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!
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.
Now, off you go. Try what you've learned.x ^ 4 = -4
equations on the set of complex numbers?
Thanks in advance!
See also this discussion at Dr. Math Forum.
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 $\displaystyle \sqrt{-2i}$.