Question

Salut! Ma ajutati si pe mine cu acest exercitiu? Va rog mult. a+bi+radical din a la patrat+b la patrat =8+4i

Answer 1

$\displaystyle\\ a+bi+\sqrt{a^2+b^2} =8+4i \\\\ a^2>0;~~b^2>0~\Longrightarrow~(a^2+b^2)>0~\Longrightarrow~ \sqrt{a^2+b^2} \in R\\\\ \text{Scriem ecuatiile:}\\\\ bi=4i ~\Longrightarrow~ \boxed{\bf b=4}\\\\ a+\sqrt{a^2+b^2} =8\\\\ a+\sqrt{a^2+4^2} =8\\\\ \sqrt{a^2+4^2} =8-a~~~\Big|~\text{Ridicam la patrat.}\\\\ a^2+4^2 = (8-a)^2\\\\ a^2 + 16 = 64 - 16a + a^2\\\\ \underline{a^2} - \underline{a^2} + 16a = 64 - 16\\\\ 16a = 48\\\\ a = \frac{48}{16}\\\\ \boxed{\bf a = 3}$