Proving the Statement: (1+u^2)^2+(1+u)^2>0

[LagrangeEuler]

Homework Statement

Prove that $(1+u^2)^2+(1+u)^2>0$
where $u=exp(\frac{2\pi i}{3})$

Homework Equations

The Attempt at a Solution

$u^4=u$ so
$(1+u^2)^2+(1+u)^2=3u^2+3u+2>0$
is there any way from here to get $(1+u^2)^2+(1+u)^2>0$?

 

[cpt_carrot]

Note that $u^2=u^*$ and so $u^2+u$ is real

 

[LagrangeEuler]

Maybe this $3u^2+3u+2>0$ because discriminant is complex? Another way?

 

[tiny-tim]

Hi LagrangeEuler!

Hint: u³ - 1 = 0 and u - 1 ≠ 0, sooo … ?

(hmm … is it true?)

 

[HallsofIvy]

In fact, do you have any reason to think that $(1+ u^2)^2+ (1+ u)^2$ is even a real number?

Here is my very simple-minded, not at all sophisticated look at it:

$$u= e^{2\pi i/3}= -1- i\sqrt{3}/2$$
$$u^2= \frac{1}{4}+ i\sqrt{3}$$
$$1+ u^2= \frac{5}{4}+ i\sqrt{3}$$
$$(1+ u^2)^2= -\frac{13}{16}+ 5i\sqrt{3}/2$$
While $1+ u$ is $-i\sqrt{3}/2$ so $(1+ u)^2$ is real and the sum has non-zero imaginary part.

 

[LCKurtz]

You have $3u^2+3u+2$ and it has been noted that $u^2 = \bar u$ so you have $3u+3\bar u + 2 = 3(u + \bar u)+2 = 3\cdot 2\mathcal{Re} (u) + 2$, which is easy enough to calculate.

 

[cpt_carrot]

In fact, do you have any reason to think that $(1+ u^2)^2+ (1+ u)^2$ is even a real number?

Here is my very simple-minded, not at all sophisticated look at it:

$$u= e^{2\pi i/3}= -1- i\sqrt{3}/2$$

This is incorrect since $|u|=1$ it should be $u= -1/2- i\sqrt{3}/2$ from which the algebra follows