Finding complex roots for $x^3 - x^2 +x -1$ (hard edition)

Question

Does someone know how i can find the complex roots to the equation: $x^3 - x^2 + x - 1$

I know that I can set it up like $(x-1)(x^2+1)$, and easy see it. But if I find that $x = 1$, is that a method to find the other two roots with math. Think that is possible to do some tricks with the numbers with polar forms, but i dont remember how.

Answer 1

We have

$$x^3-x^2+x-1=(x-1)(x^2+1)=(x-1)(x+i)(x-i)$$

where $ i =\sqrt{-1}$ it the imaginary unit.

Answer 2

Alternative approach:

Let $f(x) = (x^3 - x^2 + x - 1).$

$(x^3 - x^2 + x - 1) \times (x + 1) = x^4 - 1.$

Therefore, $\displaystyle (x^3 - x^2 + x - 1) = \frac{x^4 - 1}{x+1}.$

Therefore the $3$ roots of $f(x) = 0$ must coincide with the $4$ roots of $(x^4 - 1 = 0)$, except for the elimination of the root represented by $(x + 1) = 0.$

The roots of $(x^n - 1 = 0)$ are $e^{i2k\pi/n} ~: ~n \in \Bbb{Z^+}, ~k \in \{0,1,\cdots,[n-1]\}.$

Answer 3

The equation $ \ z^3 - z^2 + z - 1 \ = \ 0 \ $ manages to stay within reach of a trigonometric calculation, if we take it that all of the roots have the same (unit) modulus. (Since we can see readily that $ \ z = 1 \ $ is a root, this is at least reasonable to try.)

If we use $ \ z \ = \ e^{i·\theta} \ = \ \cos \theta \ + \ i·\sin \theta \ \ , $ then we are looking for $$ \ e^{i·3\theta} \ - \ e^{i·2\theta} \ + \ e^{i·\theta} \ - \ 1 $$ $$ = \ \ [ \ \cos(3·\theta) \ - \ \cos(2·\theta) \ + \ \cos\theta \ - \ 1 \ ] \ + \ i·[ \ \sin(3·\theta) \ - \ \sin(2·\theta) \ + \ \sin\theta \ ] \ \ = \ \ 0 \ \ . $$

We will work with the imaginary part first, applying "multiple-angle" identities for sine:

$$ \sin(3·\theta) \ - \ \sin(2·\theta) \ + \ \sin\theta \ \ = \ \ (3·\sin \theta \ - \ 4·\sin^3 \theta) \ - \ (2·\sin \theta·\cos \theta) \ + \ \sin \theta $$ $$ = \ \ \sin \theta \ · \ ( 3 \ - \ 4·\sin^2 \theta \ - \ 2·\cos \theta \ + \ 1 ) \ \ = \ \ \sin \theta \ · \ ( 4 \ - \ 4·[1 - \cos^2 \theta] \ - \ 2·\cos \theta ) $$ $$ = \ \ \sin \theta \ · \ ( 4· \cos^2 \theta \ - \ 2·\cos \theta ) \ \ = \ \ 2·\sin \theta · \cos \theta \ · \ ( 2· \cos \theta \ - \ 1 ) \ \ = \ \ 0 \ \ . $$

The factors of this last product offer six candidate solutions: $ \ \sin \theta \ = \ 0 \ \Rightarrow \ \theta \ = \ 0 \ , \ \pi \ \ ; $ $ \ \cos \theta \ = \ 0 \ \Rightarrow \ \theta \ = \ \pm \frac{\pi}{2} \ \ ; $ $ \ 2· \cos \theta \ - \ 1 \ = \ 0 \ \Rightarrow \ \theta \ = \ \pm \frac{\pi}{3} \ \ . $

We then apply analogous identities for cosine to the real part: $$ \cos(3·\theta) \ - \ \cos(2·\theta) \ + \ \cos\theta \ - \ 1 $$ $$ = \ \ (-3·\cos \theta \ + \ 4·\cos^3 \theta) \ - \ (-1 \ + \ 2·\cos^2 \theta) \ + \ \cos \theta \ - \ 1 $$ $$ = \ \ 4·\cos^3 \theta \ - \ 2·\cos^2 \theta \ - \ 3·\cos \theta \ + \ 1 \ + \ \cos \theta \ - \ 1 $$ $$ = \ \ \cos \theta \ · \ ( 4· \cos^2 \theta \ - \ 2·\cos \theta \ - \ 2 ) \ \ = \ \ 2·\cos \theta \ · \ ( 2· \cos \theta \ + \ 1 ) \ · \ ( \cos \theta \ - \ 1 ) \ \ = \ \ 0 \ \ . $$

The factors for the real part yield $ \ \cos \theta \ = \ 0 \ \Rightarrow \ \theta \ = \ \pm \frac{\pi}{2} \ \ ; $ $ \ 2· \cos \theta \ + \ 1 \ = \ 0 \ \Rightarrow \ \theta \ = \ \pm \frac{2·\pi}{3} \ \ $ $ \ \cos \theta \ - \ 1 \ = \ 0 \ \Rightarrow \ \theta \ = \ 0 \ \ . $

The three values of $ \ \theta \ $ that cause both the real and imaginary parts to equal zero are $ \ 0 \ , \ \pm \frac{\pi}{2} \ \ . $ Hence the three zeroes of the cubic equation are
$$ \ z \ = \ e^{i·0} \ = \ 1 \ \ , \ \ e^{i·\pi/2 } \ = \ i \ \ , \ \ e^{i·(-\pi/2)} \ = \ -i \ \ . $$

Admittedly, we got a bit "lucky" with this polynomial in that it has suitable properties (low degree, resemblance to the cyclotomic polynomials) for attempting such a method of solution. One shouldn't expect this to work well as a general approach, though there are special cases (this famous one, for instance) where it has been used successfully.