\(\frac{3}{2+3i}\)

To remove \(2+3i\) from the denominator, multiply the numerator and denominator by its complex conjugate

\(\frac{3(2-3i)}{(2+3i)(2-3i)}\)

Solve the denominator seperately

\((2+3i)(2-3i)\)

\({2}^{2}+{3}^{2}\)

\(13\)

Now you have

\(\frac{3(2-3i)}{13}\)

Expand \(3(2-3i)\) by distributing

\(3(2-3i)\)

\(6-9i\)

Now you would have

\(\frac{6-9i}{13}=\frac{6}{13}-\frac{9}{13}i\)

Your answer would be

\(\frac{6}{13}-\frac{9}{13}i\)

Hope this Helps ;P