Question

50 puncte!!exercitiul 1. din culegere Iavorschi pag 208.!❤️❤️❤️❤️❤️❤️

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it a)\ (5-7i)+(-9-2i)=5-7i-9-2i=(5-9)+(-7i-2i)=-4-9i\\ \\ f)\ (2+3i)+(4-3i)=2+3i+4-3i=(2+4)+(3i-3i)=6\\ \\ g)\ (2-i)(3+i)=6+2i-3i-(-1)=7-i\\ \\ j)\ (2-i)^2=2^2-2\cdot2\cdot i+i^2=4-4i-1=3-4i\\ \\ o)\ (1+2i)^2-(1-2i)^2=(1+2i-1+2i)(1+2i+1-2i)=4i\cdot2=8i\\ \\ s)\ \dfrac{^{1+\sqrt3i)}1+\sqrt3i}{\ \ \ 1-\sqrt3i}= \dfrac{1+2\sqrt3i-3}{133\cdot(-1)} = \dfrac{-2+2\sqrt3i}{4} =-\dfrac{2^{(2}}{4}+ \dfrac{2\sqrt3i^{(2}}{4} =\\ \\ \\ =- \dfrac{1}{2}+ \dfrac{\sqrt3i}{2} \\ \\ \\ \ell)\ \dfrac{^{2+i)}3i}{\ \ 2-i}$