Hello,

Can someone help, how to solve

x ^ 3 = 1

and/or

x ^ 4 = -4

equations on the set of complex numbers?

Thanks in advance!

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- Oct 31st 2010, 10:44 AMFujiwaraEquations 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?

Thanks in advance! - Oct 31st 2010, 11:16 AMJhevon
I'll do the first, I leave figuring out the second to youWe will solve equation in polar form, and then we can convert to the more familiar $\displaystyle \displaystyle a + ib$ form using Euler's equation, $\displaystyle \displaystyle e^{i \theta} = \cos \theta + i \sin \theta$

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

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

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

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

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

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

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

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

Which, by Euler's equation, gives:

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

those are the three roots in the complex numbers.

Quote:

x ^ 4 = -4

equations on the set of complex numbers?

Thanks in advance!

- Oct 31st 2010, 11:23 AMemakarov
See also this discussion at Dr. Math Forum.

- Nov 1st 2010, 12:49 AMFujiwara
Thanks for your help, now I understand and I can solve the problem.

- Nov 1st 2010, 08:08 AMHallsofIvy
Another way to solve $\displaystyle x^3= 1$, which is equivalent to $\displaystyle x^3- 1= 0$ is to factor it as $\displaystyle (x- 1)(x^2 +x+ 1)= 0$. You can then solve $\displaystyle x^2+ x+ 1= 0$ with the quadratic formula.

$\displaystyle x^4= -4$ is the same as $\displaystyle x^4- (-4)= 0$ and you can think of that as the difference of two squares: $\displaystyle a^2- b^2= (a- b)(a+ b)$ with $\displaystyle a= x^2$ and $\displaystyle b= 2i$. $\displaystyle x^4-(-4)= (x^2- 2i)(x^2+ 2i)= 0$. But you will now need to change to polar form to find $\displaystyle \sqrt{2i}$ and $\displaystyle \sqrt{-2i}$.