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$\left(\frac{1+\cos \phi+i \sin \phi}{1+\cos \phi-i \sin \phi}\right)^n=$
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Verified Answer
The correct answer is:
$\cos n \phi+i \sin n \phi$
L.H.S. $=\left[\frac{2 \cos ^2(\phi / 2)+2 i \sin (\phi / 2) \cos (\phi / 2)}{2 \cos ^2(\phi / 2)-2 i \sin (\phi / 2) \cos (\phi / 2)}\right]^n$
$\begin{aligned}=\left[\frac{\cos (\phi / 2)+i \sin (\phi / 2)}{\cos (\phi / 2)-i \sin (\phi / 2)}\right]^n= & {\left[\frac{e^{i(\phi / 2)}}{e^{-i(\phi / 2)}}\right]^n=\left(e^{i \phi}\right)^n } \\ & =\cos n \phi+i \sin n \phi .\end{aligned}$
$\begin{aligned}=\left[\frac{\cos (\phi / 2)+i \sin (\phi / 2)}{\cos (\phi / 2)-i \sin (\phi / 2)}\right]^n= & {\left[\frac{e^{i(\phi / 2)}}{e^{-i(\phi / 2)}}\right]^n=\left(e^{i \phi}\right)^n } \\ & =\cos n \phi+i \sin n \phi .\end{aligned}$
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