# Thread: Exercises with complex Numbers

1. ## Exercises with complex Numbers

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.

Many Thanks,

Fred

PS : Sorry for my bad english

2. Hi

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

ABC being equilateral

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

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

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

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

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

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

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

Many thanks for this quick answer

4. $\displaystyle f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

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

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

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

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

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

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

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

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

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

5. 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

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

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

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

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

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

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

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

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

6. 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

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

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

comes from :s

Thanks

8. ABC being equilateral the heights and the medians are the same

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

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

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

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

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

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

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

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

I've got a last question :

Originally Posted by running-gag
$\displaystyle f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3$

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

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

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

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

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

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

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

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

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

10. 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"

11. MANY THANKS
You're a genius
T'es Francais ?

12. 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

13. Thanks VERY VERY MUCH