Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need help with finding the roots to the equation $z^6 - 2z^3 + 2 = 0$

I start with assigning $x$ as $z^3$. This gives me the equation: $x^2 - 2x + 2 = (x-1)^2 + 1 = 0$.

Further developments:

$(x-1)^2 = -1 = e^{i\pi} \leftrightarrow x = 1 \pm e^{i\pi\frac{1}{2}}$.

This means that $x_1 = 1 + e^{i\pi\frac{1}{2}}$ and $x_2 = 1 - e^{i\pi\frac{1}{2}}$

The problem is that my math book tells me that at this stage the result for $x$ should be $x = 1 \pm i\sqrt{2}e^{\pm i\pi\frac{1}{4}}$.

What have I done wrong?

Thank you kindly for your help!

share|improve this question
    
You are right. The solution of $x^2-2x+2$ is $1\pm i=1\pm e^{i\frac\pi 2}$ and not $1 \pm i\sqrt{2}e^{\pm i\frac\pi 4}$. –  P.. Feb 20 '13 at 19:14
1  
From the answer by @nbubis it appears that your textbook simply is missing an ' $ = $ ' between the $i$ and the $\sqrt2$. –  half-integer fan Feb 20 '13 at 19:46
add comment

2 Answers 2

up vote 2 down vote accepted

What's wrong with using the simple root formula? $$x = \frac{2\pm\sqrt{4-8}}{2} = 1\pm i = 1 \pm i=\sqrt{2}e^{\pm i\pi/4}$$ You then get: $$z = x^{1/3} = 2^{1/6}e^{\pm i\pi/12}e^{2n i\pi/3}$$

share|improve this answer
add comment

Note that

$$-1 = e^{\pm i\pi}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.