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Aplicatie

Alegeti raspunsurile potrivite astfel incat sa completati corect problema.

Sa se calculeze toate valorile radicalilor si sa se determine imaginile lor geometrice::

a) $\sqrt[3]{\cos120^{\circ}+i\sin120^{\circ}};$ b) $\sqrt[4]{16(\cos240^{\circ}+i\sin240^{\circ}}).$

Rezolvare:

Daca $z=r(\cos\varphi +i\sin\varphi ),$ atunci radacinile de ordin $n$ ale lui $z$ sunt numerele complexe: $z_{k}=\sqrt[n]{r}\begin{pmatrix}\cos\frac{\varphi +2k\pi}{n}+i\sin\frac{\varphi +2k\pi}{n} \end{pmatrix},k=0,1,2,...,n-1.$

a) Avem:

$z_{k}=\cos\frac{120^{\circ}+2k180^{\circ}}{3}+i\sin\frac{120^{\circ}+2k180^{\circ}}{3}, k=0,1,2.$

$z_{k}=\cos\frac{180^{\circ}+2k120^{\circ}}{3}+i\sin\frac{180^{\circ}+2k120^{\circ}}{3}, k=0,1,2.$

$z_{k}=\sin\frac{180^{\circ}+2k120^{\circ}}{3}+i\cos\frac{180^{\circ}+2k120^{\circ}}{3}, k=0,1,2.$

Solution:

Optiune corecta
Raspuns corect.

Continua problema.
Gresit
Raspuns gresit.

Folosind formula amintita mai sus pentru $z_{k}$ gasim ca:

$z_{k}=\cos\frac{120^{\circ}+2k180^{\circ}}{3}+i\sin\frac{120^{\circ}+2k180^{\circ}}{3}, k=0,1,2.$
Gresit
Raspuns gresit.

In primul rand, un numar complex se scrie sub forma agebrica $z=x+iy$ , iar sub forma trigonometrica $z=r(cost +isint)$ , unde $r=\begin{vmatrix}z \end{vmatrix}$ , iar $t=tg\frac{x}{y}, x eq0.$

In al doilea rand, folosind formula amintita mai sus pentru $z_{k}$ , gasim ca:

$z_{k}=\cos\frac{120^{\circ}+2k180^{\circ}}{3}+i\sin\frac{120^{\circ}+2k180^{\circ}}{3}, k=0,1,2.$

Deci radacinile de ordin 3 sunt numerele:

$z_{0}=\cos40^{\circ}+i\sin40^{\circ}; z_{1}=\cos120^{\circ}+i\sin120^{\circ}; z_{2}=\cos240^{\circ}+i\sin240^{\circ}.$

$z_{0}=\cos40^{\circ}+i\sin40^{\circ}; z_{1}=\cos160^{\circ}+i\sin160^{\circ}; z_{2}=\cos280^{\circ}+i\sin280^{\circ}.$

$z_{0}=\cos40^{\circ}+i\sin40^{\circ}; z_{1}=\cos160^{\circ}+i\sin160^{\circ}; z_{2}=\cos290^{\circ}+i\sin290^{\circ}.$

Solution:

Gresit
Raspuns gresit.

Inlocuind pe rand pe k cu 0,1,2 in formula aflata mai sus, veti obtine valorile:

$z_{0}=\cos40^{\circ}+i\sin40^{\circ}; z_{1}=\cos160^{\circ}+i\sin160^{\circ}; z_{2}=\cos280^{\circ}+i\sin280^{\circ}.$
Optiune corecta
Raspuns corect.
Gresit
Raspuns gresit.

Inlocuind pe rand pe k cu 0,1,2 in formula aflata mai sus, veti obtine valorile:

$z_{0}=\cos40^{\circ}+i\sin40^{\circ}; z_{1}=\cos160^{\circ}+i\sin160^{\circ}; z_{2}=\cos280^{\circ}+i\sin280^{\circ}.$

b) In acest caz radacinile de ordin 4 sunt numerele complexe:

$z_{k}=2\begin{bmatrix}\cos\frac{240^{\circ}+2k180^{\circ}}{4}+i\sin\frac{240^{\circ}+2k180^{\circ}}{4} \end{bmatrix}, k=0,1,2,3.$

$z_{k}=2\begin{bmatrix}\cos\frac{140^{\circ}+2k180^{\circ}}{4}+i\sin\frac{140^{\circ}+2k180^{\circ}}{4} \end{bmatrix}, k=0,1,2,3.$

$z_{k}=2\begin{bmatrix}\cos\frac{240^{\circ}+2k180^{\circ}}{2}+i\sin\frac{240^{\circ}+2k180^{\circ}}{2} \end{bmatrix}, k=0,1,2,3.$

Solution:

Optiune corecta
Raspuns corect.

Continua problema.
Gresit
Raspuns gresit.

Folosind formula amintita mai sus pentru $z_{k}$ gasim ca:

$z_{k}=2\begin{bmatrix}\cos\frac{240^{\circ}+2k180^{\circ}}{4}+i\sin\frac{240^{\circ}+2k180^{\circ}}{4} \end{bmatrix}, k=0,1,2,3.$
Gresit
Raspuns gresit.

Folosind formula amintita mai sus pentru $z_{k}$ gasim ca:

$z_{k}=2\begin{bmatrix}\cos\frac{240^{\circ}+2k180^{\circ}}{4}+i\sin\frac{240^{\circ}+2k180^{\circ}}{4} \end{bmatrix}, k=0,1,2,3.$

Explicit avem numerele complexe:

$z_{0}=2(\cos60^{\circ}+i\sin60^{\circ});z_{1}=2(\cos240^{\circ}+i\sin240^{\circ});$

$z_{2}=2(\cos240^{\circ}+i\sin240^{\circ});z_{3}=2(\cos280^{\circ}+i\sin280^{\circ}).$

$z_{0}=2(\cos60^{\circ}+i\sin60^{\circ});z_{1}=2(\cos150^{\circ}+i\sin150^{\circ});$

$z_{2}=2(\cos240^{\circ}+i\sin240^{\circ});z_{3}=2(\cos320^{\circ}+i\sin320^{\circ}).$

$z_{0}=2(\cos60^{\circ}+i\sin60^{\circ});z_{1}=2(\cos150^{\circ}+i\sin150^{\circ});$

$z_{2}=2(\cos240^{\circ}+i\sin240^{\circ});z_{3}=2(\cos330^{\circ}+i\sin330^{\circ}).$

Solution:

Gresit
Raspuns gresit.

Inlocuind pe rand pe k cu 0,1,2,3 in formula aflata mai sus, veti obtine valorile:

$z_{0}=2(\cos60^{\circ}+i\sin60^{\circ});z_{1}=2(\cos150^{\circ}+i\sin150^{\circ});$

$z_{2}=2(\cos240^{\circ}+i\sin240^{\circ});z_{3}=2(\cos330^{\circ}+i\sin330^{\circ}).$
Gresit
Raspuns gresit.

Inlocuind pe rand pe k cu 0,1,2,3 in formula aflata mai sus, veti obtine valorile:

$z_{0}=2(\cos60^{\circ}+i\sin60^{\circ});z_{1}=2(\cos150^{\circ}+i\sin150^{\circ});$

$z_{2}=2(\cos240^{\circ}+i\sin240^{\circ});z_{3}=2(\cos330^{\circ}+i\sin330^{\circ}).$
Optiune corecta
Raspuns corect.

Completati spatiile libere.

In continuare vom determina imaginile geometrice pentru toate valorile radicalilor de la punctele a) si b).

Imaginile numerelor complexe $z_{k}$ de la punctul a) se afla pe cercul de centru O(0,0) si de raza JXUwMDY5 si sunt varfurile unui .

Imaginile numerelor complexe $z_{k}$ de la punctul b) se afla pe cercul de centru O(0,0) si de raza JXUwMDZh si sunt varfurile unui .

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