If \(a=\cos \alpha+i \sin \alpha, b=\cos \beta+i \sin \beta c=\cos \gamma+i \sin \gamma\) and \((a / b)+(b / c)+(c / a)=1\) then \(\cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\gamma-\alpha)=\) (a) \((3 / 2)\) (b) \(-(3 / 2)\) (c) 0 (d) 1

The given equation is \((a / b)+(b / c)+(c / a)=1\). First, we will rewrite this equation involving complex numbers: \[\frac{\cos{\alpha} + i\sin{\alpha}}{\cos{\beta} + i\sin{\beta}} + \frac{\cos{\beta} + i\sin{\beta}}{\cos{\gamma} + i\sin{\gamma}} + \frac{\cos{\gamma} + i\sin{\gamma}}{\cos{\alpha} + i\sin{\alpha}} = 1\]

Next, we will separate the real and imaginary parts by multiplying each term of the equation by the corresponding complex conjugate in the denominator: \[\frac{(\cos{\alpha} + i\sin{\alpha})(\cos{\beta} - i\sin{\beta})}{\cos^2{\beta} + \sin^2{\beta}} + \frac{(\cos{\beta} + i\sin{\beta})(\cos{\gamma} - i\sin{\gamma})}{\cos^2{\gamma} + \sin^2{\gamma}} + \frac{(\cos{\gamma} + i\sin{\gamma})(\cos{\alpha} - i\sin{\alpha})}{\cos^2{\alpha} + \sin^2{\alpha}} = 1\] Now, use the Pythagorean identities \(\cos^2 x + \sin^2 x = 1\): \[(\cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta}) + i(\sin{\alpha}\cos{\beta} - \cos{\alpha}\sin{\beta}) + (\cos{\beta}\cos{\gamma} + \sin{\beta}\sin{\gamma}) + i(\sin{\beta}\cos{\gamma} - \cos{\beta}\sin{\gamma}) + (\cos{\gamma}\cos{\alpha} + \sin{\gamma}\sin{\alpha}) + i(\sin{\gamma}\cos{\alpha} - \cos{\gamma}\sin{\alpha}) = 1\]

Combine the real and imaginary parts: \[((\cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta}) + (\cos{\beta}\cos{\gamma} + \sin{\beta}\sin{\gamma}) + (\cos{\gamma}\cos{\alpha} + \sin{\gamma}\sin{\alpha})) + i((\sin{\alpha}\cos{\beta} - \cos{\alpha}\sin{\beta}) + (\sin{\beta}\cos{\gamma} - \cos{\beta}\sin{\gamma}) + (\sin{\gamma}\cos{\alpha} - \cos{\gamma}\sin{\alpha})) = 1\] From this equation, we can notice that the imaginary part must be equal to 0 to hold the equation true. From there, we can extract the desired trigonometric expression: \[\cos(\alpha-\beta) + \cos(\beta-\gamma) + \cos(\gamma-\alpha) = (\cos{\alpha}\cos{\beta} + \sin{\alpha}\sin{\beta}) + (\cos{\beta}\cos{\gamma} + \sin{\beta}\sin{\gamma}) + (\cos{\gamma}\cos{\alpha} + \sin{\gamma}\sin{\alpha})\] Comparing this expression with the given options: (a) \(\frac{3}{2}\) (b) \(-\frac{3}{2}\) (c) 0 (d) 1 As the sum of three cosines can never be greater than 3 or smaller than -3, the only plausible option is (d) 1: \[\cos(\alpha-\beta) + \cos(\beta-\gamma) + \cos(\gamma-\alpha) = 1\]