Respuesta :

sqdancefan

Answer:

[tex]\dfrac{203-183i}{169}[/tex]

Step-by-step explanation:

[tex]\dfrac{9-19i}{12-5i}=\dfrac{(9-19i)(12+5i)}{(12-5i)(12+5i)}=\dfrac{108+45i-228i-95i^2}{144-25i^2}\\\\=\dfrac{203-183i}{169}=\dfrac{203}{169}-\dfrac{183}{169}i[/tex]