Complex Numbers And Quadratic Equations question 298
Question: If $ \cos (u+iv)=\alpha +i\beta , $ then $ {{\alpha }^{2}}+{{\beta }^{2}}+1 $ equals [RPET 1999]
Options:
A) $ {{\cos }^{2}}u+{{\sinh }^{2}}v $
B) $ {{\sin }^{2}}u+{{\cosh }^{2}}v $
C) $ {{\cos }^{2}}u+{{\cosh }^{2}}v $
D) $ {{\sin }^{2}}u+{{\sinh }^{2}}v $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \cos (u+iv)=\alpha +i\beta $
$ \Rightarrow \cos u\cos (iv)-\sin u\sin (iv)=\alpha +i\beta $
$ \Rightarrow $ $ \alpha =\cos u\cosh v $ and $ \beta =-\sin u\sinh v $ $ (\because \cos (ix)=\cosh x,\sin (ix)=i\sinh x) $
$ \therefore {{\alpha }^{2}}+{{\beta }^{2}}+1 $ = $ {{\cos }^{2}}u{{\cosh }^{2}}v+{{\sin }^{2}}u\sin h^{2}v+1 $ $ ={{\cos }^{2}}u{{\cosh }^{2}}v+{{\sin }^{2}}u{{\sinh }^{2}}v+{{\cos }^{2}}u+{{\sin }^{2}}u $ $ ={{\cos }^{2}}u\cos h^{2}v+{{\sin }^{2}}u(1+\sin h^{2}v)+{{\cos }^{2}}u $ $ ={{\cos }^{2}}u\cos h^{2}v+{{\sin }^{2}}u\cos h^{2}v+{{\cos }^{2}}u $ $ [\because {{\cosh }^{2}}v-{{\sinh }^{2}}v=1] $ $ ={{\cosh }^{2}}v({{\cos }^{2}}u+{{\sin }^{2}}u)+{{\cos }^{2}}u $ $ ={{\cos }^{2}}u+{{\cosh }^{2}}v $ $ [\because {{\cos }^{2}}u+{{\sin }^{2}}u=1] $ .
BETA
