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Question: If \(\cos(u + iv) = x + iy,\) then \(x^{2} + y^{2} + 1\) is equal to...

If cos⁡(u+iv)=x+iy,\cos(u + iv) = x + iy, then x2+y2+1x^{2} + y^{2} + 1 is equal to

A

cos⁡2u+sinh⁡2v\cos^{2}u + \sinh^{2}v

B

sin⁡2u+cos⁡h2v\sin^{2}u + {\cos h}^{2}v

C

cos⁡2u+cosh⁡2v\cos^{2}u + \cosh^{2}v

D

sin⁡2u+sin⁡h2v\sin^{2}u + {\sin h}^{2}v

Answer

cos⁡2u+cosh⁡2v\cos^{2}u + \cosh^{2}v

Explanation

Solution

cos⁡(u+iv)=x+iy\cos(u + iv) = x + iy

⇒cos⁡ucos⁡(iv)−sin⁡usin⁡(iv)=x+iy\cos u\cos(iv) - \sin u\sin(iv) = x + iy

⇒ cos⁡ucosh⁡v−isin⁡usinh⁡v=x+iy∴\cos u\cosh v - i\sin u\sinh v = x + iy\therefore x=cos⁡ucos⁡hvx = \cos u{\cos h}v

y=−sin⁡usin⁡hvy = - \sin u{\sin h}v

x2+y2=cos⁡2u.cosh⁡2v+sin⁡2u.sinh⁡2vx^{2} + y^{2} = \cos^{2}u.\cosh^{2}v + \sin^{2}u.\sinh^{2}v= (1−sin⁡2u)cosh⁡2v+sin⁡2v[cosh⁡2v−1](1 - \sin^{2}u)\cosh^{2}v + \sin^{2}v\lbrack\cosh^{2}v - 1\rbrack = cos⁡h2v−sin⁡2v{\cos h}^{2}v - \sin^{2}v

∴\therefore x2+y2+1=cosh⁡2v+1−sin⁡2ux^{2} + y^{2} + 1 = \cosh^{2}v + 1 - \sin^{2}u = cos⁡h2v+cos⁡2u{\cos h}^{2}v + \cos^{2}u