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

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

• Can you factorise it as a product of complex linear polynomials? – Angina Seng Sep 26 '20 at 14:37
• How can I do that? @AnginaSeng – User160 Sep 26 '20 at 14:38
• Hint : $(z^2+\alpha z +1)(z^2-\alpha z +1)$ ... $\alpha=?$ – Donald Splutterwit Sep 26 '20 at 14:42
• And if I've got the polynomial $z^6+1$ ? @DonaldSplutterwit – User160 Sep 26 '20 at 15:08
• Similar question : math.stackexchange.com/questions/3841247/… – Peter Sep 26 '20 at 15:29

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

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

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

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

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

$$= (z^4 + 1).$$

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
$$e^{(i\pi/6)}.$$

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

• And if I've got the polynomial $z^6+1$ ? – User160 Sep 26 '20 at 15:02
• what? how can I do it? @user2661923 – User160 Sep 26 '20 at 15:27
• @User160 For $z^6+1$ there is a similar way. – Michael Rozenberg Sep 26 '20 at 15:47
• which way exactly? @MichaelRozenberg – User160 Sep 26 '20 at 15:49
• @User160 I used $a^3+b^3=(a+b)(a^2-ab+b^2),$ $(a+b)^2=a^2+2ab+b^2$ and $a^2-b^2=(a-b)(a+b).$ – Michael Rozenberg Sep 26 '20 at 16:27

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 $$\begin{eqnarray*} z^6+1=(z^2+1)(z^2+\sqrt{3}z+1)(z^2-\sqrt{3}z+1). \end{eqnarray*}$$

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.

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.