# Complex numbers

## Homework Statement

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

## 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##?

## Answers and Replies

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

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

tiny-tim
Science Advisor
Homework Helper
Hi LagrangeEuler!

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

(hmm … is it true?)

HallsofIvy
Science Advisor
Homework Helper
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
Science Advisor
Homework Helper
Gold Member
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.

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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