Do not use a calculator in answering this question.
The complex numbers \(z\) and \(w\) are such that \(w-6\alpha z=-5\alpha \mathrm{i}+3\) and \(\alpha {{z}^{2}}+\mathrm{i}w=10\alpha +3\mathrm{i}\), where \(\alpha \) is a non-zero real constant.
Given that \(\left| z \right|>1\), find \(z\)and \(w\) in terms of \(\alpha\) (where applicable).[3]
The complex numbers \({{z}_{1}}\) and \({{z}_{2}}\) are given by \(3-\mathrm{i}\sqrt{3}\) and \(2\mathrm{i}\) respectively.
Find the possible integer values of \(n\) such that \(\frac{{{z}_{1}}^{3n}}{{{z}_{2}}^{2n}}\) is purely imaginary with a positive imaginary part.[3]
Given that \({{z}_{1}}\) is a root of the equation \(p{{x}^{3}}-9{{x}^{2}}+6x+36=0\), find the value of the real number \(p\). Using the value of \(p\) found, express \(p{{x}^{3}}-9{{x}^{2}}+6x+36\) as a product of three linear factors.[4]