# Equations on the set of complex numbers

• Oct 31st 2010, 10:44 AM
Fujiwara
Equations on the set of complex numbers
Hello,
Can someone help, how to solve
x ^ 3 = 1
and/or
x ^ 4 = -4
equations on the set of complex numbers?

• Oct 31st 2010, 11:16 AM
Jhevon
I'll do the first, I leave figuring out the second to you
Quote:

Originally Posted by Fujiwara
Hello,
Can someone help, how to solve
x ^ 3 = 1

We will solve equation in polar form, and then we can convert to the more familiar $\displaystyle a + ib$ form using Euler's equation, $\displaystyle e^{i \theta} = \cos \theta + i \sin \theta$

Let $\displaystyle x = Re^{i \theta}$, then we have

$\displaystyle R^3 e^{i 3 \theta} = 1$

But by Euler's equation, $\displaystyle 1 = e^{i 2k \pi}$, hence

$\displaystyle R^3 e^{i3 \theta} = e^{i 2k \pi}$

$\displaystyle \Rightarrow R^3 = 1 \implies R = 1$, and

$\displaystyle e^{i 3 \theta} = e^{i 2k \pi}$

$\displaystyle \Rightarrow x = e^{i \theta} = e^{i \frac {2 k \pi}3}$ for $\displaystyle k = 0,~1,~2$ (after that, the solutions repeat)

Hence, $\displaystyle x = e^{0},~e^{i \frac {2 \pi}3},~e^{i \frac {4 \pi}3}$

Which, by Euler's equation, gives:

$\displaystyle x = 1, ~-\frac 12 + i \frac {\sqrt 3}2,~-\frac 12 - i \frac {\sqrt 3}2$

those are the three roots in the complex numbers.
Quote:

x ^ 4 = -4
equations on the set of complex numbers?

Another way to solve $x^3= 1$, which is equivalent to $x^3- 1= 0$ is to factor it as $(x- 1)(x^2 +x+ 1)= 0$. You can then solve $x^2+ x+ 1= 0$ with the quadratic formula.
$x^4= -4$ is the same as $x^4- (-4)= 0$ and you can think of that as the difference of two squares: $a^2- b^2= (a- b)(a+ b)$ with $a= x^2$ and $b= 2i$. $x^4-(-4)= (x^2- 2i)(x^2+ 2i)= 0$. But you will now need to change to polar form to find $\sqrt{2i}$ and $\sqrt{-2i}$.