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If $z=\frac{7-i}{3-4 i}$ then $z^{14}=$
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The correct answer is:
$-2^{7} i$
$\begin{aligned} z=\frac{7-i}{3-4 i} \times \frac{3+4 i}{3+4 i} & \\=\frac{21+25 i+4}{16+9}=\frac{25(1+i)}{25}=(1+i) \\ z^{14}=(1+i)^{14}=\left[(1+i)^{2}\right]^{7}=(2 i)^{7} \\=2^{7} i^{7}=-2^{7} i \end{aligned}$
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