Question

am nevoie de ajutor va rog urgent​

Answer 1

Răspuns:

1)

$\left(2-3i\right)\left(2+3i\right)+\frac{3-i}{3+i}=\\4-9i^{2}+\frac{4}{5}-\frac{3}{5}i=\\4-9\left(-1\right)+\frac{4}{5}-\frac{3}{5}i=\\13+\frac{4}{5}-\frac{3}{5}i\\=\frac{69}{5}-\frac{3}{5}i$

b)

$Im\left(\frac{5+i}{3-2i}\right)=Im\left(\frac{\left(5+i\right)\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}\right\\)=\frac{15+10i+3i+2i^{2}}{9-4i}-\frac{15+10i+3i+2\left(-1\right)}{9-4\left(-1\right)}\\=\frac{15+10i+3i-2}{9+4}\\=\frac{13+13i}{13}\\=\frac{13\left(1+i\right)}{13}=>\\Im\left(\frac{13\left(1+i\right)}{13}\right)=i$

c)

$\left(\sqrt{7-4\sqrt{3}}+i\sqrt{7+4\sqrt{3}}\right)^{6}\\=\left(\sqrt{\left(2-\sqrt{3}\right)^{2}}+i\sqrt{\left(2+\sqrt{3}\right)^{2}}\right)^{6}\\=\left(2-\sqrt{3}+i\left(2+\sqrt{3}\right)\right)^{6}-2-\sqrt{3}+i\left(2+\sqrt{3}\right)\right)^{6}\\=\left(2-\sqrt{3}+\left(2+\sqrt{3}\right)i\right)^{6}\\=\left(\sqrt{\left(2-\sqrt{3}\right)^{2}+\left(2+\sqrt{3}\right)^{2}}\right)^{6}\\=\sqrt{14}^{6}=14^{3}=2744$

2)

$z^{2}+2z+5=0=>\\z=\frac{-2+\sqrt{2^{2}-4\cdot1\cdot5}}{2\cdot2};\ z_{2}=\frac{-2-\sqrt{2^{2}-4\cdot1\cdot5}}{2\cdot2}=>\\z=\frac{-2+\sqrt{4-20}}{2};\ z_{2}\\=\frac{-2-\sqrt{4-20}}{2}=>\\z=\frac{-2+4i}{2}z_{2}\\=\frac{-2-4i}{2}=>\\z=-1+2i;\ z_{2}=-1-2i$

b)

$z^{4}-5z^{2}-36=0=>\\t^{2}-5t-36=0=>\\t=-4;\ t=9=>\\z^{2}=-4;\ z^{2}=9=>z=2i;\ -2i;\ -3;\ 3;$

c)$z^{2}+2z+1=0=>z=-1$ intr o ecuatie de gradu al doilea unde primul coeficient este n iar si ultimul tot n atunci soltutiile ecuatiei sunt divizorii lui n, aici avem 1 sau -1, cum 1 nu se poate ramane -1

3)

$z^{2}-4z+7=0=>\\z=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^{2}-4\cdot1\cdot7}}{2\cdot1};\ z_{2}=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^{2}-4\cdot1\cdot7}}{2\cdot1}=>\\z=\frac{4+\sqrt{16-28}}{2};\ z_{2}=\frac{4-\sqrt{16-28}}{2}=>\\z=\frac{4+\sqrt{-12}}{2};\ z_{2}=\frac{4-\sqrt{-12}}{2}=>\\z=\frac{4+2\sqrt{3}i}{2};\ z_{2}=\frac{4-2\sqrt{3}i}{2}=>\\z=2+3i;\ z_{2}=2-3i$

$\frac{2+3i}{2-3i}+\frac{2-3i}{2+3i}\\=\frac{\left(2+3i\right)^{2}+\left(2-3i\right)^{2}}{\left(2-3i\right)\left(2+3i\right)}=>\\\frac{4+12i+9i^{2}+4-12+9i^{2}}{4-9i^{2}}\\=\frac{4+12i+9\cdot-1+4-12i+9i^{2}}{4-9\left(-1\right)}\\=\frac{4+12i-9+4-12i+9i^{2}}{4+9}\\=\frac{4-9+4+9\left(-1\right)}{13}\\=\frac{4-9+4-9}{12}=\frac{-10}{13}$

4)

$x^2 - 6 x + 13$=>rezultat in 3+2i

La 5 nu stiu scuze

6)

$|z_{1} |=|z_{2} |=1=>$ z poate fii 1 sau -1=>$\frac{z_{1}+z_{2}}{1+z_{1}z_2}$ apartine R