A) \[{{2}^{7}}\]
B) \[{{2}^{7}}i\]
C) \[{{2}^{14}}i\]
D) \[-{{2}^{7}}i\]
E) \[-{{2}^{14}}\]
Correct Answer: D
Solution :
\[z=\frac{7-i}{3-4i}=\frac{(7-i)(3+4i)}{{{(3)}^{2}}-{{(4i)}^{2}}}=(1+i)\] \[\therefore \]\[{{z}^{14}}={{(1+i)}^{14}}={{\left\{ \sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \right\}}^{14}}\] \[={{2}^{7}}\left( \cos \frac{7\pi }{2}+i\sin \frac{7\pi }{2} \right)\] \[={{2}^{7}}(-i)=-{{2}^{7}}i\]
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