Question

Va rog am nevoie de rezolvate la ex 30 |V roman subpunctele 3 si 4 va rog mult

Answer 1

3)
$(i-3)z^2+(7-11i)z+4+6i=0\\ \Delta=(7-11i)^2-4(4+6i)(-3+i)=\\ =49-154i+121i^2-4(-12+4i-18i+6i^2)=\\ =49-154i-121-4(-12-14i-6)\\ =-72-154i+72+ 56i=-98i$

Nu putem calcula √Δ ,dar putem afla care sunt radacinile lui Δ:
[tex]-98i=(x+yi)^2,\ \ x,y\in R\\ 0-98i=x^2-y^2+2xyi\rightarrow \left\{\begin{array}{ll} x^2-y^2=0\\ 2xyi=-98i\rightarrow 2xy=-98\rightarrow y=\frac{-49}{x} \end{array}\right\\\\ x^2-(\frac{-49}{x})^2=0\rightarrow x^4-49^2=0\\ x^4=7^4\rightarrow x =\pm7\\ y=\frac{-49}{x}=\frac{-49}{\pm7}=\mp7\\ \text{Radacinile: }{7-7i,-7+7i}[/tex]

[tex]z_{1,2}=\frac{-7+11i\pm(\pm7\mp7i)}{2(-3+i)}=\frac{-7+11i\pm(7-7i)}{2(-3+i)}\\ z_1=\frac{-7+11i-(7-7i)}{2(-3+i)}=\frac{-14+18i}{2(-3+i)}=\frac{(-7+9i)(-3-i)}{(-3)^2+i^2}=\frac{30-20i}{8}=\boxed{\frac{15}{4}-\frac{5}{2}i}\\\\ z_2=\frac{-7+11i+(7-7i)}{2(-3+i)}=\frac{4i}{2(-3+i)}=\frac{2i(-3-i)}{9-1}=\frac{2-6i}{8}=\boxed{\frac{1}{4}-\frac{3}{4}i}[/tex]

4)
$z^2-(1+12i)z-(13+9i)=0\\ \Delta=(1+12i)^2+4(13+9i)=1-144+24i+52+36i=-91+60i$

[tex]-91+60i=(x+yi)^2\\ -91+60=x^2-y^2+2xyi\rightarrow \left\{\begin{array}{ll} x^2-y^2=-91\\ 2xy=60\rightarrow y=\frac{30}{x}\end{array}\right\\\\ x^2-(\frac{30}{x})^2=-91\\ x^4+91x^2-900=0\\ \text{Fie }t=x^2,\ \ (t\geq0)\\ t^2+91t-900=0\\ t_1=\frac{-91-109}{2}=-\frac{200}{2}\ \textless \ 0\rightarrow\text{Imposibil}\\ t_2=\frac{-91+109}{2}=9\\ x^2=t\rightarrowx=\pm3\\ y=\frac{30}{x}=\pm10\\ \text{Radacina: }\pm(3+10i)[/tex]

[tex]z_1=\frac{1+12i-(3+10i)}{2}=\boxed{-1+i}\\ z_2=\frac{1+12i+(3+10i)}{2}=\boxed{2+11i}[/tex]