Solving complex roots

Question

a) If $z = \cos x + i \sin x$, show that $z^{-1} = \cos x - i \sin x$

b) Show that $\cos (nx) = 0.5(z^n + z^{-n})$

Both of these questions are very simple and I get how to do them.

It then follows c) Hence solve $z^4 - 3z^3 + 4z^2 -3z + 1 = 0$

Not sure about the hence in this, how do I use parts a) and b)?

Well, I see that the expression factors as $(z-1)^2(z^2-z+1)$, but I'm not sure about the application of (a) and (b)... — Brian Tung, Apr 23 '19 at 17:00
https://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x — lab bhattacharjee, Apr 23 '19 at 18:15

score 0 · Answer 1 · answered Apr 23 '19 at 16:59

a) Use the fact that if $z=a+ib$ then $z^{-1}=\frac{a-ib}{a^2+b^2}$

b) Use De Moivre formula for $z^n$ and for $z^{-n}$ and add them.

c) Verifiy that $0$ is not a root of that polynomial and after divide both side by $z^2$ then use question b)

More details for c)

Dividing both side by $z^2$ yields: $$z^2-3z+4-3z^{-1}+z^{-2}=0\iff$$ $$(z^2+z^{-2})-3(z+z^{-1})+4=0$$

1 Answers1