Complex Numbers And Quadratic Equations question 214
Question: $ {{( \frac{1+\cos \varphi +i\sin \varphi }{1+\cos \varphi -i\sin \varphi } )}^{n}}= $
Options:
A) $ \cos n\varphi -i\sin n\varphi $
B) $ \cos n\varphi +i\sin n\varphi $
C) $ \sin n\varphi +i\cos n\varphi $
D) $ \sin n\varphi -i\cos n\varphi $
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Answer:
Correct Answer: B
Solution:
L.H.S. $ ={{[ \frac{2{{\cos }^{2}}(\varphi /2)+2i\sin (\varphi /2)\cos (\varphi /2)}{2{{\cos }^{2}}(\varphi /2)-2i\sin (\varphi /2)\cos (\varphi /2)} ]}^{n}} $ $ ={{[ \frac{\cos (\varphi /2)+i\sin (\varphi /2)}{\cos (\varphi /2)-i\sin (\varphi /2)} ]}^{n}} $ $ ={{[ \frac{{e^{i(\varphi /2)}}}{{e^{-i(\varphi /2)}}} ]}^{n}}={{({e^{i\varphi }})}^{n}} $ $ =\cos n\varphi +i\sin n\varphi $ .
BETA
