Question

9 -19i/12 -5i
Do not answer in standard form

Answer 1

Answer:

$\dfrac{203-183i}{169}$

Step-by-step explanation:

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