Question

1)Sa se calculeze:
a) 29 supra 5-2i - 25 supra a-3i
b) 5-i supra 3i + 2i ("+2i este dupa fractie)
c) 7 - 3i supra 3+2i
2) Sa se determine modulele numerelor complexe:
a)z=( radical din 2 - radical din 3i)
b) z=3-i supra 2+i
c) z= (radical din 7 - 3i) totul la a 8a supra (7+ radical din 15i) totul la a 6 a

Answer 1

1) a)$\frac{29}{5-2i} = \frac{29(5+2 i)}{(5-2i)(5+2i)} = \frac{29(5+2i)}{25-4i ^{2} } = \frac{29(5+2i)}{29}$=  5+2i
b)$\frac{5-i}{3i}+2i= \frac{i(5-3i)}{3*i ^{2} }+2i= \frac{5i-3i ^{2} }{-3}+2i= \frac{5i+3}{-3}+2i= \frac{5i+3-6i}{-3}= \frac{-i+3}{-3}$
c) $\frac{7-3i}{3+2i}= \frac{(7-3i)(3-2i)}{(3+2i)(3-2i)} = \frac{21-14i-9i+6i ^{2} }{9-4i^{2} } = \frac{21-23i-6}{13}= \frac{15-23i}{13}$

2) a)|z|=|√2-√3i|=$\sqrt{ \sqrt{2}^{2}+ \sqrt{3}^{2} }$=$\sqrt{2+3} = \sqrt{5}$.
b)modul din $\frac{3-i}{2+i}$=$\frac{|3-i|}{|2+i|} = \frac{ \sqrt{3^{2}+(-1)^{2} } }{ \sqrt{2^{2}+1^{2} } }= \frac{ \sqrt{10} }{ \sqrt{5} } = \sqrt{2} .$
c) rezolvarea e in poza