182. Another two nice problems with complex numbers (own).

by Virgil Nicula, Dec 2, 2010, 9:59 AM

Quote:
Proposed problem 1 (own). Ascertain $\lambda \in\mathbb R$ so that exist $\left\{z_1,z_2,z_3\right\}\subset\mathbb C$ for which

$\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1$ and $z_1^3+z_2^3+z_3^3+3\cdot \overline{z_1+z_2+z_3}=12-4\lambda\cdot (1+i)$ .

Proof. $\left\{\begin{array}{c} \sum z_1^3+3\cdot \overline{\sum z_1}=12-4\lambda\cdot (1+i)\\\\ \sum \overline z_1^3+3\cdot\sum z_1=12-4\lambda\cdot (1-i)\end{array}\right\|\ \implies$ $\left\{\begin{array}{ccc} \oplus & : &\sum \left(z_1+\overline {z_1}\right)^3=24-8\lambda\\\\ \ominus & : & \sum \left(z_1-\overline {z_1}\right)^3=-8\lambda\cdot i\end{array}\right\|$ . Denote $z_k=x_k+y_k\cdot i$ , where $k\in\overline{1,3}$ .

Therefore, $\left\{\begin{array}{c} \sum x_k^3=3-\lambda\\\\ \sum y_k^3=\lambda\end{array}\right\|$ $\implies$ $3=\sum\left(x_k^3+y_k^3\right)\le\sum\left(x_k^2+y_k^2\right)=3$ because for any $k\in\overline {1,3}$ we have $x_k^3+y_k^3\le x_k^2+y_k^2=\left|z_k\right|^2=1$ .

In conclusion, for any $k\in\overline {1,3}\ ,\ x_k^3+y_k^3=x_k^2+y_k^2$ $\iff$ $0\le x_k^2\left(1-x_k\right)+y_k^2\left(1-y_k\right)=0$ $\iff$ $\left\{z_1,z_2,z_3\right\}\subset \{1,i\}$ $\implies$ $\boxed{\ \lambda\in\overline{0,3}\ }$ .
Quote:
Proposed problem 2 (own). Consider the complex number $w=-\frac 12+\frac {\sqrt 3}{2}\cdot i$ . For each $z\in\left\{1,w,w^2\right\}$ define the set $M(z)$ of all ordered third

form $(a,b,c)\in \mathbb C^3$ which verify relations $\left\{\begin{array}{c} 0\not\in \{a,b,c\}\\\ a+b+c=0\\\ a^2+b^2z+c^2z^2=0\end{array}\right\|$ . Prove that $M(w)\cap M\left(w^2\right)=\emptyset$ and $M(w)\cup M\left(w^2\right)=M(1)$ .

Proof. Observe that $\left\|\begin{array}{c} w^3=1\ ,\ \overline w=\frac 1w=w^2\\\\ w^2+w+1=0\end{array}\right\|$ . Suppose that $M(w)\cap M\left(w^2\right)\ne\emptyset$ , i.e. exists $(a,b,c)\in\mathbb C^3$ so that $\left\|\begin{array}{c} a+b+c=0\\\ a^2+b^2w+c^2w^2=0\\\ a^2+b^2w^2+c^2w=0\end{array}\right\|$ .

Then $\left\|\begin{array}{ccc} \left(a^2+b^2w+c^2w^2\right)+\left(a^2+b^2w^2+c^2w\right)=0 & \implies & 2a^2=b^2+c^2\\\ \left(a^2+b^2w+c^2w^2\right)-\left(a^2+b^2w^2+c^2w\right)=0 & \implies & b^2=c^2\end{array}\right\|$ $\implies$ $a=b=c=0$ , absurd. I used that $a+b+c=0$

and $w^2+w+1=0$ . In conclusion, $\boxed{\ M(w)\cap M\left(w^2\right)=\emptyset\ }$ . Consider $(a,b,c)\in \mathbb C^3$ for which $(a,b,c)\in M(w)\cup M\left(w^2\right)$ ,

i.e. $a+b+c=0$ and $\left(a^2+b^2w+c^2w^2\right)\cdot\left(a^2+b^2w^2+c^2w\right)=0$ $\iff$ $a+b+c=0$ and $a^4+b^4+c^4=a^2b^2+b^2c^2+c^2a^2$ .

Observe that $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ , i.e. $a^2+b^2+c^2=-2(ab+bc+ca)$ and $\sum a^4=\sum b^2c^2$ $\iff$

$\left(\sum a^2\right)^2=3\sum b^2c^2$ . Therefore, $4(ab+bc+ca)^2=3\sum b^2c^2$ $\iff$ $\sum b^2c^2+8abc\sum a=0$ $\iff$ $\sum b^2c^2=0$ $\implies$ $\sum a^2=0$ $\implies$

$(a,b,c)\in M(1)$ . Thus, $\boxed{\ M(w)\cup M\left(w^2\right)\subset M(1)\ }$ . Consider $(a,b,c)\in\mathbb C^3$ for which $(a,b,c)\in M(1)$ , i.e. $\sum a=0$ and $\sum a^2=0$ $\implies$

$\sum a=0$ and $\sum bc=0$ $\implies$ $\sum b^2c^2=\left(\sum bc\right)^2-2abc\sum a=0$ and $\sum a^4=\left(\sum a^2\right)^2-3\sum b^2c^2=0$ $\implies$ $\sum a^4=\sum b^2c^2$ $\iff$

$\left(a^2+b^2w+c^2w^2\right)\cdot\left(a^2+b^2w^2+c^2w\right)=0$ $\iff$ $(a,b,c)\in M(w)\cup M\left(w^2\right)$ , i.e. $\boxed {M(1)\subset M(w)\cup M\left(w^2\right) \ }$ .

In conclusion, $\boxed {M(1)= M(w)\cup M\left(w^2\right) \ }$ .
This post has been edited 33 times. Last edited by Virgil Nicula, Nov 22, 2015, 6:29 PM

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18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by ryanbear, May 9, 2023, 6:10 AM

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    by OlympusHero, Sep 14, 2022, 4:44 AM

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    by tigerzhang, Aug 1, 2021, 12:02 AM

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    by GoogleNebula, Apr 14, 2021, 5:25 AM

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    by samrocksnature, Apr 14, 2021, 5:16 AM

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    by the_mathmagician, Jan 17, 2021, 7:43 PM

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    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

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  • Joined: Jun 22, 2005
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  • Total entries: 456
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