Factoring equation with real coefficients

Tags:
1. Mar 18, 2015

Nathew

1. The problem statement, all variables and given/known data
Find the roots of $$z^4+4=0$$ and use that to factor the expression into quadratic factors with real coefficients.

2. Relevant equations
DeMoivre's formula.

3. The attempt at a solution
I have been able to identify they are $$\pm 1 \pm i$$ but i have no idea how to factor the expression.
Thanks!!

2. Mar 18, 2015

Dick

No, they aren't. None of those are roots. Try them. Use deMoivre!

3. Mar 19, 2015

Nathew

$$(1+i)^4+4=0$$
$$(-1+i)^4+4=0$$
$$(1-i)^4+4=0$$
$$(-1-i)^4+4=0$$

Not sure what you're talking about.

4. Mar 19, 2015

SammyS

Staff Emeritus
It is possible to do this in the reverse order. That is:
1. Factor $\ z^4+4\$ into quadratic factors with real coefficients.
then
2. find the roots of $\ z^4+4=0\ .\$​

That's not following the instructions, but it may give some insight.

Suppose $\ z^4+4=(z^2+az+2)(z^2+bz+2)\$.
Expand the right hand side & equate coefficients.

5. Mar 19, 2015

Dick

Sorry! I read your post as saying the roots were $\pm 1$ and $\pm i$. If $r_1$ and $r_2$ are roots then $(z-r_1)(z-r_2)$ is a factor of your polynomial. Try multiplying that out when $r_1$ and $r_2$ are complex conjugates.

6. Mar 19, 2015

HallsofIvy

Staff Emeritus
Yes, 1+ i, 1- i, -1- i, and -1+ i are roots so the we can write $z^4+ 4= (z- (1+ i))(z- (1- i))(z- (-1+ i)(z- (-1- i))= (z- 1- i)(z- 1+ i)(z+ 1- i)(z+ 1+ i)$
Write $(z- 1- i)(z- 1+ i)= ((z- 1)- i)((z- 1)+ i)$ and $(z+ 1- i)(z+ 1+ i)= ((z+ 1)- i)((z+ 1)+ i)$. Now use the fact that $(a- b)(a+ b)= a^2- b^2$.

7. Mar 19, 2015

Nathew

Thanks a lot this worked great!

8. Mar 19, 2015

SammyS

Staff Emeritus
What did you get for the quadratic factors?

9. Mar 19, 2015

Nathew

$$(z^2-2z+2)(z^2+2z+2)$$

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted