Question

ajutati va rog la matematica

Answer 1

   
[tex]\displaystyle 3a) \\ \texttt{Aducem pe z la o forma simpla si rationalizam numitorul. } \\ \\ z= \frac{(1+2i)(2+i)}{1+i}= \frac{2+4i+i +2i^2}{1+i}= \frac{2+4i+i -2}{1+i}= \\ \\ = \frac{5i}{1+i}=\frac{5i(1-i)}{(1+i)(1-i)}= \frac{5i - 5i^2}{1^2-i^2}=\frac{5i + 5}{1+1}= \frac{5+5i}{2}= \boxed{\frac{5}{2}+\frac{5}{2}i} \\ \\ Re(z) =\boxed{\frac{5}{2}} \\ \\ Im(z)=\boxed{\frac{5}{2}i} [/tex]

[tex]\displaystyle \\ |z| = \sqrt{(\frac{5}{2})^2 +(\frac{5}{2})^2}= \sqrt{\frac{25}{4} +\frac{25}{4}}=\sqrt{\frac{50}{4} }= \sqrt{\frac{25 \times 2}{4} }= \boxed{\frac{5 \sqrt{2} }{2} }[/tex]


[tex]\displaystyle 3) \\ z_{_1}=2+2\sqrt{3}i \\ z_{_2}=3-3i \\ z_{_3} = -\sqrt{3} +i \\ \\ a) \\ 2\sqrt{3} z_{_1} +4z_{_3} -z_{_2} =2\sqrt{3} (2+2\sqrt{3}i)+4(-\sqrt{3} +i)-(3-3i )= \\ =\underline{4\sqrt{3}}+12i - \underline{4\sqrt{3}} +4-3+3i= \boxed{1+15i}\\ \\ b) \\ z_{_2}^2 = (3-3i)^2 =9-2 \times 3 \times 3i +9i^2 = 9-18i -9 = \boxed{-18i } \\ \\ c) \\ |z_{_1}|=|2+2\sqrt{3}i| = \sqrt{2^2 +(2\sqrt{3})^2}=\sqrt{4 +12}= \sqrt{16}=\boxed{4} [/tex]


[tex]\displaystyle d) \\ \frac{z_{_1}}{z_{_2}} = \frac{2+2\sqrt{3}i}{3-3i} = \frac{(2+2\sqrt{3}i)(3+3i)}{(3-3i)(3+3i)} = \\ \\ =\frac{6+ 6\sqrt{3}i + 6i + 6\sqrt{3}i^2 }{9-9i^2} = \\ \\ =\frac{6+ 6\sqrt{3}i + 6i - 6\sqrt{3} }{9+9} \\ \\ =\frac{(6- 6\sqrt{3}) +(6+ 6\sqrt{3})i }{18}= \boxed{\frac{1- \sqrt{3}}{3} + \frac{1+ \sqrt{3} }{3}i }[/tex]