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# $z=\large\frac{7-i}{3-4i}$ then $z^{14}$ equals

$\begin{array}{1 1}(A)\;2^7&(B)\;2^7i\\(C)\;-2^7i&(D)\;\text{None of these}\end{array}$

$z=\large\frac{7-i}{3-4i}\times \frac{(3+4i)}{(3+4i)}=\frac{25+25i}{25}$
$\Rightarrow 1+i$
$\Rightarrow z=\sqrt 2(\cos \large\frac{\pi}{4}$$+i\sin\large\frac{\pi}{4}) \Rightarrow z^{14}=2^7(\cos \large\frac{14\pi}{4}$$+i\sin \large\frac{14\pi}{4})$
$\Rightarrow z^{14}=2^7(-i)$
$\Rightarrow -i2^7$
Hence (C) is the correct answer.