Exercises with complex Numbers

**Fred_** · March 19th 2009, 08:55 AM

Dear Friends,

I've got 2 exercises I can't solve.

The first one is :

If a,b,c ∈ C and they are affix points of the plan.
Show that the triangle ABC is equilateral only if

a^2+b^2+c^2 = ab+ bc + ca

The second one is to demonstrate that :

X^4 + 2X^3 + 4X^2 + 2X + 3

has no square root in R.

I'm looking forward to your answers,

Many Thanks,

Fred

PS : Sorry for my bad english

**running-gag** · March 19th 2009, 09:26 AM

Hi

For the first one :
Let A(a), B(b) and C(c)
Let I be the middle of [AB]

ABC being equilateral

$\left(\overrightarrow{AB}+\overrightarrow{CB}\righ t)\cdot\overrightarrow{AC}=2\:\overrightarrow{IB}\ cdot\overrightarrow{AC}=0$

Therefore
$(b-a) + (b-c) = \pm i\:\sqrt{3}(c-a)$

Squaring gives
$(2b-a-c)^2 = -3(c-a)^2$

Expanding
$4b^2+a^2+c^2-4ab-4bc+2ac = -3a^2-3c^2+6ac$

$4a^2+4b^2+4c^2 = 4ab + 4bc + 4ac$

$a^2+b^2+c^2 = ab + bc + ac$

**Fred_** · March 19th 2009, 09:32 AM

Thank you very much !
Any idea for the second one ?

Many thanks for this quick answer

**running-gag** · March 19th 2009, 10:14 AM

$f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

$f'(X)=4X^3 + 6X^2 + 8X + 2$

$f"(X)=12X^2 + 12X + 8 = 4(3X^2 + 3X + 2)$

$f"(X) \geq 0$ therefore f' is increasing over R and f' has one real root $\alpha$

f' is negative over $]-\infty,\alpha]$ and positive over $[\alpha,-\infty[$

f is decreasing over $]-\infty,\alpha]$ and increasing over $[\alpha,-\infty[$

We just need to prove that $f(\alpha) > 0$

$f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

$f(X)=\left(\frac12 X + \frac14\right)f'(X) + \left(\frac54 X^2 + \frac12 X+\frac14\right)$

$f(\alpha)=\frac54 \alpha^2 + \frac12 \alpha + \frac14$ which is positive because the discriminant is negative

**Fred_** · March 19th 2009, 01:09 PM

Thanks !!! Many Thanks.
But I don't understand how you found this ...
Can you just detail how you found it ?

Thanks

Originally Posted by running-gag

Hi

For the first one :
Let A(a), B(b) and C(c)
Let I be the middle of [AB]

ABC being equilateral

$\left(\overrightarrow{AB}+\overrightarrow{CB}\righ t)\cdot\overrightarrow{AC}=2\:\overrightarrow{IB}\ cdot\overrightarrow{AC}=0$
Where does this comes from ?

Therefore
$(b-a) + (b-c) = \pm i(c-a)$
And this ?

Squaring gives
$(2b-a-c)^2 = -(c-a)^2$

Expanding
$4b^2+a^2+c^2-4ab-4bc+2ac = -a^2-c^2+2ac$

$a^2+2b^2+c^2 = 2ab + 2 bc$

This is also true when changing a into b, a into c, ...

$a^2+2b^2+c^2 = 2ab + 2 bc$
$2a^2+b^2+c^2 = 2ab + 2 ac$
$a^2+b^2+2c^2 = 2bc + 2 ac$

Summation gives
$4a^2+4b^2+4c^2 = 4ab + 4 bc + 4ac$
$a^2+b^2+c^2 = ab + bc + ac$

**running-gag** · March 19th 2009, 01:10 PM

Originally Posted by Fred_

Thank you very much !
Any idea for the second one ?

Many thanks for this quick answer

I have corrected an error in my first post

**Fred_** · March 19th 2009, 01:15 PM

Yes I saw it.
But I still don't understand where does that
[tex]
$ \left(\overrightarrow{AB}+\overrightarrow{CB}\righ t)\cdot\overrightarrow{AC}=2\:\overrightarrow{IB}\ cdot\overrightarrow{AC}=0 $

and that $ (b-a) + (b-c) = \pm i\:\sqrt{3}(c-a) $

comes from :s

Thanks

**running-gag** · March 19th 2009, 01:23 PM

ABC being equilateral the heights and the medians are the same

$\left(\overrightarrow{AB}+\overrightarrow{CB}\righ t)\cdot\overrightarrow{AC}=\left(\overrightarrow{A I}+\overrightarrow{IB}+\overrightarrow{CI}+\overri ghtarrow{IB}\right)\cdot\overrightarrow{AC}=2\:\ov errightarrow{IB}\cdot\overrightarrow{AC}=0$

$\overrightarrow{AB}$ affix is (b-a)
$\overrightarrow{CB}$ affix is (b-c)

Therefore $\overrightarrow{AB}+\overrightarrow{CB}=2\:\overri ghtarrow{IB}$ affix is (b-a)+(b-c)

$2\:\overrightarrow{IB}\cdot\overrightarrow{AC}=0$ means that there exists a rotation of angle $\frac{\pi}{2}$ between (IB) and (AC)

Therefore
$(b-a) + (b-c) = \pm i\:\alpha(c-a)$ where $\alpha$ is a real

ABC being equilateral $IB = AC \frac{\sqrt{3}}{2}$

$2\:IB = AC \:\sqrt{3}$

$|(b-a) + (b-c)| = |\alpha|\: |c-a| = \sqrt{3} \:|c-a|$ therefore $\alpha = \sqrt{3}$

**Fred_** · March 19th 2009, 01:26 PM

Thank you very much for your previous reply's.

I've got a last question :

Originally Posted by running-gag

$f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

$f'(X)=4X^3 + 6X^2 + 8X + 2$

$f"(X)=12X^2 + 12X + 8 = 4(3X^2 + 3X + 2)$ This is the derivative isn't it ?

$f"(X) \geq 0$ therefore f' is increasing over R and f' has one real root $\alpha$ Why are you taking the derivative of a derivative ?

f' is negative over $]-\infty,\alpha]$ and positive over $[\alpha,-\infty[$

f is decreasing over $]-\infty,\alpha]$ and increasing over $[\alpha,-\infty[$

We just need to prove that $f(\alpha) > 0$

$f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

$f(X)=\left(\frac12 X + \frac14\right)f'(X) + \left(\frac54 X^2 + \frac12 X+\frac14\right)$

$f(\alpha)=\frac54 \alpha^2 + \frac12 \alpha + \frac14$ which is positive because the discriminant is negative

**running-gag** · March 19th 2009, 01:29 PM

Yes f"(X) is the second derivative of f, or the derivative of f'

To find the variations of f, you need to know the sign of f' but since f' is a cubic function you need to take the second derivative f"

**Fred_** · March 19th 2009, 01:31 PM

MANY THANKS

You're a genius

T'es Francais ?

**running-gag** · March 19th 2009, 01:33 PM

Originally Posted by Fred_

MANY THANKS

You're welcome

Originally Posted by Fred_

You're a genius

Surely not

Originally Posted by Fred_

T'es Francais ?

Oui

**Fred_** · March 19th 2009, 01:37 PM

Thanks VERY VERY MUCH

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