A) \[-1\]
B) \[1\]
C) \[2\]
D) \[-2\]
Correct Answer: B
Solution :
We have, \[{{\left[ \cos \left( \frac{\pi }{3} \right)+i\sin \left( \frac{\pi }{3} \right) \right]}^{3/4}}\] \[={{(\cos \pi +i\sin \pi )}^{1/4}}\] \[={{[\cos (2k\pi +\pi )i\sin (2k\pi +\pi )]}^{1/4}}\] \[=\cos \left\{ \frac{(2k+1)\pi }{4} \right\}+i\sin \left\{ \frac{(2k+1)\pi }{4} \right\}\] \[=cis\left\{ \frac{(2k+1)\pi }{4} \right\}(\because \,\,\cos \theta +i\sin \theta =cis\theta )\] where \[k=0,\,\,1,\,\,2,\,\,3,....\] Now, continued product of four values \[\left\{ cis\left( \frac{\pi }{4} \right) \right\}\cdot \left\{ cis\left( \frac{3\pi }{4} \right) \right\}\cdot \left\{ cis\left( \frac{5\pi }{4} \right) \right\}\cdot \left\{ cis\left( \frac{7\pi }{4} \right) \right\}\] \[=cis\left( \frac{\pi }{4}+\frac{3\pi }{4}+\frac{5\pi }{4}+\frac{7\pi }{4} \right)\] \[=cis(4\pi )\] \[=\cos 4\pi +i\sin 4\pi \] \[=1+0=1\]
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