i})(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})}=\frac{-1+\sqrt{3}}{2}-\frac{1+\sqrt{3}}{2}\mathrm{i}$$$$\mid z\mid =\sqrt{{\left(\frac{-1+\sqrt{3}}{2}\right)}^2+{(-\frac{1+\sqrt{3}}{2})}^2}=\sqrt{2}$$$$\arg \left(z\right)={\tan }^{-1}\frac{-\frac{1+\sqrt{3}}{2}}{\frac{-1+\sqrt{3}}{2}}=-{\tan }^{-1}(2+\sqrt{3})=-\frac{5\pi }{12}$$$$z={\mathrm{e}}^{-\frac{5\pi }{12}\mathrm{i}}\sqrt{2}$$

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