Podemos afirmar que o número complexo z, tal que 
, é:
A) 2 - 5i
(B) 2 + 5i
(C) 5 - 2i
(D) 5 + 2i
(E) -2 + 5i
Soluções para a tarefa
Respondido por
1
Boa tarde Yoda!
Solução!
![z=(a+bi)\\\\\
\overline{z}=(a-bi)](https://tex.z-dn.net/?f=z%3D%28a%2Bbi%29%5C%5C%5C%5C%5C%0A%0A%5Coverline%7Bz%7D%3D%28a-bi%29 "z=(a+bi)\\\\\
\overline{z}=(a-bi)")
![3(a+bi)+2(a-bi)=25-2i\\\\\ 3a+3bi+2a-2bi=25-2i\\\\ 3a+2a=25\\\\\ 5a=25\\\\\ a= \dfrac{25}{5}\\\\\ a=5 \\\\\\\\\\ 3bi-2bi=-2i\\\\\ bi=-2i\\\\
b= \dfrac{-2i}{i}\\\\\
b=-2](https://tex.z-dn.net/?f=3%28a%2Bbi%29%2B2%28a-bi%29%3D25-2i%5C%5C%5C%5C%5C+3a%2B3bi%2B2a-2bi%3D25-2i%5C%5C%5C%5C+3a%2B2a%3D25%5C%5C%5C%5C%5C+5a%3D25%5C%5C%5C%5C%5C+a%3D+%5Cdfrac%7B25%7D%7B5%7D%5C%5C%5C%5C%5C+a%3D5+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C+3bi-2bi%3D-2i%5C%5C%5C%5C%5C+bi%3D-2i%5C%5C%5C%5C%0Ab%3D+%5Cdfrac%7B-2i%7D%7Bi%7D%5C%5C%5C%5C%5C%0Ab%3D-2+ "3(a+bi)+2(a-bi)=25-2i\\\\\ 3a+3bi+2a-2bi=25-2i\\\\ 3a+2a=25\\\\\ 5a=25\\\\\ a= \dfrac{25}{5}\\\\\ a=5 \\\\\\\\\\ 3bi-2bi=-2i\\\\\ bi=-2i\\\\
b= \dfrac{-2i}{i}\\\\\
b=-2")
Forma algebrica.
![z'=a+bi\\\\ z'=5-2i\\\\\
Alternativa: C](https://tex.z-dn.net/?f=z%27%3Da%2Bbi%5C%5C%5C%5C+z%27%3D5-2i%5C%5C%5C%5C%5C%0AAlternativa%3A+C "z'=a+bi\\\\ z'=5-2i\\\\\
Alternativa: C")
Boa tarde!
Bons estudos!
Solução!
Forma algebrica.
Boa tarde!
Bons estudos!
Yoda:
JBK, muito obrigado pela ajuda, valeu mesmo!!
Obrigado! Valeu.
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