Answer:
[tex]\bold{2+\frac{5i}3-2i=2-\frac{1}3i}[/tex]
or if you mean (2 + 5i)/(3 - 2i) at complex numbers
[tex]\bold{\dfrac{2+5i}{3-2i}=\dfrac{-4+19i}{13}=-\frac{4}{13}+1\frac{6}{13}\,i}[/tex].
Step-by-step explanation:
[tex]\bold{2+\frac{5i}3-2i=2+\frac{5i}3-\frac{6i}3=2-\frac{1}3i}[/tex]
But I think you mean: (2 + 5i)/(3 - 2i) and complex numbers.
Then:
[tex]\dfrac{2+5i}{3-2i}=\dfrac{(2+5i)(3+2i)}{(3-2i)(3+2i)}=\dfrac{6+4i+15i+10i^2}{9+6i-6i-4i^2}=\dfrac{6+19i-10}{9+4}=\dfrac{-4+19i}{13}[/tex]
