What is the real part of \(\displaystyle\frac{3+2i}{2-i}\)? Express your answer as a decimal.
The answer is \(0.8\).
\( (3+2i)(2-i)^{-1} = (3+2i)\left(\frac{2}{2^2+(-1)^2} - \frac{-1}{2^2+(-1)^2}i\right) = (3+2i)(0.4 + 0.2i) = 0.8 + 1.4i\).