If $z(2-i)=(3+i)$, then $z^{38}=,($ where $z=x+$ iy $)$

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Aicharya · Accepted Answer

We have $(x+i y)(2-i)=3+i$ $	herefore 2 x+(2 y-x) i+y=3+i$ Solving eq. (1) and (2), we get $\begin{aligned} & x=1, y=1 \Rightarrow z=1+i=\sqrt{2}\left(\frac{1}{\sqrt{2}}+\mathrm{i} \frac{1}{\sqrt{2}}ight) \ & 	herefore \mathrm{z}=\sqrt{2}\left(\cos \frac{\pi}{4}+\mathrm{i} \sin \frac{\pi}{4}ight) \ & 	herefore \mathrm{z}^{38}=(\sqrt{2})^{38}\left(\cos \frac{38 \pi}{4}+\mathrm{i} \sin \frac{38 \pi}{4}ight) \ & =(2)^{19}\left[\cos \left(9 \pi+\frac{\pi}{2}ight)+\mathrm{i} \sin \left(9 \pi+\frac{\pi}{2}ight)ight] \ & =(2)^{19}\left(-\cos \frac{\pi}{2}-\mathrm{i} \sin \frac{\pi}{2}ight)=2^{19}(0-i) \ & =-\left(2^{19}ight) \mathrm{i} \end{aligned}$

If $z(2-i)=(3+i)$, then $z^{38}=,($ where $z=x+$ iy $)$

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