Gọi $z=a+bi,a,b \in \mathbb R$. Ta có PT
$\Leftrightarrow 4(a+bi)+\left(3i+1\right)(a-bi)=25+21i$
$\Leftrightarrow 4a+4bi+3ai-bi+a+3b=25+21i$
$\Leftrightarrow 5a+3b+(3a+3b)i=25+21i$
$\Leftrightarrow \begin{cases} 5a+3b=25 \\ 3a+3b=21 \end{cases}$
$\Leftrightarrow \begin{cases} a=2 \\ b=5 \end{cases}$