[tex]\frac{4+6i}{2-3i}=\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)}[/tex]
[tex]\frac{8+12i+12i+18i^2}{4+6i-6i-9i^2}[/tex]
[tex]\frac{-10+24i}{13}[/tex]
[tex]\boxed{\frac{-10}{13}+\frac{24}{13}i}[/tex]
Answer:
- [tex]\frac{10}{13}[/tex] + [tex]\frac{24}{13}[/tex] i
Step-by-step explanation:
Given
[tex]\frac{4+6i}{2-3i}[/tex]
To rationalise the denominator, multiply the numerator/ denominator by the complex conjugate of the denominator.
The conjugate of 2 - 3i is 2 + 3i, thus
[tex]\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)}[/tex] ← expand factors
= [tex]\frac{8+24i+18i^2}{4-9i^2}[/tex] → i² = - 1
= [tex]\frac{8+24i-18}{4+9}[/tex]
= [tex]\frac{-10+24i}{13}[/tex]
= - [tex]\frac{10}{13}[/tex] + [tex]\frac{24}{13}[/tex] i
