Solving Complex Trigo Equations with Cosine Inverse | Step-by-Step Guide

[Mentallic]

Homework Statement

Solve $$z^4-2z^3+5z^2-2z+1=0$$

Homework Equations

let $$z=cos\theta +isin\theta$$

$$z^n+z^{-n}=2cos(n\theta)$$

$$cos2\theta=2cos^2\theta -1$$

The Attempt at a Solution

Since $z\neq 0$ dividing through by $z^2$ yields:

$$(z^2+z^{-2}) -2(z+z^{-1})+5=0$$

Thus, $$2cos2\theta - 4cos\theta +5=0$$

Simplified: $$4cos^2\theta -4cos\theta+3=0$$

This is a quadratic in $cos\theta$ that doesn't have any real solutions:

$$cos\theta=\frac{1}{2}\left(1\pm \sqrt{2}i \right)$$

I've checked through my working thoroughly so I'm quite sure there aren't any mistakes so far, but I wouldn't know how to actually solve this equation's complex roots. I guess what I'm asking is how do I solve:

$$cos^{-1}\left[ \frac{1}{2}\left(1\pm \sqrt{2}i \right) \right]$$

 

[HallsofIvy]

You know, I presume, that $e^{i\theta}= cos(\theta)+ i sin(\theta)$ and, since cosine is an even function and sine an odd function, that $e^{-i\theta}= cos(\theta)- i sin(\theta)$. Adding those two equations, $e^{i\theta}+ e^{-i\theta}= 2 cos(\theta)$ and, finally, $cos(\theta)= (e^{i\theta}+ e^{-i\theta})/2$

Solve the equation $(e^{i\theta}+ e^{-i\theta})/2= \frac{1}{2}(1\pm i\sqrt{2})$. It would probably be best to let $x= i\theta$ first so that equation becomes $x+ x^{-1}= 1\pm i\sqrt{2}$ which reduces to a quadratic for x.

 

[Mentallic]

Ahh thanks Hallsofivy, nice tip

I end up with the result $$z=\frac{1 \pm _1 \sqrt {-3 \pm _2 4\sqrt{2}i}{2}$$

But I cannot simplify it to remove the imaginary unit from the surd. I'm sure an answer like this wouldn't be satisfactory. Maybe converting it into mod-arg form by approximation and reconverting back into $a+ib$ form by another approximation? of course, exact answers would be nicer