Let the two values of z = sqrt{frac{1-i}{1+i} | vedclass

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(D) Given $z = \sqrt{\frac{1-i}{1+i}}$.Multiply numerator and denominator by $(1-i)$:$z = \sqrt{\frac{(1-i)^2}{1^2+1^2}} = \sqrt{\frac{(1-i)^2}{2}} =

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$\frac{\pi}{2}$

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(D) Given $z = \sqrt{\frac{1-i}{1+i}}$.Multiply numerator and denominator by $(1-i)$:$z = \sqrt{\frac{(1-i)^2}{1^2+1^2}} = \sqrt{\frac{(1-i)^2}{2}} = \pm \frac{1-i}{\sqrt{2}}$.Thus,$z_1 = \frac{1-i}{\sqrt{2}}$ and $z_2 = \frac{-1+i}{\sqrt{2}}$.For $z_1 = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}$,the argument is $\arg(z_1) = 	an^{-1}\left(\frac{-1/\sqrt{2}}{1/\sqrt{2}}ight) = -\frac{\pi}{4}$.For $z_2 = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$,the argument is $\arg(z_2) = \pi + 	an^{-1}\left(\frac{1/\sqrt{2}}{-1/\sqrt{2}}ight) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Since $-\frac{\pi}{2} Therefore,$\arg(z_1) + \arg(z_2) = -\frac{\pi}{4} + \frac{3\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$.

Let the two values of $z = \sqrt{\frac{1-i}{1+i}}$ be $z_1$ and $z_2$. If $-\frac{\pi}{2} < \operatorname{Arg}(z_1) < \operatorname{Arg}(z_2) < \pi$,then $\arg(z_1) + \arg(z_2) = $

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