Home/ Intrebari/Q 84416
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

phoenix09
Pe: 2013-12-01T13:14:00+02:00 In: In:

Algebra

Buna ziua ! As aprecia daca ar putea cineva sa ma ajute cu rezolvarea acestui exercitiu:

a)Sa se rezolve ecuatia: (1-i)z^3-1+i*radical(3)=0
b)Sa se arate ca daca z1,z2 apartin C/R cu |z1|=|z2|=1 si z1^2 diferit de z2^2 atunci:

Im(1-z1^2*z^2/z1^2-z2^2)=0

Multumesc frumos ! O zi buna!

Similare

1 raspuns

  1. ghioknt
    profesor

    a)z^3=\frac{1-i\sqrt{3}}{1-i}=\frac{2(\frac{1}{2}-i\frac{\sqrt{3}}{2})}{\sqrt{2}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})}=\frac{2(cos\,\frac{\pi}{3}-isin\frac{\pi}{3})}{\sqrt{2}(cos\,\frac{\pi}{4}-isin\frac{\pi}{4})}=\frac{2(cos\,(-\frac{\pi}{3})+isin(-\frac{\pi}{3}))}{\sqrt{2}(cos\,(-\frac{\pi}{4})+isin(-\frac{\pi}{4}))}=\sqrt{2}(cos(-\frac{\pi}{12})+isin(-\frac{\pi}{12})) (am scăzut argumentele).
    Argumentul redus este -\frac{\pi}{12}+2\pi=\frac{23\pi}{12}, deci ecuaţia se scrie: z^3=\sqrt{2}(cos\,\frac{23\pi}{12}+isin\,\frac{23\pi}{12}) şi are rădăcinile
    z_k=\sqrt[3]{\sqrt{2}}(cos\,\frac{\frac{23\pi}{12}+2k\pi}{3}+isin\,\frac{\frac{23\pi}{12}+2k\pi}{3})=\sqrt[6]{2}(cos\,\frac{(24k+23)\pi}{36}+isin\,\frac{(24k+23)\pi}{36}),\;k\,\in\,\{0,\,1,\,2\}.

    b) |z_1|=|z_2|=1\,\Rightarrow\,\bar{z}_{1}=\frac{1}{z_{1}},\;\bar{z}_{2}=\frac{1}{z_{2}}.
    Dac\check{a}\; Z=\frac{1-z_{1}z_{2}}{z_{1}^{2}-z_{2}^{2}}\Rightarrow \bar{Z}=\frac{1-\bar{z}_{1}\bar{z}_{2}}{\bar{z}_{1}^{2}-\bar{z}_{2}^{2}}=\frac{1-\frac{1}{z_{1}z_{2}}}{\frac{1}{z_1^2}-\frac{1}{z_2^2}}=\frac{z_1z_2-1}{z_2^2-z_1^2}=Z.\;Pentru\;c\check{a}\;\bar{Z}=Z\Rightarrow Z\in \mathbb{R}\Rightarrow Im(Z)=0.

    Cu bine, ghioknt.

Raspunde

Raspunde