If \(z=x-\) iy and \(z^{(1 / 3)}=p+i\) q then \(\left[\\{(x / p)+(y / q)\\} /\left(p^{2}+q^{2}\right)\right]\) (a) 2 (b) \(-1\) (c) 1 (d) \(-2\)

Given, \(z = x - iy\). If \(z^{(1/3)} = p + iq\), we can convert the cubic root of \(z\) to the form \(r (\cos θ + i \sin θ)\) and apply de Moivre's theorem. This theorem states that: \[ (r (\cos θ + i \sin θ))^n = r^n (\cos nθ + i \sin nθ) \] where \(n = 1/3\). Let's find the value of \(r\) and \(θ\). Here, \(r = \sqrt{x^2 + y^2}\) and \(θ = \tan^{-1}(-y/x)\). So, \(z^{(1/3)} = \sqrt[3]{r} (\cos(θ/3) + i \sin(θ/3)) \)

Substitute this expression for \(z^{(1/3)} = p + iq\) to get: \[ \sqrt[3]{r} (\cos(θ/3) + i \sin(θ/3)) = p + iq \] Equating the real (p) and imaginary parts (q), we get: \[ \sqrt[3]{r} \cos(θ/3) = p \] and \[ \sqrt[3]{r} \sin(θ/3) = q \] From the above equations, we can rewrite \(x/p\) and \(y/q\) as: \[ x/p = \sqrt[3]{r}\cos(θ/3) / p = \cos(θ/3) \] and \[ y/q = \sqrt[3]{r}\sin(θ/3) / q = \sin(θ/3) \]

Now we substitute \(x/p = \cos(θ/3)\) and \(y/q = \sin(θ/3)\) into \(\left[(x/p) + (y/q)\right]/ \left(p^2 + q^2\right)\), to find our answer. After substitution, we get: \(\left[\cos(θ/3) + \sin(θ/3)\right]/ \left(p^2 + q^2\right)\). As \(p^2 + q^2 = r^{2/3}\), thus we have: \(\left[\cos(θ/3) + \sin(θ/3)\right]/ r^{2/3}\). As \(r = \sqrt{x^2 + y^2}\), it won't be zero, so the value of the expression depends only on \(\cos(θ/3) + \sin(θ/3)\) which could be between -2 and 2. So, the Given options are \(a) 2\), \(b) -1\), \(c) 1\), \(d) -2\). We can see from the options that all except \(b) -1\) satisfy the range of \(\cos(θ/3) + \sin(θ/3)\) Thus, \(b) -1\) is the incorrect value, as it lies in the range. So The correct choice is \(b) -1\).