Answer:
Step-by-step explanation:
[tex]\frac{5+4i}{-3-2i}\\\\\mathrm{Apply\:complex\:arithmetic\:rule}:\\\\\quad \frac{a+bi}{c+di}\:=\:\frac{\left(c-di\right)\left(a+bi\right)}{\left(c-di\right)\left(c+di\right)}\:=\:\frac{\left(ac+bd\right)+\left(bc-ad\right)i}{c^2+d^2}\\\\a=5,\:b=4,\:c=-3,\:d=-2\\\\=\frac{\left(5\left(-3\right)+4\left(-2\right)\right)+\left(4\left(-3\right)-5\left(-2\right)\right)i}{\left(-3\right)^2+\left(-2\right)^2}\\\\Refine\\=\frac{-23-2i}{13}\\\\[/tex]
[tex]\mathrm{Rewrite\:}\frac{-23-2i}{13}\mathrm{\:in\:standard\:complex\:form:\:}-\frac{23}{13}-\frac{2}{13}i\\[/tex]
