Find the values of a and b in 1/(a+bi) = (2+3i)/3i.

The given equation is:

$$\frac{1}{a+bi} = \frac{2+3i}{3i}$$

Take the reciprocal of both sides:

$$a+bi = \frac{3i}{2+3i}$$

Multiply the numerator and denominator by the conjugate of the denominator, which is $2-3i$.

$$a+bi = \frac{3i}{2+3i} \times \frac{2-3i}{2-3i}$$

Numerator:

$$3i(2-3i) = 3i(2) - 3i(3i) = 6i - 9i^2$$

Since $i^2 = -1$:

$$6i - 9(-1) = 6i + 9 = 9+6i$$

Denominator:

$$(2+3i)(2-3i) = 2^2 - (3i)^2 = 4 - 9i^2$$

Since $i^2 = -1$:

$$4 - 9(-1) = 4 + 9 = 13$$

So, the equation becomes:

$$a+bi = \frac{9+6i}{13}$$ $$a+bi = \frac{9}{13} + \frac{6}{13}i$$

By comparing the real and imaginary parts of the complex numbers on both sides of the equation, we get:

$$a = \frac{9}{13}$$ $$b = \frac{6}{13}$$

The values are:

$$a = \frac{9}{13}, b = \frac{6}{13}$$