If cos (u + iv) = alpha + ibeta , then {alpha | vedclass

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(C) Given $\cos (u + iv) = \alpha + i\beta$. Using the identity $\cos (A + B) = \cos A \cos B - \sin A \sin B$,we have: $\cos u \cos (iv) - \sin u \si

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$\cos ^2 u + \cosh ^2 v$

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(C) Given $\cos (u + iv) = \alpha + i\beta$. Using the identity $\cos (A + B) = \cos A \cos B - \sin A \sin B$,we have: $\cos u \cos (iv) - \sin u \sin (iv) = \alpha + i\beta$. Since $\cos (iv) = \cosh v$ and $\sin (iv) = i \sinh v$,we get: $\cos u \cosh v - i \sin u \sinh v = \alpha + i\beta$. Comparing real and imaginary parts: $\alpha = \cos u \cosh v$ and $\beta = - \sin u \sinh v$. Now,calculate ${\alpha ^2} + {\beta ^2} + 1$: ${\alpha ^2} + {\beta ^2} + 1 = (\cos u \cosh v)^2 + (- \sin u \sinh v)^2 + 1$ $= \cos ^2 u \cosh ^2 v + \sin ^2 u \sinh ^2 v + 1$. Using the identity $\cosh ^2 v = 1 + \sinh ^2 v$,we substitute $1 = \cosh ^2 v - \sinh ^2 v$: $= \cos ^2 u \cosh ^2 v + \sin ^2 u \sinh ^2 v + (\cosh ^2 v - \sinh ^2 v)$ $= \cosh ^2 v (\cos ^2 u + 1) + \sinh ^2 v (\sin ^2 u - 1)$ Alternatively,using $\sin ^2 u = 1 - \cos ^2 u$: $= \cos ^2 u \cosh ^2 v + (1 - \cos ^2 u) \sinh ^2 v + 1$ $= \cos ^2 u \cosh ^2 v + \sinh ^2 v - \cos ^2 u \sinh ^2 v + 1$ $= \cos ^2 u (\cosh ^2 v - \sinh ^2 v) + (\sinh ^2 v + 1)$ $= \cos ^2 u (1) + \cosh ^2 v = \cos ^2 u + \cosh ^2 v$.

List-$I$	List-$II$
$(i)$ $a \bar{a}$	$(A)$ $-\frac{\pi}{3}$
$(ii)$ $\arg \left(\frac{1}{\bar{a}}\right)$	$(B)$ $-i \sqrt{3}$
$(iii)$ $a - \bar{a}$	$(C)$ $2i / \sqrt{3}$
$(iv)$ $\operatorname{Im}\left(\frac{4}{3a}\right)$	$(D)$ $1$
	$(E)$ $\pi / 3$
	$(F)$ $\frac{2}{\sqrt{3}}$

If $\cos (u + iv) = \alpha + i\beta ,$ then ${\alpha ^2} + {\beta ^2} + 1$ equals

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