Problem 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 $$0\leq m <n$$.