1. We are given the problem: Solve the equations (a) $x^3 - 1 = 0$, (b) $x^4 + 1 = 0$, (c) $x^5 + 1 = 0$, (d) $x^6 - i = 0$, (e) $x^5 = 1 + i$, and then find the roots $\alpha, \alpha^2, \alpha^3, \alpha^4$ of $x^5 - 1 = 0$ and show that $(1-\alpha)(1-\alpha^2)(1-\alpha^3)(1-\alpha^4) = 5$. 2. (a) Solve $x^3 - 1 = 0$. This can be written as $x^3 = 1$. The roots of unity are given by $x = e^{2\pi i k / 3}$ for $k=0,1,2$. Thus, the solutions are: $$x_0 = 1,\quad x_1 = e^{2\pi i/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2},\quad x_2 = e^{4\pi i/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}.$$ 3. (b) Solve $x^4 + 1=0$, or $x^4 = -1 = e^{i\pi}$. The fourth roots are: $$x_k = e^{i(\pi + 2\pi k)/4} = e^{i(\pi/4 + \pi k/2)},\quad k=0,1,2,3.$$ Explicitly: $$x_0 = e^{i\pi/4} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2},$$ $$x_1 = e^{i3\pi/4} = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2},$$ $$x_2 = e^{i5\pi/4} = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2},$$ $$x_3 = e^{i7\pi/4} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}.$$ 4. (c) Solve $x^5 + 1 = 0$, or $x^5 = -1 = e^{i\pi}$. The roots are: $$x_k = e^{i(\pi + 2\pi k)/5} = e^{i(\pi/5 + 2\pi k/5)}, \quad k=0,...,4.$$ 5. (d) Solve $x^6 - i = 0$ or $x^6 = i = e^{i\pi/2}$. The roots are: $$x_k = e^{i(\pi/12 + \pi k/3)}, \quad k=0,...,5.$$ 6. (e) Solve $x^5 = 1 + i = \sqrt{2}e^{i\pi/4}$. Roots are: $$x_k = (\sqrt{2})^{1/5} e^{i(\pi/20 + 2\pi k/5)} = 2^{1/10} e^{i(\pi/20 + 2\pi k/5)}, \quad k=0,...,4.$$ 7. Next, find roots $\alpha, \alpha^2, \alpha^3, \alpha^4$ of $x^5 -1 = 0$: $$\alpha = e^{2\pi i /5}.$$ The roots are $1, \alpha, \alpha^2, \alpha^3, \alpha^4$. 8. Prove $(1-\alpha)(1-\alpha^2)(1-\alpha^3)(1-\alpha^4) = 5$. Note that $x^5 -1=(x-1)(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4)$. Consider the polynomial:\ $$\frac{x^5 -1}{x-1} = x^4 + x^3 + x^2 + x +1.$$ Evaluating at $x=1$ gives: $$1^4 + 1^3 + 1^2 +1 +1 = 5.$$ The left side can be written as: $$\prod_{k=1}^4 (1 - \alpha^k)$$ because if the roots of denominator are $\alpha^k$, then $\prod (1-\alpha^k)$ is the value of the numerator polynomial at $1$. 9. Hence, the identity is shown: $$(1 - \alpha)(1 - \alpha^2)(1 - \alpha^3)(1 - \alpha^4) = 5.$$