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If $\mathrm{i}=\sqrt{-1}$ then $\operatorname{Arg}\left[\frac{(1+\mathrm{i})^{2025}}{(1-\mathrm{i})^{2022}}\right]=$
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Verified Answer
The correct answer is:
$\frac{-\pi}{4}$
$\arg \left[\frac{(1+i)^{2025}}{(1-i)^{2022}}\right]$
$\frac{(1+i)^{2025}}{(1-i)^{2022}}=\frac{(\sqrt{2})^{2025}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{2025}}{(\sqrt{2})^{2022}\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{2022}}$
$\begin{aligned} & =(\sqrt{2})^3 \frac{\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{2025}}{\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)^{2022}} \\ & =\sqrt{2} e^{i\left(\frac{2025 \pi}{4}+\frac{2022 \pi}{4}\right)}=\sqrt{2} e^{i\left(\frac{-\pi}{4}\right)} \\ & \therefore \arg (Z)=-\frac{\pi}{4}\end{aligned}$
$\frac{(1+i)^{2025}}{(1-i)^{2022}}=\frac{(\sqrt{2})^{2025}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{2025}}{(\sqrt{2})^{2022}\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{2022}}$
$\begin{aligned} & =(\sqrt{2})^3 \frac{\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{2025}}{\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)^{2022}} \\ & =\sqrt{2} e^{i\left(\frac{2025 \pi}{4}+\frac{2022 \pi}{4}\right)}=\sqrt{2} e^{i\left(\frac{-\pi}{4}\right)} \\ & \therefore \arg (Z)=-\frac{\pi}{4}\end{aligned}$
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