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Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 06 Noiembrie, 2014

EXERCITIUL 6

Suport teoretic:

Inductie matematica,ecuatii algebrice,ecuatii binome,numere complexe,forma trigonometrica.

Enunt:

Pe multimea C, a numerelor complexe, se defineste legea de compozitie:

x * y =  xy - 4x - 4y + 20.

Sa se rezolve in C ecuatia

z^{(3n)} - 12z^{(2n)} + 49z^{(n)} - 68 = 0,\;n\in{\mathbb{N^*}},z^{(3n)} - 12z^{(2n)} + 49z^{(n)} - 68 = 0,\;n\in{\mathbb{N^*}},

unde, prin definitie: 

z^{(m)} = \begin{matrix}\underbrace{{z}\star{z}\star{z}\star{\cdots}\star{z}}\\de\;m\;ori\end{matrix},\;unde\;m\in{\mathbb{N^*}}.z^{(m)} = \begin{matrix}\underbrace{{z}\star{z}\star{z}\star{\cdots}\star{z}}\\de\;m\;ori\end{matrix},\;unde\;m\in{\mathbb{N^*}}.

Raspuns:

S=\{4;\;4+cos{\frac{\frac{3\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{3\pi}{2}+2k\pi}{n}};\;4+cos{\frac{\frac{\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{\pi}{2}+2k\pi}{n}}|k=\overline{0,n-1}\}.S=\{4;\;4+cos{\frac{\frac{3\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{3\pi}{2}+2k\pi}{n}};\;4+cos{\frac{\frac{\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{\pi}{2}+2k\pi}{n}}|k=\overline{0,n-1}\}.

Rezolvare:

Luand x = y = z, se obtinesuccesiv: 

z * z = z² - 4z - 4z + 20 = z² -  8z + 20 = (z - 4)² + 4.

z * z * z = (z * z) * z = [(z - 4)² + 4] * z =

= [(z - 4)² + 4]·z - 4·[(z - 4)² + 4] - 4· z + 20 = ... = (z - 4)³ + 4.

Se prefigureaza, astfel, formula:

{z}^{(m)} ={(z-4)}^{m}+4,\;\forall{m}\in{\mathbb{N^*}},{z}^{(m)} ={(z-4)}^{m}+4,\;\forall{m}\in{\mathbb{N^*}},

care se verifica usor prin inductie matematica.

Folosind notatia

z^{(n)}=t,z^{(n)}=t,  

ecuatia devine:

t^3-12t^2+49t-68=0.t^3-12t^2+49t-68=0.

Cu ajutorul schemei lui Horner si tinand cont de faptul ca aceasta

ecuatie algebrica are coeficienti intregi, se gaseste usor ca tЄ{4,4-i,4+i}. 

Distingem cazurile:

1)\;{t=4}\Leftrightarrow{z^{(n)}}=4\Leftrightarrow{{(z-4)}^n+4=4}\Leftrightarrow{z=4};1)\;{t=4}\Leftrightarrow{z^{(n)}}=4\Leftrightarrow{{(z-4)}^n+4=4}\Leftrightarrow{z=4};  

(radacina reala, multipla de ordinul n).

2)\;{t=4-i}\Leftrightarrow{z^{(n)}}=4-i\Leftrightarrow{{(z-4)}^n+4=4-i}\Leftrightarrow{(z-4)^n}=-i;2)\;{t=4-i}\Leftrightarrow{z^{(n)}}=4-i\Leftrightarrow{{(z-4)}^n+4=4-i}\Leftrightarrow{(z-4)^n}=-i;  

S-a obtinut o ecuatie binoma si avem succesiv:

{{(z-4)}^n=cos{\frac{3\pi}{2}}+isin{\frac{3\pi}{2}}}\Leftrightarrow{z=4+cos{\frac{\frac{3\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{3\pi}{2}+2k\pi}{n}}},\;k=\overline{0,n-1};{{(z-4)}^n=cos{\frac{3\pi}{2}}+isin{\frac{3\pi}{2}}}\Leftrightarrow{z=4+cos{\frac{\frac{3\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{3\pi}{2}+2k\pi}{n}}},\;k=\overline{0,n-1};

(n radacini complexe diferite).

3)\;{t=4+i}\Leftrightarrow{z^{(n)}}=4+i\Leftrightarrow{{(z-4)}^n+4=4+i}\Leftrightarrow{(z-4)^n}=i;3)\;{t=4+i}\Leftrightarrow{z^{(n)}}=4+i\Leftrightarrow{{(z-4)}^n+4=4+i}\Leftrightarrow{(z-4)^n}=i;  

S-a obtinut o ecuatie binoma si avem succesiv:

{{(z-4)}^n=cos{\frac{\pi}{2}}+isin{\frac{\pi}{2}}}\Leftrightarrow{z=4+cos{\frac{\frac{\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{\pi}{2}+2k\pi}{n}}},\;k=\overline{0,n-1};{{(z-4)}^n=cos{\frac{\pi}{2}}+isin{\frac{\pi}{2}}}\Leftrightarrow{z=4+cos{\frac{\frac{\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{\pi}{2}+2k\pi}{n}}},\;k=\overline{0,n-1};

(n radacini complexe distincte).

Deci solutia finala este:

S=\{4;\;4+cos{\frac{\frac{3\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{3\pi}{2}+2k\pi}{n}};\;4+cos{\frac{\frac{\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{\pi}{2}+2k\pi}{n}}|k=\overline{0,n-1}\}.S=\{4;\;4+cos{\frac{\frac{3\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{3\pi}{2}+2k\pi}{n}};\;4+cos{\frac{\frac{\pi}{2}+2k\pi}{n}}+isin{\frac{\frac{\pi}{2}+2k\pi}{n}}|k=\overline{0,n-1}\}.


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