If i=sqrt{-1},then operatorname{Arg}left[frac | vedclass

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(A) Let $Z = \frac{(1+i)^{2025}}{(1-i)^{2022}}$.We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.Substituting these values:$Z

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$\frac{-\pi}{4}$

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(A) Let $Z = \frac{(1+i)^{2025}}{(1-i)^{2022}}$.We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.Substituting these values:$Z = \frac{(\sqrt{2} e^{i\pi/4})^{2025}}{(\sqrt{2} e^{-i\pi/4})^{2022}}$$Z = \frac{(\sqrt{2})^{2025} e^{i(2025\pi/4)}}{(\sqrt{2})^{2022} e^{-i(2022\pi/4)}}$$Z = (\sqrt{2})^3 e^{i(2025\pi/4 + 2022\pi/4)}$$Z = 2\sqrt{2} e^{i(4047\pi/4)}$Since $4047\pi/4 = 1011\pi + 3\pi/4$,the principal argument is $\operatorname{Arg}(Z) = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$.

If $i=\sqrt{-1}$,then $\operatorname{Arg}\left[\frac{(1+i)^{2025}}{(1-i)^{2022}}\right]=$

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