Question

Cum se rezolva exercitiile urmatoare (nu vreau doar raspunsul, vreau si modalitatea de rezolvare pentru clasa a 10-a)?
1) Sa se rezolve ecuatia 4z² +8|z|² -3=0.
2) Sa se rezolve in ℂ ecuatiile:
z⁵=-1; z³=conjugat de z.
3) Sa se calculeze zⁿ + z-ⁿ stiind ca z + 1/z=-2cosx.

Answer 1

1) Consideram z=a+bi ,a,b∈R
                        |z| =√(a²+b²);
4z²+8|z|²-3=0
4(a+bi)²+8(a²+b)²-3=0
4(a²+2abi-b²)+8(a²+b²)-3=0
4a²+8abi-4b²+8a²+8b²-3=0
12a²+4b²+8abi=3
Obtinem sistemul:
{ 8ab=0⇒ a=0 sau b=0
{                    
{12a²+4b²=3 
Daca a=0:
4b²=3
b²=3/4⇒ b∈{+- √3/2}

Daca b=0:
12a²=3
a²=1/4 ⇒a∈{+- 1/2}
Asadar z∈{1/2 , -1/2, -i√3/2 , i√3/2}

[tex]2)Folosim\ forma\ trigonometrica\ a\ numarului\ complex:\\ z^5=-1\\ r=|z|= 1\\ cos \Phi= \frac{x}{r}=-1\\ sin \Phi= \frac{y}{r}=0\\ Din\ ambele\ rezulta: \Phi=\pi\ radiani\\ Z\z_k=cos\frac{\Phi+2k\pi}{5}+i\cdot sin\frac{\Phi+2k\pi}{5},k=\overline{0,4}\\ k=0\Rightarrow z_0=cos \frac{\pi}{5}+i\cdot sin\frac{\pi}{5}\\ k=1\Rightarrow z_1=cos \frac{3\pi}{5}+i\cdot sin \frac{3\pi}{5}\\ k=2\Rightarrow z_2=cos \pi+i\cdot sin \pi\\ k=3\Rightarrow z_3=cos \frac{7\pi}{5}+i\cdot sin \frac{7\pi}{5}\\ [/tex]
[tex]k=4\Rightarrow z_4=cos \frac{9\pi}{5}+i\cdot sin \frac{9\pi}{5}\\ [/tex]
[tex]z^3=\overline{z}\\ Notam : z=a+bi, a,b\in R\\ (a+bi)^3=a-bi\\ a^3-ib^3+3abi(a+bi)=a-bi\\ a^3-ib^3+3a^2bi-3ab^2=a-bi\\ Obtinem\ sistemul:\\ \left \{ {{a^3-3ab^2=a} \atop {b^3+3a^2b=-b}} \right.\\ In\ mod\ normal\ ai\ putea\ sa\ il\ rezolvi\ si\ singur. [/tex]

Scz,dar 3 nu stiu sa il fac.