# Thread: Roots of a complex number

1. ## Roots of a complex number

Solve the following equation. Leave your solutions in trigonometric form. (Enter your answers from smallest to largest value of the argument of the number.)

x^4 + 2x^2 + 4= 0

i did the quad formula and got

-1 (plus/minus) 1i √3

after this everything kept going down hill. tried putting it in trig form etc etc.. help plz.

2. Originally Posted by Nismo
Solve the following equation. Leave your solutions in trigonometric form. (Enter your answers from smallest to largest value of the argument of the number.) after this everything kept going down hill. tried putting it in trig form etc etc.. help plz.

x4 + 2x2 + 4 = 0

i did the quad formula and got

-1 (plus/minus) 1i √3

after this everything kept going down hill. tried putting it in trig form etc etc.. help plz.
I'd use a dummy variable to start with.

Let $X = x^2$

so that the equation becomes

$X^2 + 2X + 4 = 0$

$X^2 + 2X + 1^2 - 1^2 + 4 = 0$

$(X + 1)^2 + 3 = 0$

$(X + 1)^2 = -3$

$X + 1 = \pm i\sqrt{3}$

$X = -1 \pm i\sqrt{3}$.

This means that

$x^2 = -1 + i\sqrt{3}$ or $x^2 = -1 - i\sqrt{3}$

You would need to convert to polars and use DeMoivre's Theorem to find $x$.

Case 1: $x^2 = -1 + i\sqrt{3}$.

$|-1 + i\sqrt{3}| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$.

Note that $-1 + i\sqrt{3}$ is in Quadrant 2.

So $\theta = \pi - \arctan{\frac{\sqrt{3}}{1}}$

$= \pi - \arctan{\sqrt{3}}$

$= \pi - \frac{\pi}{3}$

$= \frac{2\pi}{3}$.

So $x^2 = 2\,\textrm{cis}\,\frac{2\pi}{3}$

Using DeMoivre's Theorem:

$x = 2^{\frac{1}{2}}\,\textrm{cis}\,\frac{2\pi}{2\cdot 3}$

$= \sqrt{2}\,\textrm{cis}\,\frac{\pi}{3}$.

It is also important to note that there is another solution, and the two solutions are equally spaced along the unit circle. Can you find it?

Case 2: $x^2 = -1 - i\sqrt{3}$.

Follow the same procedure as above. Note that $-1 - i\sqrt{3}$ is in quadrant 4.