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

Răspuns :

[tex] \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} [/tex]