Find the complex number z that satisfies
(1+i)z−2¯z=−11+25i.
Set z=a+biSet ¯z=a−bi
(1+i)z−2¯z=−11+25i(1+i)(a+bi)−2(a−bi)=−11+25i(1+i)a+(1+i)bi−2a+2bi=−11+25ia+ia+bi+bi2−2a+2bi=−11+25i|i2=−1a+ia+bi−b−2a+2bi=−11+25i−(a+b)+i(a+3b)=−11+25icompare−(a+b)=−11a+b=11a=11−b(a+3b)=2511−b+3b=252b=14b=7a=11−ba=11−7a=4z=a+biz=4+7i