91Ó°ÊÓ

(In this step, we express statements a) and b) in terms of the properties of the \(n\)th roots of unity, which are the complex numbers \(z\) such that \(z^n = 1\). a) There exists an \(n\)th root of unity, \(\alpha\), such that \(1+\alpha\) is also an \(n\)th root of unity. In other words, there exists an \(\alpha\) such that \(\alpha^n=1\) and \((1+\alpha)^n=1\). b) There exists an \(n\)th root of unity, \(\beta\), such that \(1-\beta\) is also an \(n\)th root of unity. In other words, there exists an \(\beta\) such that \(\beta^n=1\) and \((1-\beta)^n=1\).)

(Assume that statement a) holds, i.e., there exists an \(\alpha\) such that \(\alpha^n=1\) and \((1+\alpha)^n=1\). We want to show that there exists a \(\beta\) such that \(\beta^n=1\) and \((1-\beta)^n=1\). From the complex conjugate property, we know that if a complex number \(z\) is a root of unity, then its complex conjugate \(z^*=1/z\) is also a root of unity. Since the conjugate of \(\alpha\), denoted by \(\alpha^*\), is also an \(n\)th root of unity, using the property of complex conjugates, we have: $$ (\alpha^*)^n = \left( \frac{1}{\alpha} \right)^n = \frac{1}{\alpha^n} = \frac{1}{1} = 1. $$ Now, let's consider the complex number \(1-\alpha^*\): $$ (1-\alpha^*)^n = 1^n - \binom{n}{1}\alpha^{*n} + \binom{n}{2}\alpha^{*2n} -\ldots = 1 - n\alpha^* + \binom{n}{2}(\alpha^*)^2 -\ldots $$ Since \(\alpha^n=1\) and \((1+\alpha)^n=1\), we can rewrite this as: $$ (1-\alpha^*)^n = 1 - n\alpha^* + \binom{n}{2}(1+\alpha)^2 -\ldots = 1 - n\alpha^* + \binom{n}{2}(0)^2 -\ldots = 1 - n\alpha^* $$ We see that \((1-\alpha^*)^n = 1 - n\alpha^*\), which satisfies statement b) for \(\beta = \alpha^*\). Thus, if a) is true, then b) must also be true.)

(Now, assume that statement b) holds, i.e., there exists a \(\beta\) such that \(\beta^n=1\) and \((1-\beta)^n=1\). We want to show that there exists an \(\alpha\) such that \(\alpha^n=1\) and \((1+\alpha)^n=1\). Repeat the reasoning from the previous step but reverse the roles of a) and b). Let \(\beta^*\) be the complex conjugate of \(\beta\), we know that: $$ (\beta^*)^n = \left( \frac{1}{\beta} \right)^n = \frac{1}{\beta^n} = 1. $$ Now, consider the complex number \(1+\beta^*\): $$ (1+\beta^*)^n = 1^n + \binom{n}{1}\beta^{*n} + \binom{n}{2}\beta^{*2n} +\ldots = 1 + n\beta^* + \binom{n}{2}(\beta^*)^2 +\ldots $$ Since \(\beta^n=1\) and \((1-\beta)^n=1\), we can rewrite this as: $$ (1+\beta^*)^n = 1 + n\beta^* + \binom{n}{2}(1-\beta)^2 +\ldots = 1 + n\beta^* + \binom{n}{2}(0)^2 +\ldots = 1 + n\beta^* $$ We see that \((1+\beta^*)^n = 1 + n\beta^*\), which satisfies statement a) for \(\alpha = \beta^*\). Thus, if b) is true, then a) must also be true.)

(We have shown that a) implies b) and b) implies a). Therefore, the statements a) and b) are equivalent. Theorem is proven.)