- #1

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## 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##?

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- Thread starter LagrangeEuler
- Start date

- #1

- 661

- 16

Prove that ##(1+u^2)^2+(1+u)^2>0##

where ##u=exp(\frac{2\pi i}{3})##

##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##?

- #2

- 29

- 4

Note that [itex]u^2=u^*[/itex] and so [itex]u^2+u[/itex] is real

- #3

- 661

- 16

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

- #4

tiny-tim

Science Advisor

Homework Helper

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Hi LagrangeEuler!

Hint: u^{3} - 1 = 0 and u - 1 ≠ 0, sooo … ?

(hmm …*is* it true?)

Hint: u

(hmm …

- #5

HallsofIvy

Science Advisor

Homework Helper

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

[tex]u= e^{2\pi i/3}= -1- i\sqrt{3}/2[/tex]

[tex]u^2= \frac{1}{4}+ i\sqrt{3}[/tex]

[tex]1+ u^2= \frac{5}{4}+ i\sqrt{3}[/tex]

[tex](1+ u^2)^2= -\frac{13}{16}+ 5i\sqrt{3}/2[/tex]

While [itex]1+ u[/itex] is [itex]-i\sqrt{3}/2[/itex] so [itex](1+ u)^2[/itex] is real and the sum has non-zero imaginary part.

- #6

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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.

Last edited:

- #7

- 29

- 4

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

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

[tex]u= e^{2\pi i/3}= -1- i\sqrt{3}/2[/tex]

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

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