91Ó°ÊÓ

Recall that if \(z = a + bi\), where \(a, b \in \mathbb{R}\), then its complex conjugate is denoted as \(\bar{z}\) which is the complex number \(a - bi\).

First, we need to find the complex conjugate of \(z^{n}\) (denoted as \(\overline{z^n}\)). We know that \(z^n = (a + bi)^n\). By expanding this term using the binomial theorem, we will have a real part and an imaginary part, say \(a_{n} + b_{n}i\). Then, the complex conjugate of \(z^n\) is \(\overline{z^n} = a_{n} - b_{n}i\).

Next, we will find \((\bar{z})^n\). We know that \(\bar{z} = a - bi\), so \((\bar{z})^n = (a - bi)^n\). Expanding this term using the binomial theorem, we will have a real part and an imaginary part, say \(c_{n} + d_{n}i\).

Now, we want to show that \(a_{n} - b_{n}i = c_{n} + d_{n}i\). Notice that \(z^n = (a+bi)^n = a^n+\binom{n}{1}a^{n-1}(bi)+\binom{n}{2}a^{n-2}(bi)^2+\cdots+\binom{n}{n-1}a(bi)^{n-1}+b^n(bi)^n\). From \((a+bi)(a-bi) = a^2 + b^2\), we see that any power of \((bi)\) times any power of \((-bi)\) will be real, and any odd power \((bi)\) times any odd power of \((-bi)\) will be imaginary. As a result, we can observe that the real parts and imaginary parts are equal coefficients in the expansions of \(z^n\) and its conjugate \((\bar{z})^n\). So, we have shown that both the real and imaginary parts of \(\overline{z^{n}}\) and \((\bar{z})^{n}\) are equal, thus proving that \(\overline{z^{n}}=(\bar{z})^{n}\) for every complex number \(z\) and every positive integer \(n\).