Fie \( z_1,z_2\in\mathbb{C},|z_1|=|z_2|=|z_1^2+z_1z_2+z_2^2|=1 \).
Demonstrati ca \( |z_1^n+z_2^n|\in\{0,\sqrt{2},2\},\forall n\in\mathbb{N} \).
Numere complexe
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Numere complexe
n-ar fi rau sa fie bine 
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Re: numere complexe
\( \left|z_1\right|=\left|z_2\right|=1\ \Longleftrightarrow\ \overline {z_1}=\frac {1}{z_1} \) , \( \overline {z_2}=\frac {1}{z_2} \) si in acest caz \( |z_1^2+z_1z_2+z_2^2|=1\ \Longleftrightarrow \)
\( \left(z_1^2+z_1z_2+z_2^2\right)^2=z^2_1z_2^2\ \Longleftrightarrow\ z_1+z_2=0\ \ \vee\ \ z_1^2+z_2^2=0\ \Longleftrightarrow \)
\( \ z_2=-z_1\ \ \vee\ \ z_2=\pm iz_1\ . \) In concluzie, pentru orice \( n\in\mathbb N \) avem
\( \left|z_1^n+z_2^n\right|\ \in\ \left\{\ \left|1+(-1)^n\right|\ ,\ \left|1+(\pm i)^n\right|\ \right\}\ \subset \left\{\ 0\ ,\ 2\ ,\ \sqrt 2\ \right\}\ . \)
\( \left(z_1^2+z_1z_2+z_2^2\right)^2=z^2_1z_2^2\ \Longleftrightarrow\ z_1+z_2=0\ \ \vee\ \ z_1^2+z_2^2=0\ \Longleftrightarrow \)
\( \ z_2=-z_1\ \ \vee\ \ z_2=\pm iz_1\ . \) In concluzie, pentru orice \( n\in\mathbb N \) avem
\( \left|z_1^n+z_2^n\right|\ \in\ \left\{\ \left|1+(-1)^n\right|\ ,\ \left|1+(\pm i)^n\right|\ \right\}\ \subset \left\{\ 0\ ,\ 2\ ,\ \sqrt 2\ \right\}\ . \)