A) \[n\cos \phi \]
B) \[\cos n\phi \]
C) \[\cos \left( \frac{n\phi }{2} \right)\]
D) \[\sin \left( \frac{n\phi }{2} \right)\]
Correct Answer: B
Solution :
\[{{\left( \frac{1+\cos \phi +i\sin \phi }{1+\cos \phi -i\sin \phi } \right)}^{n}}=u+iv\] \[\Rightarrow \] \[\left( \frac{2{{\cos }^{2}}\frac{\phi }{2}+2i\sin \frac{\phi }{2}\cos \frac{\phi }{2}}{2{{\cos }^{2}}\frac{\phi }{2}-2i\sin \frac{\phi }{2}\cos \frac{\phi }{2}} \right)=u+iv\] \[\Rightarrow \]\[{{\left( \frac{{{e}^{i\frac{\phi }{2}}}}{{{e}^{-i\frac{\phi }{2}}}} \right)}^{n}}=u+iv\Rightarrow ({{e}^{in\phi }})=n+iv\] \[\Rightarrow \] \[\cos n\phi +i\sin n\phi =u+iv\] \[\Rightarrow \] \[u=\cos n\phi v=\sin n\phi \]
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