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

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(A) Given $z(2-i) = 3+i$. $z = \frac{3+i}{2-i} = \frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6+3i+2i+i^2}{4+1} = \frac{6+5i-1}{5} = \frac{5+5i}{5} = 1+i$. N

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$-2^{19} i$

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(A) Given $z(2-i) = 3+i$. $z = \frac{3+i}{2-i} = \frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6+3i+2i+i^2}{4+1} = \frac{6+5i-1}{5} = \frac{5+5i}{5} = 1+i$. Now,express $z$ in polar form: $z = 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$. Using De Moivre's Theorem,$z^{38} = (\sqrt{2})^{38} \left( \cos \frac{38\pi}{4} + i \sin \frac{38\pi}{4} \right)$. $z^{38} = 2^{19} \left( \cos \frac{19\pi}{2} + i \sin \frac{19\pi}{2} \right)$. Since $\frac{19\pi}{2} = 9\pi + \frac{\pi}{2}$,we have $\cos \frac{19\pi}{2} = 0$ and $\sin \frac{19\pi}{2} = -1$. $z^{38} = 2^{19} (0 + i(-1)) = -2^{19} i$.

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

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