Complex roots

Every complex polynomial factors completely into degree 1 complex polynomials. Since real polynomials are a special case of complex polynomials, they can also be factored into degree 1 complex polynomials.

Finding the $n^\text{th}$ complex roots of a number

We want to solve $z^n = m$ for $z$ , where $z,n,m \in \mathbb C$ . First write $z$ and $m$ in polar form, $z = re^{i\theta}$ , $m = r_k e ^{i \theta_m}$ .

\begin{align*} (re^{i\theta})^n &= r_m e^{i\theta_m} \\ r^n e^{in\theta} &= r_m e^{i\theta_m}. \end{align*}

Then the solutions are $z=re^{i\theta}$ such that $r$ satisfies $r^n = r_m$ and $\theta$ satisfies $n\theta = \theta_m + 2\pi k$ for some $k \in \Z$ .

Example

Let’s solve $z = \sqrt[4]{i}$ .

\begin{align*} (r e^{i\theta})^4 &= i \\ r^4 e^{4i\theta} &= e^{-i\pi/2} \\ z &= e^{-i(\pi/8 - \pi k/2)} \\ &= e^{-i\pi/8} e^{i \pi k/2}. \end{align*}

Complex roots

Finding the nthn^\text{th}nth complex roots of a number

Finding the $n^\text{th}$ complex roots of a number