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Question Number 208398 by mokys last updated on 14/Jun/24

$$writez=\frac{1}{\sqrt{3}+i}in{e}^{i\theta }$$

Answered by A5T last updated on 14/Jun/24

$$z=\frac{\sqrt{3}-i}{{\left(\sqrt{3}\right)}^2-{\left(i\right)}^2}=\frac{\sqrt{3}-i}{4},\mid z\mid =\sqrt{{\left(\frac{\sqrt{3}}{4}\right)}^2+{\left(\frac{-1}{4}\right)}^2}=\frac{1}{4}$$$$tan\theta =\frac{\frac{-1}{4}}{\frac{\sqrt{3}}{4}}=\frac{-1}{\sqrt{3}}\Rightarrow \theta =\frac{-\pi }{6}$$$$\Rightarrow z=\frac{e^{\frac{-i\pi }{6}}}{2}$$

Answered by mathzup last updated on 15/Jun/24

$$z=\frac{\sqrt{3}-i}{4}=\frac{\sqrt{3}}{4}-\frac{1}{4}i(=x+iy)$$$$\mid z\mid =\sqrt{{\left(\frac{\sqrt{3}}{4}\right)}^2+{(-\frac{1}{4})}^2}=\sqrt{\frac{3}{16}+\frac{1}{16}}$$$$=\frac{1}{2}wehavez=\mid z\mid e^{iarctan\left(\frac{y}{x}\right)}$$$$=\frac{1}{2}{e}^{iarctan(-\frac{1}{.\sqrt{3}})}=\frac{1}{2}{e}^{-\frac{i\pi }{6}}$$

Answered by lepuissantcedricjunior last updated on 15/Jun/24

$$\mathit{z}=\frac{1}{\sqrt{3}+\mathit{i}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}+\mathit{i}\frac{1}{2}}=\frac{\frac{1}{2}}{{\mathit{e}}^{\mathit{i}\frac{\mathit{\pi }}{6}}}=\frac{1}{2}{\mathit{e}}^{-\frac{\mathit{\pi }}{6}}$$$$=>{\mathit{e}}^{\mathit{i}\mathit{\theta }}=\frac{1}{2}{\mathit{e}}^{-\mathit{i}\frac{\mathit{\pi }}{6}}$$$$=>\mathit{\theta }=-\frac{\mathit{\pi }}{6}$$

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