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Being $z=x+yi$ how can I factorize the polynomial $z^4+1$ as a product of real quadratic polynomials?


I don't really understand what I am asked to do. How can I start with it?

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  • $\begingroup$ Can you factorise it as a product of complex linear polynomials? $\endgroup$ – Angina Seng Sep 26 '20 at 14:37
  • $\begingroup$ How can I do that? @AnginaSeng $\endgroup$ – User160 Sep 26 '20 at 14:38
  • $\begingroup$ Hint : $(z^2+\alpha z +1)(z^2-\alpha z +1)$ ... $ \alpha=?$ $\endgroup$ – Donald Splutterwit Sep 26 '20 at 14:42
  • $\begingroup$ And if I've got the polynomial $z^6+1$ ? @DonaldSplutterwit $\endgroup$ – User160 Sep 26 '20 at 15:08
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    $\begingroup$ Similar question : math.stackexchange.com/questions/3841247/… $\endgroup$ – Peter Sep 26 '20 at 15:29
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I agree with the other answers but prefer an intuitive approach, which uses the idea that $e^{(i\theta)} = \cos \theta + i\sin \theta$.

You want all values $e^{(i\alpha)}$ such that
$\left[e^{(i\alpha)}\right]^4 = e^{(i4\alpha)} = -1 = e^{(i\pi)}.$

The easiest way to do that is to pretend that
$e^{(i\pi)}$ can actually be represented by the 4 elements $\{e^{(i\pi)}, e^{(i3\pi)}, e^{(i5\pi)}, e^{(i7\pi)}\}.$

Then, with the argument of each of the 4 elements divided by 4,
you see that the 4 distinct roots are
$\{e^{(i\pi/4)}, e^{(i3\pi/4)}, e^{(i5\pi/4)}, e^{(i7\pi)/4}\}.$

Having identified the 4 roots, you need to combine them into conjugate pairs, and then use each pair of roots to form a quadratic.

This results in $\left[\left(z - e^{(i\pi/4)}\right) \left(z - e^{(i7\pi/4)}\right)\right] \times \left[\left(z - e^{(i3\pi/4)}\right) \left(z - e^{(i5\pi/4)}\right)\right] $

$=\left[ \left(z - \langle\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\rangle\right) \left(z - \langle\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\rangle\right) \right]$

$\times \left[ \left(z - \langle-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\rangle\right) \left(z - \langle-\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\rangle\right) \right] $

$= \left(z^2 -\sqrt{2}z + \frac{1}{2} + \frac{1}{2}\right) \times \left(z^2 +\sqrt{2}z + \frac{1}{2} + \frac{1}{2}\right) $

$= \left(z^2 -\sqrt{2}z + 1\right) \times \left(z^2 +\sqrt{2}z + 1\right) $

$= (z^4 + 1).$

Addendum
Per OP's request : attacking $(z^6 + 1) = 0.$

As in the original answer, pretend that
$e^{(i\pi)}$ can actually be represented by the 6 elements $\{e^{(i\pi)}, e^{(i3\pi)}, e^{(i5\pi)}, e^{(i7\pi)}, e^{(i9\pi)}, e^{(i11\pi)}\}.$

Then, with the argument of each of the 6 elements divided by 6,
you see that the 6 distinct roots are
$\{e^{(i\pi/6)}, e^{(i3\pi/6)}, e^{(i5\pi/6)}, e^{(i7\pi)/6}, e^{(i9\pi/6)}, e^{(i11\pi)/6}\}.$

Having identified the 6 roots, you need to combine them into conjugate pairs, and then use each pair of roots to form a quadratic.

This results in $\left[\left(z - e^{(i\pi/6)}\right) \left(z - e^{(i11\pi/6)}\right)\right]$

$\times \left[\left(z - e^{(i3\pi/6)}\right) \left(z - e^{(i9\pi/6)}\right)\right] $

$\times \left[\left(z - e^{(i5\pi/6)}\right) \left(z - e^{(i7\pi/6)}\right)\right]. $

The rest of the conversion into real quadratics would follow the same method as in the original answer, simply multiplying everything out.

What makes this answer convenient is that all 6 roots of
$\left[e^{i\pi}\right]^{(1/6)}$ are special angles
each of whose sin and cosine key off of
$e^{(i\pi/6)}.$

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I think, the following is better. $$z^4+1=z^4+2z^2+1-2z^2=(z^2+1)^2-(\sqrt2z)^2=$$ $$=(z^2-\sqrt2z+1)(z^2+\sqrt2z+1).$$

Also, $$z^6+1=(z^2+1)(z^4-z^2+1)=(z^2+1)((z^2+1)^2-3z^2)=$$ $$=(z^2+1)(z^2-\sqrt3z+1)(z^2+\sqrt3z+1).$$

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  • $\begingroup$ And if I've got the polynomial $z^6+1$ ? $\endgroup$ – User160 Sep 26 '20 at 15:02
  • $\begingroup$ what? how can I do it? @user2661923 $\endgroup$ – User160 Sep 26 '20 at 15:27
  • $\begingroup$ @User160 For $z^6+1$ there is a similar way. $\endgroup$ – Michael Rozenberg Sep 26 '20 at 15:47
  • $\begingroup$ which way exactly? @MichaelRozenberg $\endgroup$ – User160 Sep 26 '20 at 15:49
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    $\begingroup$ @User160 I used $a^3+b^3=(a+b)(a^2-ab+b^2),$ $(a+b)^2=a^2+2ab+b^2$ and $a^2-b^2=(a-b)(a+b).$ $\endgroup$ – Michael Rozenberg Sep 26 '20 at 16:27
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So you understood the hint \begin{eqnarray*} (z^2+\alpha z +1)(z^2-\alpha z +1)=z^4+\underbrace{(2-\alpha^2)}_{2-\alpha^2=0}z^2+1. \end{eqnarray*} To do the next one in your comment ... Factorise $z^6+1=(z^2+1)(z^4-z^2+1)$ \begin{eqnarray*} (z^2+\alpha z +1)(z^2-\alpha z +1)=z^4+\underbrace{(2-\alpha^2)}_{2-\alpha^2=-1}z^2+1. \end{eqnarray*} So \begin{eqnarray*} z^6+1=(z^2+1)(z^2+\sqrt{3}z+1)(z^2-\sqrt{3}z+1). \end{eqnarray*}

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Direct factorization treating imaginary $ (i^2=-1) $ number algebraically

$$ z^4+1= (z^2-i)(z^2+i)=(z-\sqrt i)(z+\sqrt i)(z-i\sqrt i)(z+i\sqrt i)$$

The arguments in the complex plane are odd multiples of $\pi/4$ because exponent directly multiplies/divides the argument, tips of radius vector are at $ (2k-1) \pi/2$ on unit circle.

enter image description here

If $\sin \pi/4= \cos \pi/4=\dfrac{1}{\sqrt2}=q,\;$ then the four factors are $$=(z+(-1-q))(z+(-1+q))(z+(1-q))(z+(1+q)).$$

The equation is encountered as elastic foundation ode for plates.

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