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

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