Finding the fourth roots of a complex number

Question

How would I find the fourth roots of $-81i$ in the complex numbers?

Here is what I currently have:

$w = -81i$

$r = 9$

$\theta = \arctan (-81)$?

Although I am not sure it's correct or if I am on the right track. May I have some help please?

Answer 1

$$w:= -81i = 81 e^{-i\pi/2},81 e^{3i\pi/2}, 81e^{7i\pi/2},81 e^{11i\pi/2},$$ So the 4th roots of $w$ are: $$\sqrt[4]{81} e^{-i\pi/8}, \sqrt[4]{81}e^{3i\pi/8}, \sqrt[4]{81} e^{7i\pi/8}, \sqrt[4]{81} e^{11i\pi/8},$$ Where $\sqrt[4]{81}$ denotes the unique real positive 4th root of 81. By algebraic considerations, we know there are exactly 4 4th roots. So these are all of them.

Answer 2

Hints:

1) Rewrite $-81i$ as $0-81i$.

2) Find the modulus $|z|$ and argument of $0-81i$, using the formulas $$|z| = \sqrt {a^2 + b^2}$$ and $$\arg \theta = \arctan \dfrac {b}{a}$$

3) Use Demoivre's formula $z^{1/n} = |z|^{1/n} \left(\cos \dfrac {\theta k}{n} + i \sin \dfrac {\theta k}{n}\right)$, $k \in (0,3)$ with the values you've found.

Answer 3

Hint:

It is much simpler: the (real) fourth root of $81$ is $3$. So you simply have to determine the fourth roots of $-i$. For that, use the complex exponential notation: $$\mathrm e^{4i\theta}=\mathrm e^{\tfrac{3i\pi}2},\;\text{ so }\;4\theta\equiv \frac{3\pi}2\bmod 2\pi.$$ Can you proceed?

Answer 4

Use Euler's formula: If the complex number is $z = \rho e^{i \theta} = \rho (\cos \theta + i \sin \theta)$ (polar coordinates; $\rho, \theta$ are reals), then:

$\begin{align*} z^\alpha &= \rho^\alpha \cdot e^{i \alpha \theta} \end{align*}$

In the particular case that $\alpha = 1 / n$ for a natural number $n$, as $e^{i \theta} = e^{i (\theta + 2 k \pi)}$ :

$\begin{align*} z^{1/n} &= \rho^{1/n} \cdot e^{i (\theta + 2 \pi) / n} \end{align*}$

I.e., the $n$-th roots are situated on a circle of radius $\rho^{1/n}$ around 0, distributed evenly one at an angle $\theta /n$ and the others $2 \pi / n$ apart. For $n = 4$, they form a square.