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  • Problem 1.19

    [1.19] Fix a positive integer \(n\) and a complex number \(w\). Find all solutions to \(z^{n}=w\).

    Let \(n\in\mathbb{Z}\) and \(w\in\mathbb{C}\).
    So \(w=re^{i\theta}\) for some non-negative \(r\in\mathbb{R}\) and \(\theta\in\mathbb{R}\).
    Then, \[\begin{aligned} z^{n}&=re^{i\theta}\\ &=re^{i\theta}e^{(2\pi m)i}\text{ For $m\in\mathbb{Z}$}\\ &=re^{i(\theta+2\pi m)}\end{aligned}\] Thus, \[\begin{aligned} z=\sqrt[n]{z^{n}}=\sqrt[n]{r}e^{\frac{i(\theta+2\pi m)}{n}}\end{aligned}\] If \(m>n\), then by the division algorithm, \(m=qn+k\) for some \(q\in\mathbb{Z}\) and some positive integer \(0\leq k <n\).
    So, \[\begin{aligned} e^{\frac{i(\theta+2\pi m)}{n}} &=e^{\frac{i\theta}{n}}e^{\frac{i2\pi (qn+k)}{n}}\\ &=e^{\frac{i\theta}{n}}e^{\frac{i2\pi qn+i2\pi k}{n}}\\ &=e^{\frac{i\theta}{n}}e^{\frac{i2\pi qn}{n}}e^{\frac{i2\pi k}{n}}\\ &=e^{\frac{i\theta}{n}}e^{i2\pi q}e^{\frac{i2\pi k}{n}}\\ &=e^{\frac{i\theta}{n}}e^{\frac{i2\pi k}{n}}\\ &=e^{\frac{i(\theta +2\pi k)}{n}}\end{aligned}\] Since \(0\leq k < n\), we need only to consider the solutions where \(0\leq m < n\). The final solution then is:
    \(z=\sqrt[n]{z^{n}}=\sqrt[n]{r}e^{\frac{i(\theta+2\pi m)}{n}}\) for all \(m\in\mathbb{Z}\) such that \(0\leq m <n\).

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