Question

Write the quotient as a +bi 4+6i/ 2-3i​

Answer 1

$\frac{4+6i}{2-3i}=\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)}$

$\frac{8+12i+12i+18i^2}{4+6i-6i-9i^2}$

$\frac{-10+24i}{13}$

$\boxed{\frac{-10}{13}+\frac{24}{13}i}$

Answer 2

Answer:

- $\frac{10}{13}$ + $\frac{24}{13}$ i

Step-by-step explanation:

Given

$\frac{4+6i}{2-3i}$

To rationalise the denominator, multiply the numerator/ denominator by the complex conjugate of the denominator.

The conjugate of 2 - 3i is 2 + 3i, thus

$\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)}$ ← expand factors

= $\frac{8+24i+18i^2}{4-9i^2}$ → i² = - 1

= $\frac{8+24i-18}{4+9}$

= $\frac{-10+24i}{13}$

= - $\frac{10}{13}$ + $\frac{24}{13}$ i