Question

Seja o número complexo Z, tal que:

A) 3z + 2z (conjugado) = 10 + 5i

B) z (z + z (conjugado) ) = 50 + 40i

C) iz - 2z (conjugado) = -14 + 16i

Answer 1

Seja   Z = a + b.i, temos:

a)

$3Z+2\bar{Z}~=~10+5i\\\\\\3~.~(a+bi)+2~.~(a-bi)~=~10+5i\\\\\\3a+2a+3bi-2bi~=~10+5i\\\\\\\boxed{5a+bi~=~10+5i}\\\\\\Igualando~as~partes~reais~e~igualando~as~partes~imaginarias:\\\\\\\left \{ {{5a~=~10} \atop {bi~=~5i}} \right. \\\\\boxed{a~=~2}\\\\\boxed{b=5}\\\\\\Sendo~assim,~~\boxed{Z~=~2+5i}$

b)

$Z.(Z+\bar{Z})~=~50+40i\\\\\\(a+bi)~.~(a+bi~+~a-bi)~=~50+40i\\\\\\(a+bi)~.~(2a)~=~50+40i\\\\\\2a^2+2abi~=~50+40i\\\\\\\boxed{a^2+abi~=~25+20i}\\\\\\Igualando~as~partes~reais~e~igualando~as~partes~imaginarias:\\\\\\\left \{ {{a^2~=~25} \atop {abi~=~20i}} \right. \\\\\\a~=~\sqrt{25}\\\\\boxed{a~=~5}\\\\\\5~.~bi~=~20i\\\\\boxed{b=4}\\\\\\Sendo~assim,~~\boxed{Z~=~5+4i}$

c)

$iZ-2\bar{Z}~=~-14+16i\\\\\\(ai+bi^2)~-~(2a-2bi)~=~-14+16i\\\\\\-b-2a+(a+2b)i~=~-14+16i\\\\\\\boxed{-(2a+b)+(a+2b)i~=~-14+16i}\\\\\\Igualando~as~partes~reais~e~igualando~as~partes~imaginarias:\\\\\\\left \{ {{-(2a+b)~=~-14} \atop {(a+2b)i~=~16i}} \right. \\\\\\-(2a+b)+2.(a+2b)~=~-14+2~.~16\\\\b~=~\frac{18}{3}\\\\\boxed{b~=~6}\\\\\\\\-(2a+b)~=~-14\\\\\boxed{a=4}\\\\\\Sendo~assim,~~\boxed{Z~=~4+6i}$