A) \[{{\cos }^{2}}u+{{\sinh }^{2}}v\]
B) \[{{\sin }^{2}}u+{{\cosh }^{2}}v\]
C) \[{{\cos }^{2}}u+{{\cosh }^{2}}v\]
D) \[{{\sin }^{2}}u+{{\sinh }^{2}}v\]
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\[\beta =-\sin u\,\sinh v\]\[(\because \,\cos (ix)\,=\cosh x,\,\sin (ix)=i\sinh x)\] \[\therefore \,\,{{\alpha }^{2}}+{{\beta }^{2}}+1\] = \[{{\cos }^{2}}u\,{{\cosh }^{2}}v+{{\sin }^{2}}u\,\sin \,{{h}^{2}}v+1\] \[={{\cos }^{2}}u\,{{\cosh }^{2}}v+{{\sin }^{2}}u\,{{\sinh }^{2}}v+{{\cos }^{2}}u+{{\sin }^{2}}u\]\[={{\cos }^{2}}u\,\,\cos \,{{h}^{2}}v+{{\sin }^{2}}u\,(1+\sin \,{{h}^{2}}v)+{{\cos }^{2}}u\] \[={{\cos }^{2}}u\,\cos \,{{h}^{2}}v+{{\sin }^{2}}\,u\,\cos \,{{h}^{2}}v+{{\cos }^{2}}u\]\[[\because {{\cosh }^{2}}v-{{\sinh }^{2}}v=1]\] \[={{\cosh }^{2}}v\,({{\cos }^{2}}u+{{\sin }^{2}}u)+{{\cos }^{2}}u\] \[={{\cos }^{2}}u+{{\cosh }^{2}}v\,\]\[[\because \,{{\cos }^{2}}u+{{\sin }^{2}}u=1]\].
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