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