### Instruction: Simplify $\displaystyle\frac{1-i}{2+3i}$, where $i^2 = -1.$.### Response: To simplify a complex fraction, I want to get rid of the imaginary part in the denominator.
I can do this by multiplying both the numerator and denominator by the conjugate of the denominator, which is $2-3i$.
This way, the denominator will become a difference of squares with $i^2 = -1$, and the numerator will be a product of two binomials.
So I have:

\begin{align*}
\frac{1-i}{2+3i} &= \frac{(1-i)(2-3i)}{(2+3i)(2-3i)}\\
&= \frac{2 - 3i - 2i + 3i^2}{4 - 6i + 6i - 9i^2}\\
&= \frac{2 - 5i - 3}{4 + 9}\\
&= \frac{-1 - 5i}{13}
\end{align*}