- With \(\omega = s e ^ { i \varphi }\) , where \(s \geq 0\) and \(\varphi \in \mathbb { R }\) , solve the equation \(z ^ { n } = \omega\) in C where n is a natural number. How many solutions are there?
Solution
Let \[ \omega=se^{i\varphi},\qquad s\ge0. \] If \(s=0\), then \(z^n=0\) has the unique solution \(z=0\).
Assume \(s>0\). Writing \(z=re^{i\theta}\), the equation \(z^n=\omega\) gives \[ r^n=s, \qquad n\theta=\varphi+2\pi k \] for some \(k\in\mathbb Z\). Hence \[ z_k=s^{1/n}\exp\!\left(i\frac{\varphi+2\pi k}{n}\right), \qquad k=0,1,\ldots,n-1. \] These \(n\) values are distinct, since equality of \(z_k\) and \(z_\ell\) would imply \((k-\ell)/n\in\mathbb Z\), impossible for distinct \(k,\ell\in\{0,\ldots,n-1\}\). Every integer \(k\) is congruent modulo \(n\) to one of these, so these are all solutions.
Thus there is one solution when \(\omega=0\), and exactly \(n\) solutions when \(\omega\ne0\).