# Using the complex plane, how do you evaluate and locate the distinct solutions to (w)^4 = -1?

Mar 5, 2016

$\left\{\frac{\sqrt{2}}{2} \left(1 + i\right) , \frac{\sqrt{2}}{2} \left(- 1 + i\right) , \frac{\sqrt{2}}{2} \left(- 1 - i\right) , \frac{\sqrt{2}}{2} \left(1 - i\right)\right\}$

#### Explanation:

For this, we will use Euler's Formula ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$
Using this, we can express any complex number as $R {e}^{i \theta}$ for some $R , \theta \in \mathbb{R}$

First, note that $- 1 = \cos \left(\pi\right) + i \sin \left(\pi\right) = {e}^{i \pi}$

However, we can add or subtract $2 \pi$ from an angle without changing the value of the complex number, so as a more general form, we have

$- 1 = {e}^{i \left(\pi + 2 \pi k\right)}$ for some integer $k$

As $R = 1$ in this case, we know that any roots of $- 1$ will also have $R = 1$. Then, we can write the general form of the roots as follows:

Let $w = {e}^{i \theta}$

$\implies {e}^{i \left(\pi + 2 \pi k\right)} = {\left({e}^{i \theta}\right)}^{4} = {e}^{i \left(4 \theta\right)}$

$\implies 4 \theta = \pi + 2 \pi k$

$\implies \theta = \frac{\pi}{4} + \frac{\pi}{2} k$

Restricting theta to $\left[0 , 2 \pi\right)$ this gives us

$\theta \in \left\{\frac{\pi}{4} , \frac{3 \pi}{4} , \frac{5 \pi}{4} , \frac{7 \pi}{4}\right\}$

$\implies w \in \left\{{e}^{i \frac{\pi}{4}} , {e}^{i \frac{3 \pi}{4}} , {e}^{i \frac{5 \pi}{4}} , {e}^{i \frac{7 \pi}{4}}\right\}$

Thus, converting back to to $a + b i$ form using Euler's formula gain, we end with the solution set

$\left\{\frac{\sqrt{2}}{2} \left(1 + i\right) , \frac{\sqrt{2}}{2} \left(- 1 + i\right) , \frac{\sqrt{2}}{2} \left(- 1 - i\right) , \frac{\sqrt{2}}{2} \left(1 - i\right)\right\}$