Let the two values of $z=\sqrt{\frac{1-i}{1+i}}$ be $z_1$ and $z_2$. If $-\frac{\pi}{2}$ $ <…
Let the two values of $z=\sqrt{\frac{1-i}{1+i}}$ be $z_1$ and $z_2$. If $-\frac{\pi}{2}$
$ < \operatorname{Arg}\left(\mathrm{z}_1\right) < \operatorname{Arg}\left(\mathrm{z}_2\right) < \pi$, then $\arg \left(\mathrm{z}_1\right)+\arg \left(\mathrm{z}_2\right)=$
$\frac{\pi}{4}$
$\frac{3 \pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
Solution
Given
$z=\sqrt{\frac{(1-i)}{(1+i)}}=\sqrt{\frac{(1-i)}{(1+i)} \times \frac{(1-i)}{(1-i)}}$
$\Rightarrow z=\sqrt{\frac{(1-i)^2}{1+1}}=\sqrt{\frac{(1-i)^2}{2}} \Rightarrow z=\frac{ \pm(1-i)}{\sqrt{2}}$
So,we have
$z_1=\frac{1-i}{\sqrt{2}}$ and $z_2=\frac{-(1-i)}{\sqrt{2}}=\frac{-1+i}{\sqrt{2}}$
$\Rightarrow \arg \left(z_1\right)=\arg \left(\frac{1-i}{\sqrt{2}}\right)=\arg \left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)$
$\Rightarrow \arg \left(z_1\right)=-\tan ^{-1}\left(\frac{\left(\frac{1}{\sqrt{2}}\right)}{\left(\frac{1}{\sqrt{2}}\right)}\right)=-\frac{\pi}{4}$ ...(i)
And $\arg \left(z_2\right)=\arg \left(\frac{-1+i}{\sqrt{2}}\right)=\arg \left(-\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)$
$\Rightarrow \quad \arg \left(z_2\right)=\pi+\tan ^{-1}\left(\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right)=\pi-\frac{\pi}{4}$
$\Rightarrow \quad \arg \left(z_2\right)=\frac{3 \pi}{4}$ ...(ii)
From (i) and (ii)
$\arg \left(z_1\right)+\arg \left(z_2\right)=-\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{2 \pi}{4}=\frac{\pi}{2}$