1) a)[tex] \frac{29}{5-2i} = \frac{29(5+2 i)}{(5-2i)(5+2i)} = \frac{29(5+2i)}{25-4i ^{2} } = \frac{29(5+2i)}{29} [/tex]= 5+2i
b)[tex] \frac{5-i}{3i}+2i= \frac{i(5-3i)}{3*i ^{2} }+2i= \frac{5i-3i ^{2} }{-3}+2i= \frac{5i+3}{-3}+2i= \frac{5i+3-6i}{-3}= \frac{-i+3}{-3} [/tex]
c) [tex] \frac{7-3i}{3+2i}= \frac{(7-3i)(3-2i)}{(3+2i)(3-2i)} = \frac{21-14i-9i+6i ^{2} }{9-4i^{2} } = \frac{21-23i-6}{13}= \frac{15-23i}{13} [/tex]
2) a)|z|=|√2-√3i|=[tex] \sqrt{ \sqrt{2}^{2}+ \sqrt{3}^{2} } [/tex]=[tex] \sqrt{2+3} = \sqrt{5} [/tex].
b)modul din [tex] \frac{3-i}{2+i} [/tex]=[tex] \frac{|3-i|}{|2+i|} = \frac{ \sqrt{3^{2}+(-1)^{2} } }{ \sqrt{2^{2}+1^{2} } }= \frac{ \sqrt{10} }{ \sqrt{5} } = \sqrt{2} .[/tex]
c) rezolvarea e in poza
