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