Answer:
[tex]\dfrac{9}{13}-\dfrac{19}{13}i[/tex]
Step-by-step explanation:
Remember [tex]i^2=-1[/tex]
You are given the fraction [tex]\dfrac{3+5i}{-2+3i}[/tex]
First, multiply the numerator and the denominator by -2-3i:
[tex]\dfrac{3+5i}{-2+3i}=\dfrac{(3+5i)(-2-3i)}{(-2+3i)(-2-3i)}=\dfrac{(3+5i)(-2-3i)}{(-2)^2-(3i)^2}=\dfrac{(3+5i)(-2-3i)}{4-9i^2}=\dfrac{(3+5i)(-2-3i)}{4+9}[/tex]
This gives you 13 in denominator, now multiply two complex numbers in numerator:
[tex](3+5i)(-2-3i)=-6-9i-10i-15i^2=-6-19i+15=9-19i[/tex]
Thus, the initial fraction is
[tex]\dfrac{9-19i}{13}=\dfrac{9}{13}-\dfrac{19}{13}i[/tex]
